Proposed Formal Definition of a Concrete Category

[Dan Christensen]

"The categories that naturally come to mind when trying to think up
examples of a category (such as sets with functions, groups with
homomorphisms, topological spaces with continuous maps, …) are
sometimes called concrete categories. These are categories where the
objects are sets, usually with some additional structure (group
structure, a topology, etc.), and the morphisms are well-defined
functions between those sets that preserve the structure."

Source: "Concrete and Non-Concrete Categories (Informally)"
http://mathprelims.wordpress.com/2009/02/24/concrete-and-non-concrete-categories-informally/

Formally:

1 ALL(a):[a @ ob <=> P(a)]

where
@ = epsilon (is an element of)
ob = the class whose elements have property P

2 ALL(a):ALL(b):[a @ ob & b @ ob => EXIST(hom):[ALL(f):[f @ hom
<=> ALL(c):[c @ a => f(c) @ b] <-- f is a function
& ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]] <-- f is a
surjection

where
hom = the set (class?) of all morphisms (surjections) from a to b

3 ALL(a):[a @ ob => EXIST(i):ALL(b):[b @ a => i(b)=b]] <-- identity
morphism

where
i = the required identity morphism for each element of ob

The required composition of morphisms is just usual composition of
functions.

Comments?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com

[Alan Eaton]

On 12/10/2012 03:31, Dan Christensen wrote:
> "The categories that naturally come to mind when trying to think up
> examples of a category (such as sets with functions, groups with

> homomorphisms, topological spaces with continuous maps, �) are

The morphisms need not be surjections.
Why are you back on that horse again?

[Dan Christensen]

On Oct 11, 2:01 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> On 12/10/2012 03:31, Dan Christensen wrote:
>
>
>
>
>
>
>
>
>
> > "The categories that naturally come to mind when trying to think up
> > examples of a category (such as sets with functions, groups with

> > homomorphisms, topological spaces with continuous maps, ) are

> > sometimes called concrete categories.  These are categories where the
> > objects are sets, usually with some additional structure (group
> > structure, a topology, etc.), and the morphisms are well-defined
> > functions between those sets that preserve the structure."
>
> > Source: "Concrete and Non-Concrete Categories (Informally)"

> >http://mathprelims.wordpress.com/2009/02/24/concrete-and-non-concrete...

>
> > Formally:
>
> > 1  ALL(a):[a @ ob <=> P(a)]
>
> > where
> > @ = epsilon (is an element of)
> > ob = the class whose elements have property P
>
> > 2  ALL(a):ALL(b):[a @ ob & b @ ob => EXIST(hom):[ALL(f):[f @ hom
> >     <=> ALL(c):[c @ a => f(c) @ b]   <-- f is a function
> >     & ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]     <-- f is a
> > surjection
>
> > where
> > hom = the set (class?) of all morphisms (surjections) from a to b
>
> > 3  ALL(a):[a @ ob => EXIST(i):ALL(b):[b @ a => i(b)=b]]  <-- identity
> > morphism
>
> > where
> > i = the required identity morphism for each element of ob
>
> > The required composition of morphisms is just usual composition of
> > functions.
>
> > Comments?
>
> > Dan

> > Download my DC Proof 2.0 software athttp://www.dcproof.com

>
> The morphisms need not be surjections.
> Why are you back on that horse again?

You are right -- I do keep vacillating. Apologies in advance.

I have been trying to piece together the essence of category theory
from several, not necessarily consistent informal accounts. Nothing
seems to be exactly what I was looking for. The above represents my
latest thinking on this matter.

I think you have to start with morphisms themselves. From Wiki:

"In mathematics, a morphism is an abstraction derived from structure-
preserving mappings between two mathematical structures. The notion of
morphism recurs in much of contemporary mathematics. In set theory,
morphisms are functions; in linear algebra, linear transformations; in
group theory, group homomorphisms; in topology, continuous functions,
and so on.
The study of morphisms and of the structures (called objects) over
which they are defined, is central to category theory. Much of the
terminology of morphisms, as well as the intuition underlying them,
comes from concrete categories, where the objects are simply sets with
some additional structure, and morphisms are structure-preserving
functions....

"There are two operations which are defined on every morphism, the
domain (or source) and the codomain (or target). If a morphism f has
domain X and codomain Y, we write f : X -> Y. Thus a morphism is
represented by an arrow from its domain to its codomain. The
collection of all morphisms from X to Y is denoted homC(X,Y) or simply
hom(X, Y) and called the hom-set between X and Y."

Source: "Morphisms" http://en.wikipedia.org/wiki/Morphism

In category theory (for concrete categories), the "arrow" seems to
connect the domain and codomain of some morphism that preserves some
property (P in my axioms). We want to refer to those arrows
collectively. We want all of those ordered pairs of sets (in ob) that
are "connected" by one or more morphisms.

Given a pair of sets x and y in ob, how do we specify the morphisms
that connect them? Let x be the domain, and y the codomain of these
morphisms. These morphisms are a subset of all functions mapping the
elements of set x to the elements of set y. We want ONLY the
surjections, because a function mapping x to y that is not a
surjection cannot be a morphism connecting x and y, with x as the
domain and y as the codomain. Your thoughts?

[Dan Christensen]

Another, more "visual" approach making use of an "Arrow" predicate...

Define the class ob:

1 ALL(a):[a @ ob <=> P(a)]

where
@ = epsilon (is an element of)

Define the Arrow predicate: For all a and b in ob, Arrow(a,b) means
there exists a surjection from a to b

2 ALL(a):ALL(b):[a @ ob & b @ ob

=> [Arrow(a,b) <=> EXIST(f):[ALL(c):[c @ a => f(c) @ b]
& ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]]


The Arrow predicate is reflexive on ob

3 ALL(a):[a @ ob => Arrow(a,a)]

[Alan Eaton]

On 12/10/2012 05:57, Dan Christensen wrote:
> On Oct 11, 2:01 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:

<snipped for brevity>

The arrow does not '...connect the domain and codomain of some
morphism...'. The arrow *is* the morphism.

> Given a pair of sets x and y in ob, how do we specify the morphisms
> that connect them? Let x be the domain, and y the codomain of these
> morphisms. These morphisms are a subset of all functions mapping the
> elements of set x to the elements of set y. We want ONLY the
> surjections, because a function mapping x to y that is not a
> surjection cannot be a morphism connecting x and y, with x as the
> domain and y as the codomain. Your thoughts?
>
> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com

Forget about your notion of connection.


Perhaps an explicit example is required.

Consider the class of vector spaces over R(the real numbers).

Let objects = the class of vector spaces over R.

Let morphisms = the class of R-linear maps between the objects.

For any two objects X and Y, let hom(X,Y) be the class of
all R-linear maps from X to Y.

For any object X, let id(X) be the R-linear identity map on X.

For any two morphisms that are composable as R-linear maps let
their composition be their composition as R-linear maps.


The above definitions of objects, morphisms, hom, id and composition
collectively satisfy the definition of a category.


Consider a particular pair of vector spaces over R:

X = R^3
Y = R^2

and a particular map between them:

f:X->Y
f((x,y,z)) = (x,0)

then we have:

X and Y are objects in this category because they are vector
spaces over R.

f is a morphism in this category because f is an R-linear map
between the objects X and Y.

f is an element of hom(X,Y) because f is an R-linear map from X to Y.

f is not injective.
f is not surjective.

[LudovicoVan]

"Dan Christensen" <Dan_Chr...@sympatico.ca> wrote in message
news:29016cf2-2880-4d29...@i14g2000yqe.googlegroups.com...

> test

failed

-LV

[Dan Christensen]

On Oct 11, 11:37 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

> Another, more "visual" approach making use of an "Arrow" predicate...
>
> Define the class ob:
>
> 1  ALL(a):[a @ ob <=> P(a)]
>
> where
> @ = epsilon (is an element of)
>
> Define the Arrow predicate: For all a and b in ob, Arrow(a,b) means
> there exists a surjection from a to b
>
> 2  ALL(a):ALL(b):[a @ ob & b @ ob
>    => [Arrow(a,b) <=> EXIST(f):[ALL(c):[c @ a => f(c) @ b]
>    & ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]]
>
> The Arrow predicate is reflexive on ob
>
> 3  ALL(a):[a @ ob => Arrow(a,a)]
>

Developing this idea a little further, I define Category and Arrow
predicates:

Define: Category

1 ALL(a):[Category(a)
<=> ALL(b):[b @ a => EXIST(f):ALL(c):[c @ b => f(c)=c]]]

Define: Arrow

2 ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob
& EXIST(f):[ALL(c):[c @ a => f(c) @ b]

& ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]

It is then easy to prove the Arrow relation is reflexive:

ALL(x):[Category(x) => ALL(a):[a @ x => Arrow(x,a,a)]]

The Arrow relation is transitive:

ALL(x):[Category(x) => ALL(a):ALL(b):ALL(c):[a @ x & b @ x & c @ x
=> [Arrow(x,a,b) & Arrow(x,b,c) => Arrow(x,a,c)]]]

[Alan Eaton]

On 13/10/2012 06:54, Dan Christensen wrote:

<snip>

> Developing this idea a little further, I define Category and Arrow
> predicates:
>
> Define: Category
>
> 1 ALL(a):[Category(a)
> <=> ALL(b):[b @ a => EXIST(f):ALL(c):[c @ b => f(c)=c]]]
>
> Define: Arrow
>
> 2 ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob
> & EXIST(f):[ALL(c):[c @ a => f(c) @ b]
> & ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]
>

Still obsessing over surjective functions?

> It is then easy to prove the Arrow relation is reflexive:
>
> ALL(x):[Category(x) => ALL(a):[a @ x => Arrow(x,a,a)]]
>
> The Arrow relation is transitive:
>
> ALL(x):[Category(x) => ALL(a):ALL(b):ALL(c):[a @ x & b @ x & c @ x
> => [Arrow(x,a,b) & Arrow(x,b,c) => Arrow(x,a,c)]]]
>
> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com
>

In category theory, 'arrow' and 'morphism' are synonyms.

If your intention was for the relation Arrow to indicate the
presence of arrows between objects of a category then
consider the following:

dom(f) =df {x| some y.((x,y) e f)}

rng(f) =df {y| some x.((x,y) e f)}

f:func =df (
all p.( p e f -> some x,y.( p=(x,y) ) )
& all x,y,z.( (x,y) e f & (x,z) e f -> y=z )
)

f:x->y =df ( f:func & dom(f)=x & rng(f) c y )


0 = {}
1 = {0}
2 = {0,1}

object = {1,2}

arrow = {f| some x,y.(x e object & y e object & f:x->y) }

Let composition of arrows be the usual composition of functions.
For any objects x,y let hom(x,y) = {f| f:x->y}
For any object x let id(x) = {(a,a)| a e x}


We have, then, a category. A very simple one.

According to your definitions we have:

Category(object)
~Arrow(object, 1, 2)

but we have:

{(0,0)} e hom(1,2)

so we *should* have had:

Arrow(object, 1, 2)


If your intention for the relation Arrow was otherwise then either
you are misusing the terminology or you are confused as to what a
category is.

[Dan Christensen]

On Oct 12, 8:44 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> On 13/10/2012 06:54, Dan Christensen wrote:
>
> <snip>
>
> > Developing this idea a little further, I define Category and Arrow
> > predicates:
>
> > Define: Category
>
> > 1  ALL(a):[Category(a)
> >     <=> ALL(b):[b @ a => EXIST(f):ALL(c):[c @ b => f(c)=c]]]
>
> > Define: Arrow
>
> > 2  ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob
> >     & EXIST(f):[ALL(c):[c @ a => f(c) @ b]
> >     & ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]
>
> Still obsessing over surjective functions?
>
> > It is then easy to prove the Arrow relation is reflexive:
>
> > ALL(x):[Category(x) => ALL(a):[a @ x => Arrow(x,a,a)]]
>
> > The Arrow relation is transitive:
>
> > ALL(x):[Category(x) => ALL(a):ALL(b):ALL(c):[a @ x & b @ x & c @ x
> > => [Arrow(x,a,b) & Arrow(x,b,c) => Arrow(x,a,c)]]]
>
> > Dan

> > Download my DC Proof 2.0 software athttp://www.dcproof.com

It is true that I am improvising to some extent and using "category"
is a slightly different way than is usual in mathematics. It seems
unavoidable, however. I will summarize my thinking on morphisms, etc.
Perhaps you can tell me where you think I am wrong.

Every class of objects is completely determined by a property that is
common to all elements of that class.

Morphisms are structure-preserving functions. Suppose, for example,
that we have a class X with some defined structure, that is it has
some property P that defines certain relationships between elements of
X. If a function maps each element of X to EACH of the elements of
another class Y that also has property P, then that function is a
morphism with respect to P.

Consider the class C of all classes that have the some common property
P. A function mapping any element of C to another element of C is a
morphism with respect to P. And that function must be a surjection. An
arrow from A to B indicates the existence of a morphism wrt P.

Now, an identity function on any element of C is, of course, a
morphism wrt P. Therefore, our system of arrows must also be
reflexive. Using ordinary composition of functions, we can prove that
our system of arrows must also be transitive.

[Graham Cooper]

On Oct 13, 2:14 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

go for the hat trick and add in symmetric and you get a partition!

Herc

[Alan Eaton]

On 13/10/2012 14:14, Dan Christensen wrote:
> On Oct 12, 8:44 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
>> On 13/10/2012 06:54, Dan Christensen wrote:
>>
>> <snip>
>>

<snip>

Your deviant term usage is avoidable.

> Every class of objects is completely determined by a property that is
> common to all elements of that class.
>
> Morphisms are structure-preserving functions. Suppose, for example,
> that we have a class X with some defined structure, that is it has
> some property P that defines certain relationships between elements of
> X. If a function maps each element of X to EACH of the elements of
> another class Y that also has property P, then that function is a
> morphism with respect to P.
>
> Consider the class C of all classes that have the some common property
> P. A function mapping any element of C to another element of C is a
> morphism with respect to P. And that function must be a surjection. An
> arrow from A to B indicates the existence of a morphism wrt P.

There is an arrow from A to B for each morphism from A to B because
each arrow is a morphism and each morphism is an arrow.

As I said already: in category theory, 'arrow' and 'morphism' are synonyms.

>
> Now, an identity function on any element of C is, of course, a
> morphism wrt P. Therefore, our system of arrows must also be
> reflexive. Using ordinary composition of functions, we can prove that
> our system of arrows must also be transitive.
>
> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com

I gave you an explicit example of a category that is small enough
for the details to be worked out by hand.

hom(1,1) = { {(0,0)} }
hom(1,1) contains a single morphism. It is a surjection.

hom(1,2) = { {(0,0)}, {(0,1)} }
hom(1,2) contains two morphisms, neither of which is a surjection.

hom(2,1) = { {(0,0),(1,0)} }
hom(2,1) contains a single morphism. It is a surjection.

hom(2,2) = {
{(0,0),(1,0)},
{(0,1),(1,0)},
{(0,0),(1,1)},
{(0,1),(1,1)}
}
hom(2,2) contains four morphisms, two of which are not surjections.


Two more examples follow.

In the category Set:

Every object is a set.
Every set is an object.

Every morphism is a total function from one set to another.
Every total function from one set to another is a morphism.

For any pair of sets X and Y, hom(X,Y) is the class of all
total functions from X to Y.

Composition of morphisms is function composition.


For any set X, hom(X,pow(X)) is non-empty and contains no surjections.
(where pow(X) is the power set of X)


For any partially ordered set (P, <=) there is a category such that:

The objects are the elements of P.

The morphisms are the pairs (p,q) such that p <= q.

For pair of morphisms ((p,q),(q,r)), the composition is (p,r).

None of the morphisms are functions.
Therefore, none of the morphisms are surjections.

[Dan Christensen]

Maybe I should use another name for my category predicate. How about
"Identities?" Would that address your concern?

> > Every class of objects is completely determined by a property that is
> > common to all elements of that class.
>
> > Morphisms are structure-preserving functions. Suppose, for example,
> > that we have a class X with some defined structure, that is it has
> > some property P that defines certain relationships between elements of
> > X. If a function maps each element of X to EACH of the elements of
> > another class Y that also has property P, then that function is a
> > morphism with respect to P.
>
> > Consider the class C of all classes that have the some common property
> > P. A function mapping any element of C to another element of C is a
> > morphism with respect to P. And that function must be a surjection. An
> > arrow from A to B indicates the existence of a morphism wrt P.
>
> There is an arrow from A to B for each morphism from A to B because
> each arrow is a morphism and each morphism is an arrow.
>
> As I said already: in category theory, 'arrow' and 'morphism' are synonyms.
>

So, we agree.

>
>
> > Now, an identity function on any element of C is, of course, a
> > morphism wrt P. Therefore, our system of arrows must also be
> > reflexive. Using ordinary composition of functions, we can prove that
> > our system of arrows must also be transitive.
>
> > Dan

> > Download my DC Proof 2.0 software athttp://www.dcproof.com

>
> I gave you an explicit example of a category that is small enough
> for the details to be worked out by hand.
>
> hom(1,1) = { {(0,0)} }
> hom(1,1) contains a single morphism. It is a surjection.
>
> hom(1,2) = { {(0,0)}, {(0,1)} }
> hom(1,2) contains two morphisms, neither of which is a surjection.
>
> hom(2,1) = { {(0,0),(1,0)} }
> hom(2,1) contains a single morphism. It is a surjection.
>
> hom(2,2) = {
>    {(0,0),(1,0)},
>    {(0,1),(1,0)},
>    {(0,0),(1,1)},
>    {(0,1),(1,1)}}
>
> hom(2,2) contains four morphisms, two of which are not surjections.
>
> Two more examples follow.
>

Good.

> In the category Set:
>
>    Every object is a set.
>    Every set is an object.
>
>    Every morphism is a total function from one set to another.
>    Every total function from one set to another is a morphism.
>

Yes, they preserve the property of being a set.

>    For any pair of sets X and Y, hom(X,Y) is the class of all
>    total functions from X to Y.
>
>    Composition of morphisms is function composition.
>
>    For any set X, hom(X,pow(X)) is non-empty and contains no surjections.
>    (where pow(X) is the power set of X)
>
> For any partially ordered set (P, <=) there is a category such that:
>
>    The objects are the elements of P.
>
>    The morphisms are the pairs (p,q) such that p <= q.
>

Morphisms are structure-preserving functions. No structure is being
preserved. These are not even functions. You have only a relation on
P.

>    For pair of morphisms ((p,q),(q,r)), the composition is (p,r).
>
>    None of the morphisms are functions.
>    Therefore, none of the morphisms are surjections.

There are no morphisms.

[Frederick Williams]

If you look up the definition of category you will see that Alan Eaton's
example satisfies it.

--
Where are the songs of Summer?--With the sun,
Oping the dusky eyelids of the south,
Till shade and silence waken up as one,
And morning sings with a warm odorous mouth.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> Morphisms are structure-preserving functions. Suppose, for example,
> that we have a class X with some defined structure, that is it has
> some property P that defines certain relationships between elements of
> X. If a function maps each element of X to EACH of the elements of
> another class Y that also has property P, then that function is a
> morphism with respect to P.

That's not the definition of morphism[1]. In fact, I'm not sure *what*
your last sentence is supposed to mean, but I suspect you're trying to
require that morphisms are onto.

Take, for example, the category Monoid. A monoid is a tuple

<S, e, *>

where e in S and *: S x S -> S is associative.

A morphism

f: <S, e, *> -> <S', e', *'>

is a set function f:S -> S' such that

f(e) = e'

f(s * t) = f(s) *' f(t)

There is *NO REQUIREMENT THAT f BE ONTO*.

Example: Let S = {0}, e = 0 and *:S x S -> S be the identity. It is
trivial to note that <S, e, *> and <N, 0, +> are both monoids (where N
is the set of natural numbers, of course).

It is also obvious that the map f:S -> N mapping 0 to 0 is a monoid
homomorphism (i.e., a morphism in the category Monoid). This map is not
onto.

Footnotes:
[1] I don't see that your notion of structure is coherent either.

--
"And I'll reinforce the point that you are an enemy of humanity, that
my predecessors are people like Gauss, Euler, Newton, Archimedes and
others who you are spitting upon as you do it to me by trying to keep
their discipline trashed as it is now." -- James S. Harris

[Alan Eaton]

On 14/10/2012 00:10, Dan Christensen wrote:
> On Oct 13, 6:32 am, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
>> On 13/10/2012 14:14, Dan Christensen wrote:

>>> It is true that I am improvising to some extent and using "category"
>>> is a slightly different way than is usual in mathematics. It seems
>>> unavoidable, however. I will summarize my thinking on morphisms, etc.
>>> Perhaps you can tell me where you think I am wrong.
>>
>> Your deviant term usage is avoidable.
>>
>
> Maybe I should use another name for my category predicate. How about
> "Identities?" Would that address your concern?

Call it whatever you like.
Are you going to change the thread title as well?
My concern is your misapprehension of categories. Renaming the predicate
does not address that concern.

>>
>> I gave you an explicit example of a category that is small enough
>> for the details to be worked out by hand.
>>
>> hom(1,1) = { {(0,0)} }
>> hom(1,1) contains a single morphism. It is a surjection.
>>
>> hom(1,2) = { {(0,0)}, {(0,1)} }
>> hom(1,2) contains two morphisms, neither of which is a surjection.
>>
>> hom(2,1) = { {(0,0),(1,0)} }
>> hom(2,1) contains a single morphism. It is a surjection.
>>
>> hom(2,2) = {
>> {(0,0),(1,0)},
>> {(0,1),(1,0)},
>> {(0,0),(1,1)},
>> {(0,1),(1,1)}}
>>
>> hom(2,2) contains four morphisms, two of which are not surjections.
>>
>> Two more examples follow.
>>
>
> Good.

What is good? That there are more examples following?

I notice that you did not comment on the fact that there are
non-surjective morphisms in the preceding example.

>
>> In the category Set:
>>
>> Every object is a set.
>> Every set is an object.
>>
>> Every morphism is a total function from one set to another.
>> Every total function from one set to another is a morphism.
>>
>
> Yes, they preserve the property of being a set.
>
>
>> For any pair of sets X and Y, hom(X,Y) is the class of all
>> total functions from X to Y.
>>
>> Composition of morphisms is function composition.
>>
>> For any set X, hom(X,pow(X)) is non-empty and contains no surjections.
>> (where pow(X) is the power set of X)
>>

I notice, again, that you did not comment on the fact that there are
non-surjective morphisms in the preceding example.

>> For any partially ordered set (P, <=) there is a category such that:
>>
>> The objects are the elements of P.
>>
>> The morphisms are the pairs (p,q) such that p <= q.
>>
>
> Morphisms are structure-preserving functions. No structure is being
> preserved. These are not even functions. You have only a relation on
> P.

The morphisms of a category need not be functions.
I indicated, below, that that is the case in this example.

That some categories have objects without non-trivial structure and
morphisms that are not functions is the reason that some people
(myself included) prefer the term 'arrow' over 'morphism'.

>> For pair of morphisms ((p,q),(q,r)), the composition is (p,r).
>>
>> None of the morphisms are functions.
>> Therefore, none of the morphisms are surjections.
>
> There are no morphisms.

Yes there are.

As Frederick indicated, the previous example satisfies the definition
of a category.

[Dan Christensen]

On Oct 13, 12:55 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> On 14/10/2012 00:10, Dan Christensen wrote:
>
> > On Oct 13, 6:32 am, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> >> On 13/10/2012 14:14, Dan Christensen wrote:
> >>> It is true that I am improvising to some extent and using "category"
> >>> is a slightly different way than is usual in mathematics. It seems
> >>> unavoidable, however. I will summarize my thinking on morphisms, etc.
> >>> Perhaps you can tell me where you think I am wrong.
>
> >> Your deviant term usage is avoidable.
>
> > Maybe I should use another name for my category predicate. How about
> > "Identities?" Would that address your concern?
>
> Call it whatever you like.
> Are you going to change the thread title as well?
> My concern is your misapprehension of categories. Renaming the predicate
> does not address that concern.
>

OK.

> >> I gave you an explicit example of a category that is small enough
> >> for the details to be worked out by hand.
>
> >> hom(1,1) = { {(0,0)} }
> >> hom(1,1) contains a single morphism. It is a surjection.
>
> >> hom(1,2) = { {(0,0)}, {(0,1)} }
> >> hom(1,2) contains two morphisms, neither of which is a surjection.
>
> >> hom(2,1) = { {(0,0),(1,0)} }
> >> hom(2,1) contains a single morphism. It is a surjection.
>
> >> hom(2,2) = {
> >>     {(0,0),(1,0)},
> >>     {(0,1),(1,0)},
> >>     {(0,0),(1,1)},
> >>     {(0,1),(1,1)}}
>
> >> hom(2,2) contains four morphisms, two of which are not surjections.
>
> >> Two more examples follow.
>
> > Good.
>
> What is good?

It's not your fault -- I just couldn't make sense of the previous
example.

> That there are more examples following?
>
> I notice that you did not comment on the fact that there are
> non-surjective morphisms in the preceding example.
>
> >> In the category Set:
>
> >>     Every object is a set.
> >>     Every set is an object.
>
> >>     Every morphism is a total function from one set to another.
> >>     Every total function from one set to another is a morphism.
>
> > Yes, they preserve the property of being a set.
>
> >>     For any pair of sets X and Y, hom(X,Y) is the class of all
> >>     total functions from X to Y.
>
> >>     Composition of morphisms is function composition.
>
> >>     For any set X, hom(X,pow(X)) is non-empty and contains no surjections.
> >>     (where pow(X) is the power set of X)
>
> I notice, again, that you did not comment on the fact that there are
> non-surjective morphisms in the preceding example.
>

Interesting point, but isn't a morphism wrt to some property (or
structure) P just a transformation that preserves property P? That is
my intuitive sense of it. You cannot transform any set into its power
set since, in a transformation, one element is transforms into exactly
one other element. Can you then have morphism from a set to its power
set?

> >> For any partially ordered set (P, <=) there is a category such that:
>
> >>     The objects are the elements of P.
>
> >>     The morphisms are the pairs (p,q) such that p <= q.
>
> > Morphisms are structure-preserving functions. No structure is being
> > preserved. These are not even functions. You have only a relation on
> > P.
>
> The morphisms of a category need not be functions.

An odd notion of "morphism" to my mind.

> I indicated, below, that that is the case in this example.
>
> That some categories have objects without non-trivial structure and
> morphisms that are not functions is the reason that some people
> (myself included) prefer the term 'arrow' over 'morphism'.
>
> >>     For pair of morphisms ((p,q),(q,r)), the composition is (p,r).
>
> >>     None of the morphisms are functions.
> >>     Therefore, none of the morphisms are surjections.
>
> > There are no morphisms.
>
> Yes there are.
>

A morphism has to mean something more than a single ordered pair. It
has to include the notion of a structure preserving transformation,
does it not? Otherwise, you are just talking about a binary relation
defined on a class of objects that may be nothing more atomic
individuals.

> As Frederick indicated, the previous example satisfies the definition
> of a category.

The usual definitions of category theory seem rather vague to me. It
seems to me that there are too many words, and too much hand waving. I
am open to suggestions -- something stated in the notation of first-
order logic and set theory, something of the form of the axioms in my
original posting if that's not asking too much.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> Interesting point, but isn't a morphism wrt to some property (or
> structure) P just a transformation that preserves property P? That is
> my intuitive sense of it. You cannot transform any set into its power
> set since, in a transformation, one element is transforms into exactly
> one other element. Can you then have morphism from a set to its power
> set?

The function f:N -> P(N) mapping n to {n} is a morphism in the category
Set.

So is the function g:N -> P(N) mapping n to N \ {n}.

And the function mapping n to N.

And the function mapping n to n u {0,...,47}.

And so on.

You see that *sets* of elements are elements of the powerset. Thus, a
function S -> P(S) takes elements of S to subsets of S -- because
subsets of S *are* elements of P(S).

--
Jesse F. Hughes

"My name is Apusta Malusta Cadeau and I fight bad guys. And I'm a
knight." -- A. M. Cadeau (nee Quincy P. Hughes), age 4

[Dan Christensen]

On Oct 13, 4:13 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > Interesting point, but isn't a morphism wrt to some property (or
> > structure) P just a transformation that preserves property P? That is
> > my intuitive sense of it. You cannot transform any set into its power
> > set since, in a transformation, one element is transforms into exactly
> > one other element. Can you then have morphism from a set to its power
> > set?
>
> The function f:N -> P(N) mapping n to {n} is a morphism in the category
> Set.
>
> So is the function g:N -> P(N) mapping n to N \ {n}.
>
> And the function mapping n to N.
>
> And the function mapping n to n u {0,...,47}.
>
> And so on.
>
> You see that *sets* of elements are elements of the powerset.  Thus, a
> function S -> P(S) takes elements of S to subsets of S -- because
> subsets of S *are* elements of P(S).
>

As I understand, a morphism on is a structure-preserving
transformation. I don't see the point of category theory otherwise.

[Jesse F. Hughes]

In the case of the category Set, there is no structure to be preserved.
It is a degenerate case.

In any case, whether you see the point or not, you ought to be looking
at actual textbooks and working through their examples rather than just
trying to draw inferences from your own intuitions of what category
theory should be.

Because, you see, people have already defined terms like "category" and
even "concrete category" and you don't have to guess what these ought to
mean. You can learn what they actually do mean.

So, start with the definition of category, learn what a functor is,
learn what it means for a functor to be faithful and then you can figure
out what a concrete category is. It's a category C with a faithful
functor C -> Set. (Oops! You'll have to learn what the category Set
is, too.)

That's how you learn what existing terminology means. Much simpler than
just guessing.

--
Jesse F. Hughes
"If the car stops and you're not getting out, then you have to start
it again." -- Quincy P. Hughes (age 3) on his father's skills
with a manual transmission.

[Rotwang]

Clearly, then, it doesn't have one. All the mathematicians who think
they've found applications of category theory as usually understood must
have been mistaken. There can't possibly be any application of
categories whose so-called morphisms are something other than
structure-preserving transformations, because any such application would
be immediately apparent to you, Dan Christensen, upon sort-of reading
several informal explanations of what a category is.

Don't let Jesse or Alan or Frederick or any of those other mediocre
minds who were foolish enough to waste their time finding out the actual
definition of 'category' tell you otherwise - mathematical terms such as
'morphism' mean exactly what you guess them to mean, and any
mathematician or mathematical textbook that says they mean something
else is wrong.


--
I have made a thing that superficially resembles music:

http://soundcloud.com/eroneity/we-berated-our-own-crapiness

[Graham Cooper]

How many of these texts have computer programs that support these
definitions?

It's very hard to find computer generated definitions that will fit
into a formal parser.

IMO group theory and category theory look like AI reruns of Search
Operands and Semantic Networks.

the ole

X -isa-> Y

Beware the HARD PROBLEM!

Herc

[Dan Christensen]

On Oct 13, 8:07 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Oct 14, 9:34 am, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
>
>
>
>
>
>
>
> > On 13/10/2012 22:31, Dan Christensen wrote:
>
> > > As I understand, a morphism on is a structure-preserving
> > > transformation. I don't see the point of category theory otherwise.
>
> > Clearly, then, it doesn't have one. All the mathematicians who think
> > they've found applications of category theory as usually understood must
> > have been mistaken. There can't possibly be any application of
> > categories whose so-called morphisms are something other than
> > structure-preserving transformations, because any such application would
> > be immediately apparent to you, Dan Christensen, upon sort-of reading
> > several informal explanations of what a category is.
>
> > Don't let Jesse or Alan or Frederick or any of those other mediocre
> > minds who were foolish enough to waste their time finding out the actual
> > definition of 'category' tell you otherwise - mathematical terms such as
> > 'morphism' mean exactly what you guess them to mean, and any
> > mathematician or mathematical textbook that says they mean something
> > else is wrong.
>
> How many of these texts have computer programs that support these
> definitions?
>
> It's very hard to find computer generated definitions that will fit
> into a formal parser.
>

Odd, isn't it? I found that even for something as ubiquitous as
Peano's Axioms, it took me quite a bit of tinkering to piece together
a workable and truly formal set of axioms. I don't know if my
definition of a concrete category here is "correct" or workable, but
the experts here seem content with the verbose and, to me, vague
axioms/definitions usually given as the basis for category theory.

Could there actually be anything to the popular (mis?)perception:
"Abstract nonsense is a popular term used by mathematicians to
describe certain kinds of arguments and concepts in category theory?"

Source: "Introduction to Category Theory"
http://en.wikiversity.org/wiki/Introduction_to_Category_Theory

[Jesse F. Hughes]

Yes, it really is remarkable that you have greater insights in every
mathematical topic than the experts have. And the definition of
category really is remarkably complex, isn't it?

Part of your problem is that you want to leap straight for the concept
of concrete category, but it would be best if you learned the simpler
concepts first, namely, category in its usual generality. That is the
easier concept to understand.

On the other hand, you could instead pretend to be revolutionizing the
field with your half-baked ideas free from the clutter of actual
understanding. That's probably more fun. Do that.

But if you want to actually *learn* what a category is, I'd be happy to
help.

>
> Could there actually be anything to the popular (mis?)perception:
> "Abstract nonsense is a popular term used by mathematicians to
> describe certain kinds of arguments and concepts in category theory?"
>
> Source: "Introduction to Category Theory"
> http://en.wikiversity.org/wiki/Introduction_to_Category_Theory

Category theory is abstract nonsense by design, but it's darned useful
abstract nonsense.

--
Jesse F. Hughes

"Wikipedia [...] is not a passive observer but a active and aggressive
fascist dictator of nicknames." -- Archimedes ("Arky") Plutonium

[Dan Christensen]

On Oct 13, 5:43 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Oct 13, 4:13 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> >> > Interesting point, but isn't a morphism wrt to some property (or
> >> > structure) P just a transformation that preserves property P? That is
> >> > my intuitive sense of it. You cannot transform any set into its power
> >> > set since, in a transformation, one element is transforms into exactly
> >> > one other element. Can you then have morphism from a set to its power
> >> > set?
>
> >> The function f:N -> P(N) mapping n to {n} is a morphism in the category
> >> Set.
>
> >> So is the function g:N -> P(N) mapping n to N \ {n}.
>
> >> And the function mapping n to N.
>
> >> And the function mapping n to n u {0,...,47}.
>
> >> And so on.
>
> >> You see that *sets* of elements are elements of the powerset.  Thus, a
> >> function S -> P(S) takes elements of S to subsets of S -- because
> >> subsets of S *are* elements of P(S).
>
> > As I understand, a morphism on is a structure-preserving
> > transformation. I don't see the point of category theory otherwise.
>
> In the case of the category Set, there is no structure to be preserved.
> It is a degenerate case.
>

There is a rudimentary structure defined by set membership.

> In any case, whether you see the point or not, you ought to be looking
> at actual textbooks and working through their examples rather than just
> trying to draw inferences from your own intuitions of what category
> theory should be.
>

I have been looking for a while, for a truly formal definition of the
concept of a category. Unable to find one ready-made, I have been
trying to piece together my own from various informal definitions (see
original posting). Call it a high intolerance for ambiguity in things
mathematical.

> Because, you see, people have already defined terms like "category" and
> even "concrete category" and you don't have to guess what these ought to
> mean.  You can learn what they actually do mean.

I shouldn't have to "learn" what they actually mean. Ideally, in
mathematics, we should have a formal, stand-alone definition. Apart
from some well chosen, informal commentary, that is all the "meaning"
you should need. Why can't anyone provide such a definition? Is it
impossible in the case of category theory?

[Dan Christensen]

On Oct 14, 12:03 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

So complex that a formal definition seems to elude all the experts
here.

> Part of your problem is that you want to leap straight for the concept
> of concrete category, but it would be best if you learned the simpler
> concepts first, namely, category in its usual generality.  That is the
> easier concept to understand.

The formal definition of a category then...

>
> On the other hand, you could instead pretend to be revolutionizing the
> field with your half-baked ideas free from the clutter of actual
> understanding.

You don't like my formal definition? It could well be flawed, so let's
see yours. No words, no hand waving, just mathematical symbols if you
don't mind. My tiny brain can't seem to handle much more than that.

[Graham Cooper]

On Oct 14, 2:03 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> On the other hand, you could instead pretend to be revolutionizing the
> field with your half-baked ideas free from the clutter of actual

> understanding.  That's probably...


...what formalism means?

Herc

[Alan Eaton]

On 14/10/2012 14:45, Dan Christensen wrote:

<snip>

> So complex that a formal definition seems to elude all the experts
> here.

For a characterisation of small categories see the following post:
https://groups.google.com/d/msg/sci.logic/xeTxnHvzNLg/QwmCns3L94oJ

<snip>

> You don't like my formal definition? It could well be flawed, so let's
> see yours. No words, no hand waving, just mathematical symbols if you
> don't mind. My tiny brain can't seem to handle much more than that.

Your definition is most certainly flawed.

The following characterises my understanding of categories.
You might note that I make use only of predicate notions.


Notational notes:

I write '==' for the biconditional connective.
I write 'some' where others might write 'exists'.


Primitive Predicate Symbols:

cat
ob
mor
dom
cod
hom
comp
id


Intended Interpretations of the Primitive Predicate Symbols:

(just ignore this section if you really don't want any words)

cat(c)
c is a category.

ob(c,x)
x is an object in the category c.

mor(c,f)
f is a morphism in the category c.

dom(c,f,x)
x is the domain of the morphism f in the category c.

cod(c,f,x)
x is the codomain of the morphism f in the category c.

hom(c,x,y,f)
f is a morphism from the object x to the object y in the category c.

comp(c,f,g,h)
h is the composition of the morphisms f and g in the category c.

id(c,x,f)
f is the identity morphism for the object x in the category c.


Rules:

all c,x.( ob(c,x) -> cat(c) )
all c,f.( mor(c,f) -> cat(c) )
all c,f,x.( dom(c,f,x) -> mor(c,f) and ob(c,x) )
all c,f,x.( cod(c,f,x) -> mor(c,f) and ob(c,x) )
all c,x,y,f.( hom(c,x,y,f) -> ob(c,x) & ob(c,y) & mor(c,f) )
all c,f,g,h.( comp(c,f,g,h) -> mor(c,f) & mor(c,g) & mor(c,h) )
all c,f,x.( id(c,x,f) -> ob(c,x) & mor(c,f)

all c,f.( mor(c,f) -> some x.dom(c,f,x) )
all c,f,x1,x2.( dom(c,f,x1) & dom(c,f,x2) -> x1=x2)

all c,f.( mor(c,f) -> some y.cod(c,f,y) )
all c,f,y1,y2.( cod(c,f,y1) & cod(c,f,y2) -> y1=y2)

all c,x,y,f.( hom(c,x,y,f) == dom(c,f,x) & cod(c,f,y) )

all c,f,g,x,y,z.(
( dom(c,g,x) & cod(c,g,y)
& dom(c,f,y) & cod(c,f,z)
)-> some h.comp(c,f,g,h)
)
all c,f,g,h1,h2.( comp(c,f,g,h1) & comp(c,f,g,h2) -> h1=h2 )
all c,f,g,h.(
comp(c,f,g,h) -> some x,y,z.(
dom(c,g,x) & cod(c,g,y)
& dom(c,f,y) & cod(c,f,z)
& dom(c,h,x) & cod(c,h,z)
)
)
all c,f,g,h,fg,gh,fgh1,fgh2.(
( comp(c,f,g,fg)
& comp(c,fg,h,fgh1)
& comp(c,g,h,gh)
& comp(c,f,gh,fgh2)
)-> fgh1=fgh2
)

all c,x,f.( id(c,x,f) -> all g,h.( comp(c,f,g,h) -> g=h ) )
all c,x,f.( id(c,x,f) -> all g,h.( comp(c,g,f,h) -> g=h ) )

[Alan Eaton]

On 14/10/2012 18:58, Alan Eaton wrote:

<snip>

> all c,f,g,x,y,z.(
> ( dom(c,g,x) & cod(c,g,y)
> & dom(c,f,y) & cod(c,f,z)
> )-> some h.comp(c,f,g,h)
> )
> all c,f,g,h1,h2.( comp(c,f,g,h1) & comp(c,f,g,h2) -> h1=h2 )
> all c,f,g,h.(
> comp(c,f,g,h) -> some x,y,z.(
> dom(c,g,x) & cod(c,g,y)
> & dom(c,f,y) & cod(c,f,z)
> & dom(c,h,x) & cod(c,h,z)
> )
> )

A slight simplification:

all c,f,g,x,y,z.( hom(c,x,y,g) & hom(c,y,z,f) -> some h.comp(c,f,g,h) )

all c,f,g,h1,h2.( comp(c,f,g,h1) & comp(c,f,g,h2) -> h1=h2 )

all c,f,g,h.(
comp(c,f,g,h) -> some x,y,z.(

hom(c,x,y,g) & hom(c,y,z,f) & hom(c,x,z,h)
)
)

[Alan Eaton]

On 14/10/2012 18:58, Alan Eaton wrote:

<snip>

> all c,x,f.( id(c,x,f) -> all g,h.( comp(c,f,g,h) -> g=h ) )
> all c,x,f.( id(c,x,f) -> all g,h.( comp(c,g,f,h) -> g=h ) )

Whoops!
I forgot something important:

all c,x,f.( id(c,x,f) -> hom(c,x,x,f) )

[Frederick Williams]

Dan Christensen wrote:

> A morphism has to mean something more than a single ordered pair. It
> has to include the notion of a structure preserving transformation,
> does it not?

A morphism in a category is just what the definition says it is (and the
definition doesn't say "a morphism is a single ordered pair"). If you
want to discuss some other thing why not give it a different name?

> Otherwise, you are just talking about a binary relation
> defined on a class of objects that may be nothing more atomic
> individuals.
>
> > As Frederick indicated, the previous example satisfies the definition
> > of a category.
>
> The usual definitions of category theory seem rather vague to me. It
> seems to me that there are too many words, and too much hand waving. I
> am open to suggestions -- something stated in the notation of first-

> order logic and set theory, [...]

Are sets big enough? Consider the category of categories.

[Frederick Williams]

Dan Christensen wrote:

>
> As I understand, a morphism on is a structure-preserving
> transformation.

Which is to say that you don't understand.

> I don't see the point of category theory otherwise.

So leave it alone.

[Frederick Williams]

Dan Christensen wrote:

>
> I shouldn't have to "learn" what they actually mean.

The battle cry of the crank.

> Ideally, in
> mathematics, we should have a formal, stand-alone definition. Apart
> from some well chosen, informal commentary, that is all the "meaning"
> you should need. Why can't anyone provide such a definition?

What's wrong with
http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C_and_morphisms?

> Is it
> impossible in the case of category theory?

If you want to know more see the references at the end of the Wikipedia
article, in particular Mac Lane, Saunders (1998). Categories for the
Working Mathematician. Graduate Texts in Mathematics 5 (2nd ed.).
Springer-Verlag. ISBN 0-387-98403-8. That used to be the Bible, maybe
it still is, you should ask one the modern young people.

There's also http://katmat.math.uni-bremen.de/acc/acc.pdf.

[Dan Christensen]

I have no problem with interpretations of what each variable might
mean.

Thanks, Alan! This is precisely the sort of thing I have been looking
for. I will review it in detail and post my comments at a later time.

[Dan Christensen]

On Oct 14, 7:13 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> Dan Christensen wrote:
>
> > I shouldn't have to "learn" what they actually mean.
>
> The battle cry of the crank.
>
> > Ideally, in
> > mathematics, we should have a formal, stand-alone definition. Apart
> > from some well chosen, informal commentary, that is all the "meaning"
> > you should need. Why can't anyone provide such a definition?
>

> What's wrong with http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C...
>

As an informal introduction, there is nothing wrong with it. I used it
as one of the inspirations for creating my formal definitions. The
author describes morphisms as "structure-preserving functions" (for
concrete categories). That notion seems to be open to various
interpretations here -- is it any function (not necessarily a
surjection), just a binary relation, or (my choice) a transformation
(a surjective function)?

> > Is it
> > impossible in the case of category theory?
>
> If you want to know more see the references at the end of the Wikipedia
> article, in particular Mac Lane, Saunders (1998). Categories for the
> Working Mathematician. Graduate Texts in Mathematics 5 (2nd ed.).
> Springer-Verlag. ISBN 0-387-98403-8.  That used to be the Bible, maybe
> it still is, you should ask one the modern young people.
>
> There's also http://katmat.math.uni-bremen.de/acc/acc.pdf.
>

Here they say that morphisms can be structure preserving functions as
well (for concrete categories), but they need even be not functions at
all (for abstract categories). Here, I focus on concrete categories.

For structure preserving functions, I assumed they meant a
transformation. Nothing else makes sense to me.

[Jesse F. Hughes]

It sure seems that some people think so.

--
Jesse F. Hughes
"If you hadn't noticed, basically every result I have destroys some
precious belief of mathematicians and they have from what I've gathered
basically gone collectively bonkers." -- James S. Harris

[Jesse F. Hughes]

Alan has given one formal definition. I'll give another, rather
different than his. Like his, I'll deal with small categories for
simplicity's sake.

Cat(O,M,d,cod,c,id) is intended to stand for the claim that the pair
<O,M,d,cod,c,id> is a category, where

O is the collection of objects,
M the collection of morphisms,
dom the domain function,
cod the codomain function,
c the composition function and
id the function picking out identities.

Let Delta(O) be the set { <o,o> | o in O }.

Let f:X ->_partial Y express the relation "f is a partial function from
X to Y."


Cat(O,M,c) <=> dom:M -> O & cod:M -> O & c subset M x M x M & id:O -> M &
(A o in O)(dom(id(o)) = O & cod(id(o)) = O) &
(A m,m' in M)( (E m'')(A m''')( <m,m',m'''> in c ->
m''' = m'' ) &
(A m,m' in M)((E m'')( <m,m',m''> in c ) <->
cod(m') = dom(m)) &
(A m,m',m'' in M)( <m,m',m''> in c -> (*)
dom(m'') = dom(m') ) &
cod(m'') = cod(m) )) &
(A m in M)( <m, id(dom(m)), m> in c & (**)
<id(cod(m)), m, m> in c ) &
Assoc(c)

where Assoc(c) is the usual formula expressing that a given partial
function as associative. This is trivial but tedious to write out and
I'll do so only if pressed.

My formalization is somewhat simpler than Alan's, his is more
appropriate for actually doing work I'd wager.

The first line of mine is just getting the types of the components
right. (NOTE for Dan: I'm working in ZF, so it would be redundant to
write "Set(M)".)

The second line is just stating that the domain and codomain of id_o is
o.

The next two clauses specify that composition is a partial function
M x M -> M defined on all and only compatible morphisms (morphisms with
appropriately matching domain and codomain).

The clause marked (*) asserts that the domain and codomain of the
composite function is appropriate, i.e., that (in traditional notation)
dom(f o g) = dom(g) & cod(f o g) = cod(f).

The clause marked (**) asserts that composition with the identity
changes nothing.

The final clause (unexpanded) is that composition is associative.

There may be minor errors in the above, which I tossed off with my first
cup of coffee, but that looks about right to me.

My, how difficult that was!

>
>> Part of your problem is that you want to leap straight for the concept
>> of concrete category, but it would be best if you learned the simpler
>> concepts first, namely, category in its usual generality.  That is the
>> easier concept to understand.
>
> The formal definition of a category then...

Done.

>
>>
>> On the other hand, you could instead pretend to be revolutionizing the
>> field with your half-baked ideas free from the clutter of actual
>> understanding.
>
> You don't like my formal definition? It could well be flawed, so let's
> see yours. No words, no hand waving, just mathematical symbols if you
> don't mind. My tiny brain can't seem to handle much more than that.

Here's some help for you. In order to formalize the concept, perhaps we
should have some examples in mind. Here are a few examples.

Set:
Obj: sets
Mor: functions between sets

Monoid:
Obj: Monoids
Mor: functions preserving identity and satisfying
f(x * y) = f(x) * f(y)

Group, Field, etc.
Obj: the algebraic structure
Mor: the usual notion of homomorphism for that structure.

PSet:
Obj: <S,s>, where S is a set and s in S (these are called pointed
sets)
Mor: an arrow <S,s> -> <T,t> is a function f:S -> T such that
f(s) = t.

PartialOrder:
Obj: partially ordered sets
Mor: monotonic maps

BotOrder1: (Not the traditional name, but I forget)
Obj: partially ordered sets with least elements
Mor: monotonic maps

BotOrder2: (Not the traditional name, but I forget)
Obj: partially ordered sets with least elements
Mor: monotonic maps which take the least element to the least element

Top:
Obj: topological spaces
Mor: continuous maps

N:
Obj: natural numbers
Mor: between any two objects n, n', there is at most one arrow. There
is an arrow iff n <= n'

Note that the category N generalizes to arbitrary partial orders. Any
partial order *is* a category. In fact, category is a generalization of
partial order.

There. Now however you want to define category, all of those have to be
categories (where the identity arrows and composition should be obvious
for each). If not, whatever you've done, you have *not* defined
category properly.



--
Jesse F. Hughes
"It's your choice though, if you do not believe in mathematics, in the
importance of its healthiness and correctness, then you can just walk
away now." -- James S Harris, on the Pythagorean Oath

[Jesse F. Hughes]

There really is no ambiguity in the definition. It's quite simple and
clear.

>> Because, you see, people have already defined terms like "category" and
>> even "concrete category" and you don't have to guess what these ought to
>> mean.  You can learn what they actually do mean.
>
> I shouldn't have to "learn" what they actually mean. Ideally, in
> mathematics, we should have a formal, stand-alone definition. Apart
> from some well chosen, informal commentary, that is all the "meaning"
> you should need. Why can't anyone provide such a definition? Is it
> impossible in the case of category theory?

What is simpler than the following?

A category is a tuple <O,M,dom,cod,o,id> satisfying the following
conditions:

dom:M -> O
cod:M -> O
o is a partial function M x M -> M written in infix form
id:O -> M (we write id(x) as id_x)
f o g is defined iff dom(f) = cod(g)
whenever f o g is defined, dom(f o g) = dom(g) and
cod(f o g) = cod(f)
whenever f o g and g o h are defined,
(f o g) o h = f o (g o h)
f o id_(dom f) = f & id_(cod(g)) o g = g

This is not formal, but it's not at all ambiguous. I really am
uncertain why you can't read this simple definition and formalize it
properly on the first go.

--
Jesse F. Hughes
"But regardless of my goofs, my reality of journals is different from
ANY of yours, as they just treat me in a special way."
-- James S. Harris

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 14, 7:13 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>> Dan Christensen wrote:
>>
>> > I shouldn't have to "learn" what they actually mean.
>>
>> The battle cry of the crank.
>>
>> > Ideally, in
>> > mathematics, we should have a formal, stand-alone definition. Apart
>> > from some well chosen, informal commentary, that is all the "meaning"
>> > you should need. Why can't anyone provide such a definition?
>>
>> What's wrong with http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C...
>>
>
> As an informal introduction, there is nothing wrong with it. I used it
> as one of the inspirations for creating my formal definitions. The
> author describes morphisms as "structure-preserving functions" (for
> concrete categories). That notion seems to be open to various
> interpretations here -- is it any function (not necessarily a
> surjection), just a binary relation, or (my choice) a transformation
> (a surjective function)?

Ah, there's your problem. Start with category and don't bother with the
informal English crap. Just go with the definition.

The definition of concrete category is more complicated than the
definition of category. Start with the simple.

Then, learn the actual components of the definition of concrete
category. It is *not* defined as a set with structure-preserving
morphisms. That's a vague description. A concrete category is a

category C with a faithful functor C -> Set.

Don't worry about what "structure-preserving" means. That's only
suggestive terminology. Learn what a category is, what a functor is,
and what it means for a functor to be faithful.

>
>
>> > Is it
>> > impossible in the case of category theory?
>>
>> If you want to know more see the references at the end of the Wikipedia
>> article, in particular Mac Lane, Saunders (1998). Categories for the
>> Working Mathematician. Graduate Texts in Mathematics 5 (2nd ed.).
>> Springer-Verlag. ISBN 0-387-98403-8.  That used to be the Bible, maybe
>> it still is, you should ask one the modern young people.
>>
>> There's also http://katmat.math.uni-bremen.de/acc/acc.pdf.
>>
>
> Here they say that morphisms can be structure preserving functions as
> well (for concrete categories), but they need even be not functions at
> all (for abstract categories). Here, I focus on concrete categories.

DON'T. It isn't helpful for your project.

That's rather like trying to formalize polygon by first formalizing
right triangle. It's not the way to go. Start with the simplest
definition, i.e., the most general.

> For structure preserving functions, I assumed they meant a
> transformation. Nothing else makes sense to me.

You're going about it backwards. You think that the subject is vague
because you're looking at the plain English descriptions rather than the
definitions.
--
Meaningless movies
on the screen behind the band that's blowing Waterboys,
throwing shapes "My Love is My Rock
Half of the music is on tape in the Weary Land"

[Jesse F. Hughes]

"Jesse F. Hughes" <je...@phiwumbda.org> writes:

> Cat(O,M,c) <=> dom:M -> O & cod:M -> O & c subset M x M x M & id:O -> M &
> (A o in O)(dom(id(o)) = O & cod(id(o)) = O) &
> (A m,m' in M)( (E m'')(A m''')( <m,m',m'''> in c ->
> m''' = m'' ) &
> (A m,m' in M)((E m'')( <m,m',m''> in c ) <->
> cod(m') = dom(m)) &
> (A m,m',m'' in M)( <m,m',m''> in c -> (*)
> dom(m'') = dom(m') ) &
> cod(m'') = cod(m) )) &
> (A m in M)( <m, id(dom(m)), m> in c & (**)
> <id(cod(m)), m, m> in c ) &
> Assoc(c)

Minor parentheses fixes.

Cat(O,M,c) <=> dom:M -> O & cod:M -> O & c subset M x M x M & id:O -> M &
(A o in O)(dom(id(o)) = O & cod(id(o)) = O) &

(A m,m' in M)(E m'')(A m''')( <m,m',m'''> in c ->

m''' = m'' ) &
(A m,m' in M)((E m'')( <m,m',m''> in c ) <->
cod(m') = dom(m)) &
(A m,m',m'' in M)( <m,m',m''> in c -> (*)

dom(m'') = dom(m') &
cod(m'') = cod(m) ) &

(A m in M)( <m, id(dom(m)), m> in c & (**)
<id(cod(m)), m, m> in c ) &
Assoc(c)

--
Jesse F. Hughes

"You people are the diminishment of a world."
-- James S. Harris, to mathematicians.

[Frederick Williams]

Graham Cooper wrote:

>
> go for the hat trick and add in symmetric and you get a partition!

Have you caught Nam's disease?

[Dan Christensen]

(Correction)

On Oct 14, 7:57 am, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

I think you will also need:

1. Some reference the class of all objects associated with a category.
If morphisms are to be structure preserving functions (for concrete
categories), you will need the class of all objects that exhibit that
structure.

2. A definition (using your notation) of the form

all c.(cat(c) == ( ... ))


3. A definition of the form

all c,x,y,f.(hom(c,x,y,f) == ( ... ))

[Nam Nguyen]

On 14/10/2012 8:55 AM, Frederick Williams wrote:
> Graham Cooper wrote:
>
>>
>> go for the hat trick and add in symmetric and you get a partition!
>
> Have you caught Nam's disease?

You're a liar Frederick.

And that can be shown/proved. So why don't you get lost.


--
----------------------------------------------------
There is no remainder in the mathematics of infinity.

NYOGEN SENZAKI
----------------------------------------------------

[Frederick Williams]

Nam Nguyen wrote:
>
> On 14/10/2012 8:55 AM, Frederick Williams wrote:
> > Graham Cooper wrote:
> >
> >>
> >> go for the hat trick and add in symmetric and you get a partition!
> >
> > Have you caught Nam's disease?
>
> You're a liar Frederick.

A question is not a lie.

> And that can be shown/proved.

So prove it. (Prove what, I'm not sure.)

> So why don't you get lost.

I can't. I'm actually a homing pigeon pretending to be a human. For me
to get lost is an impossibility.

[Alan Eaton]

On 15/10/2012 03:55, Dan Christensen wrote:
> (Correction)
>
> On Oct 14, 7:57 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> wrote:

<snip>

>> Thanks, Alan! This is precisely the sort of thing I have been looking
>> for. I will review it in detail and post my comments at a later time.
>>
>
> I think you will also need:
>
> 1. Some reference the class of all objects associated with a category.
> If morphisms are to be structure preserving functions (for concrete
> categories), you will need the class of all objects that exhibit that
> structure.

Why? I already have the predicate ob(c,x).
Adding classes (or sets) complicates things unnecessarily.

I intended for this definition to be as simple and as general as
possible. Hence the reliance only on predicates.

> 2. A definition (using your notation) of the form
>
> all c.(cat(c) == ( ... ))

The family of rules serves to define cat(c).

> 3. A definition of the form
>
> all c,x,y,f.(hom(c,x,y,f) == ( ... ))

I already have such a rule:

all c,x,y,f.( hom(c,x,y,f) == dom(c,f,x) & cod(c,f,y) )

Since there is no general rule characterising morphisms anything else
would limit generality.

[Dan Christensen]

On Oct 14, 2:59 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> On 15/10/2012 03:55, Dan Christensen wrote:
>
> > (Correction)
>
> > On Oct 14, 7:57 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> > wrote:
>
> <snip>
>
> >> Thanks, Alan! This is precisely the sort of thing I have been looking
> >> for. I will review it in detail and post my comments at a later time.
>
> > I think you will also need:
>
> > 1. Some reference the class of all objects associated with a category.
> > If morphisms are to be structure preserving functions (for concrete
> > categories), you will need the class of all objects that exhibit that
> > structure.
>
> Why? I already have the predicate ob(c,x).
> Adding classes (or sets) complicates things unnecessarily.
>

Then what meaning are we to attach to a structure-preserving function?

> I intended for this definition to be as simple and as general as
> possible. Hence the reliance only on predicates.
>
> > 2. A definition (using your notation) of the form
>
> > all c.(cat(c) == (  ...  ))
>
> The family of rules serves to define cat(c).
>

Given a class with the a certain structure common to each element, how
are we to formally determine whether this constitutes a category?

> > 3. A definition of the form
>
> > all c,x,y,f.(hom(c,x,y,f) == (  ...  ))
>
> I already have such a rule:
>
> all c,x,y,f.( hom(c,x,y,f) == dom(c,f,x) & cod(c,f,y) )
>

Yes, I missed that. Still, how is the idea of a structure-preserving
function brought into your definition? In my definition of Arrow, I
have the equivalent of your x and y being elements of the same class.
And instead of the codomain of f, I look at the the equivalent of the
image of x under f. Again:

ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob
& EXIST(f):[ALL(c):[c @ a => f(c) @ b]
& ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]

> Since there is no general rule characterising morphisms anything else
> would limit generality.

We are talking about so-called concrete categories that involve
morphisms that are structure-preserving functions. (Abstract
categories seem to be nothing more than a reflexive and transitive
relation defined on some class -- not very useful or interesting, I
would think.)

[Shmuel Metz]

In <877gqur...@phiwumbda.org>, on 10/13/2012

at 05:42 PM, "Jesse F. Hughes" <je...@phiwumbda.org> said:

>In the case of the category Set, there is no structure to be
>preserved.

One could view x \in y as a structure to be preserved, and define a
category whose morphisms from A to B were the mappings f:A->B such
that for all x,y \in A, f(x) \in f(y) if x \in y.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[Rotwang]

On 14/10/2012 21:44, Dan Christensen wrote:
> [...]

>
> Yes, I missed that. Still, how is the idea of a structure-preserving
> function brought into your definition?

It isn't, because the definition of a category says nothing about
structure-preserving functions. How are you not getting this?


> [...]

>
>> Since there is no general rule characterising morphisms anything else
>> would limit generality.
>
> We are talking about so-called concrete categories that involve
> morphisms that are structure-preserving functions. (Abstract
> categories seem to be nothing more than a reflexive and transitive
> relation defined on some class

No, they don't. They seem to be what the definition of a category says
they are, as given by Alan and Jesse and Mac Lane and the bit of that
Wikipedia page on category theory where the authors give the actual
definition of a category:

http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C_and_morphisms

> -- not very useful or interesting, I
> would think.)

As usual, you would think wrong.


--
I have made a thing that superficially resembles music:

http://soundcloud.com/eroneity/we-berated-our-own-crapiness

[Jesse F. Hughes]

Shmuel (Seymour J.) Metz <spam...@library.lspace.org.invalid> writes:

> In <877gqur...@phiwumbda.org>, on 10/13/2012
> at 05:42 PM, "Jesse F. Hughes" <je...@phiwumbda.org> said:
>
>>In the case of the category Set, there is no structure to be
>>preserved.
>
> One could view x \in y as a structure to be preserved, and define a
> category whose morphisms from A to B were the mappings f:A->B such
> that for all x,y \in A, f(x) \in f(y) if x \in y.

Yes, but that's not the category Set, of course.

But if you prefer, in the category Set, we don't require the morphisms
to preserve any structure at all.

In any case, Dan's tripping over informal motivations about what a
concrete category intuitively is, and thinking that this serves as a
definition. The boy insists on misunderstanding everything he touches.

--
"I arrest anybody I think needs arresting, Mr. Carter, and I'm not in
the habit of explaining why."
"There's a law about that ---"
"You're in Dodge, Mr. Carter." -- Gunsmoke radio show / John Ashcroft

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 14, 2:59 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
>> On 15/10/2012 03:55, Dan Christensen wrote:
>>
>> > (Correction)
>>
>> > On Oct 14, 7:57 am, Dan Christensen <Dan_Christen...@sympatico.ca>
>> > wrote:
>>
>> <snip>
>>
>> >> Thanks, Alan! This is precisely the sort of thing I have been looking
>> >> for. I will review it in detail and post my comments at a later time.
>>
>> > I think you will also need:
>>
>> > 1. Some reference the class of all objects associated with a category.
>> > If morphisms are to be structure preserving functions (for concrete
>> > categories), you will need the class of all objects that exhibit that
>> > structure.
>>
>> Why? I already have the predicate ob(c,x).
>> Adding classes (or sets) complicates things unnecessarily.
>>
>
> Then what meaning are we to attach to a structure-preserving function?

How many times must I say this?

Don't worry about "structure-preserving functions". Start with a simple
understanding of what a category is by the *definition* of category.
The phrase "structure-preserving function" is nothing but a bit of
suggestive language, not part of the definition.

>> I intended for this definition to be as simple and as general as
>> possible. Hence the reliance only on predicates.
>>
>> > 2. A definition (using your notation) of the form
>>
>> > all c.(cat(c) == (  ...  ))
>>
>> The family of rules serves to define cat(c).
>>
>
> Given a class with the a certain structure common to each element, how
> are we to formally determine whether this constitutes a category?

Your question is utterly meaningless.

A category consists of a collection of objects, a collection of arrows
(with domain and codomain), a partial composition function on the
arrows, and an identity arrow for each object, where these things
satisfy the various conditions that I won't repeat here.

The word "structure" does not occur in the definition of category.

You keep looking at an informal description of concrete category,
pretending that it is a definition and complaining that it is vague.

That is just not a fruitful endeavor.

>
>
>> > 3. A definition of the form
>>
>> > all c,x,y,f.(hom(c,x,y,f) == (  ...  ))
>>
>> I already have such a rule:
>>
>> all c,x,y,f.( hom(c,x,y,f) == dom(c,f,x) & cod(c,f,y) )
>>
>
> Yes, I missed that. Still, how is the idea of a structure-preserving
> function brought into your definition? In my definition of Arrow, I
> have the equivalent of your x and y being elements of the same class.
> And instead of the codomain of f, I look at the the equivalent of the
> image of x under f. Again:
>
> ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob
> & EXIST(f):[ALL(c):[c @ a => f(c) @ b]
> & ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]
>
>
>> Since there is no general rule characterising morphisms anything else
>> would limit generality.
>
> We are talking about so-called concrete categories that involve
> morphisms that are structure-preserving functions. (Abstract
> categories seem to be nothing more than a reflexive and transitive
> relation defined on some class -- not very useful or interesting, I
> would think.)

It includes all the concrete categories.

I once wanted to study geometric figures, but that includes all these
weird shapes that aren't interesting or useful, so I decided to define
polygon so that only right triangles are polygons. That seemed like a
great place to start.

Dammit, Dan, the easiest place to start studying a field is in the basic
definitions. In category theory, the most basic definition is
*category*. Why do you insist on looking at concrete categories, when
you have no idea what a functor is? (A concrete category is a category
C with a faithful functor C -> Set. That's the fucking definition, Dan,
and it doesn't involve the words "structure-preserving".)

--
"It seems to me that in wartime Americans shouldn't be attacking each
other in this way on a *worldwide* forum. Then again, I know I'm an
American, but I have no way of knowing that you are, which would
explain a lot." --James Harris, on why Yanks should accept his proof

[Dan Christensen]

On Oct 14, 5:33 pm, Rotwang <sg...@hotmail.co.uk> wrote:
> On 14/10/2012 21:44, Dan Christensen wrote:
>
> > [...]
>
> > Yes, I missed that. Still, how is the idea of a structure-preserving
> > function brought into your definition?
>
> It isn't, because the definition of a category says nothing about
> structure-preserving functions. How are you not getting this?
>

Oh, I don't know....

"Category theory emphasizes the morphisms – the structure-preserving
mappings – between these objects."
http://en.wikipedia.org/wiki/Category_theory

"Morphisms are structure-preserving functions."
http://en.wikipedia.org/wiki/Morphism

"The study of categories is an attempt to axiomatically capture what
is commonly found in various classes of related mathematical
structures by relating them to the structure-preserving functions
between them."

"Most basic categories have as objects certain mathematical
structures, and the
structure-preserving functions as morphisms."
http://www.staff.science.uu.nl/~ooste110/syllabi/catsmoeder.pdf

"Most of the examples of categories that one encounters in the
literature encode mathematical structures: the objects will be
examples of this mathematical structure and the morphisms will be the
structure-preserving maps between these."
http://www.cs.ox.ac.uk/people/bob.coecke/ctfwp1_final.pdf

etc.

> > [...]
>
> >> Since there is no general rule characterising morphisms anything else
> >> would limit generality.
>
> > We are talking about so-called concrete categories that involve
> > morphisms that are structure-preserving functions. (Abstract
> > categories seem to be nothing more than a reflexive and transitive
> > relation defined on some class
>
> No, they don't. They seem to be what the definition of a category says
> they are, as given by Alan and Jesse and Mac Lane and the bit of that
> Wikipedia page on category theory where the authors give the actual
> definition of a category:
>

> http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C...
>

Again, at this link, we have:

"Category theory emphasizes the morphisms – the structure-preserving
mappings – between these object."

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 14, 5:33 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>> On 14/10/2012 21:44, Dan Christensen wrote:
>>
>> > [...]
>>
>> > Yes, I missed that. Still, how is the idea of a structure-preserving
>> > function brought into your definition?
>>
>> It isn't, because the definition of a category says nothing about
>> structure-preserving functions. How are you not getting this?
>>
>
> Oh, I don't know....
>

> "Category theory emphasizes the morphisms - the structure-preserving
> mappings - between these objects."

> http://en.wikipedia.org/wiki/Category_theory
>
> "Morphisms are structure-preserving functions."
> http://en.wikipedia.org/wiki/Morphism

Those aren't mathematical definitions. They're informal language that
suggests the point of category theory.

Look at the definitions.

> "The study of categories is an attempt to axiomatically capture what
> is commonly found in various classes of related mathematical
> structures by relating them to the structure-preserving functions
> between them."
>
> "Most basic categories have as objects certain mathematical
> structures, and the
> structure-preserving functions as morphisms."
> http://www.staff.science.uu.nl/~ooste110/syllabi/catsmoeder.pdf

Do you know what a definition is?

Look, on that first page, there really is a definition of category.
Here:

http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C_and_morphisms

>
> "Most of the examples of categories that one encounters in the
> literature encode mathematical structures: the objects will be
> examples of this mathematical structure and the morphisms will be the
> structure-preserving maps between these."
> http://www.cs.ox.ac.uk/people/bob.coecke/ctfwp1_final.pdf
>

Yes, intuitively, the morphisms *do* preserve structure in most
examples, but this is *NOT* part of the definition of category.

> etc.
>
>
>> > [...]
>>
>> >> Since there is no general rule characterising morphisms anything else
>> >> would limit generality.
>>
>> > We are talking about so-called concrete categories that involve
>> > morphisms that are structure-preserving functions. (Abstract
>> > categories seem to be nothing more than a reflexive and transitive
>> > relation defined on some class
>>
>> No, they don't. They seem to be what the definition of a category says
>> they are, as given by Alan and Jesse and Mac Lane and the bit of that
>> Wikipedia page on category theory where the authors give the actual
>> definition of a category:
>>
>> http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C...
>>
>
> Again, at this link, we have:
>

> "Category theory emphasizes the morphisms - the structure-preserving
> mappings - between these object."

If you can't tell the difference between plain English commentary and
actual definitions, then it's no wonder you're having trouble
formalizing the definition.

Honestly, Dan, you have to really be trying to misunderstand. You claim
that the term "category" aren't clearly defined, but you refuse to read
the definition when it's on the same fucking web page as something else
you quote.

What is wrong with you? Rotwang even pointed you to the definition, but
instead you leap straight back to the loose, suggestive language of
informal discussion.

Don't you want to know what a category is?

--
"I'd step through arguments in such detail that it was like I was
teaching basic arithmetic and some poster would come back and act like
I hadn't said anything that made sense. For a while I almost started
to doubt myself." -- James S. Harris, so close and yet....

[Dan Christensen]

On Oct 14, 5:43 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Oct 14, 2:59 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> >> On 15/10/2012 03:55, Dan Christensen wrote:
>
> >> > (Correction)
>
> >> > On Oct 14, 7:57 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> >> > wrote:
>
> >> <snip>
>
> >> >> Thanks, Alan! This is precisely the sort of thing I have been looking
> >> >> for. I will review it in detail and post my comments at a later time.
>
> >> > I think you will also need:
>
> >> > 1. Some reference the class of all objects associated with a category.
> >> > If morphisms are to be structure preserving functions (for concrete
> >> > categories), you will need the class of all objects that exhibit that
> >> > structure.
>
> >> Why? I already have the predicate ob(c,x).
> >> Adding classes (or sets) complicates things unnecessarily.
>
> > Then what meaning are we to attach to a structure-preserving function?
>
> How many times must I say this?
>
> Don't worry about "structure-preserving functions".

Otherwise, we are only talking about a very uninteresting binary
relation.

> Start with a simple
> understanding of what a category is by the *definition* of category.
> The phrase "structure-preserving function" is nothing but a bit of
> suggestive language, not part of the definition.
>

On the contrary, it seems integral to the notion of a concrete
category. (See links in my reply to Rotwang.)

> >> I intended for this definition to be as simple and as general as
> >> possible. Hence the reliance only on predicates.
>
> >> > 2. A definition (using your notation) of the form
>
> >> > all c.(cat(c) == (  ...  ))
>
> >> The family of rules serves to define cat(c).
>
> > Given a class with the a certain structure common to each element, how
> > are we to formally determine whether this constitutes a category?
>
> Your question is utterly meaningless.
>

Not at all.

> A category consists of a collection of objects, a collection of arrows
> (with domain and codomain), a partial composition function on the
> arrows, and an identity arrow for each object, where these things
> satisfy the various conditions that I won't repeat here.
>

You mean a binary relation that is reflexive and transitive. <Yawn!>

> The word "structure" does not occur in the definition of category.
>

Not in abstract categories where the objects in question are just
atomic individuals, but it most certainly does on concrete categories.

> You keep looking at an informal description of concrete category,
> pretending that it is a definition and complaining that it is vague.
>

Unfortunately, the only available descriptions of concrete category
seem to be quite informal. If I want a formal one, it seems I will
have to make it up.

> That is just not a fruitful endeavor.

I am quite happy with the latest versions of my definition of Category
and Arrow predicates. The only question seems to be: surjections or
not surjections?


Does anyone know of some widely accepted, fundamental results for
concrete categories in general that I could use to test alternative
formulations?

That is precisely what I have been doing.

> In category theory, the most basic definition is
> *category*.  Why do you insist on looking at concrete categories, when
> you have no idea what a functor is?

One thing at a time, Jesse. Let's formally nail down what a category
is. I am more interested in concrete, less so in the abstract variety.

"Most of the examples of categories that one encounters in the
literature encode mathematical structures: the objects will be
examples of this mathematical structure and the morphisms will be the

structure-preserving maps between these. This kind of categories is
usually referred to as concrete categories....

"Since the key structural data of a category is its composition,
emphasis is given to the structure preserving maps rather than the
structures themselves."

Source: "Categories for the practising physicist"
http://www.cs.ox.ac.uk/people/bob.coecke/ctfwp1_final.pdf

[Dan Christensen]

On Oct 14, 10:28 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Oct 14, 5:33 pm, Rotwang <sg...@hotmail.co.uk> wrote:
> >> On 14/10/2012 21:44, Dan Christensen wrote:
>
> >> > [...]
>
> >> > Yes, I missed that. Still, how is the idea of a structure-preserving
> >> > function brought into your definition?
>
> >> It isn't, because the definition of a category says nothing about
> >> structure-preserving functions. How are you not getting this?
>
> > Oh, I don't know....
>
> > "Category theory emphasizes the morphisms - the structure-preserving
> > mappings - between these objects."
> >http://en.wikipedia.org/wiki/Category_theory
>
> > "Morphisms are structure-preserving functions."
> >http://en.wikipedia.org/wiki/Morphism
>
> Those aren't mathematical definitions.  They're informal language that
> suggests the point of category theory.
>
> Look at the definitions.
>

The ones I have seen online aren't very formal or mathematical either.
Hence, my quest here.

> > "The study of categories is an attempt to axiomatically capture what
> > is commonly found in various classes of related mathematical
> > structures by relating them to the structure-preserving functions
> > between them."
>
> > "Most basic categories have as objects certain mathematical
> > structures, and the
> > structure-preserving functions as morphisms."
> >http://www.staff.science.uu.nl/~ooste110/syllabi/catsmoeder.pdf
>
> Do you know what a definition is?
>
> Look, on that first page, there really is a definition of category.
> Here:
>

>  http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C...

>
>
>
> > "Most of the examples of categories that one encounters in the
> > literature encode mathematical structures: the objects will be
> > examples of this mathematical structure and the morphisms will be the
> > structure-preserving maps between these."
> >http://www.cs.ox.ac.uk/people/bob.coecke/ctfwp1_final.pdf
>
> Yes, intuitively, the morphisms *do* preserve structure in most
> examples, but this is *NOT* part of the definition of category.
>

That seems to be a problem with the informal definitions often given
in the literature.

> > etc.
>
> >> > [...]
>
> >> >> Since there is no general rule characterising morphisms anything else
> >> >> would limit generality.
>
> >> > We are talking about so-called concrete categories that involve
> >> > morphisms that are structure-preserving functions. (Abstract
> >> > categories seem to be nothing more than a reflexive and transitive
> >> > relation defined on some class
>
> >> No, they don't. They seem to be what the definition of a category says
> >> they are, as given by Alan and Jesse and Mac Lane and the bit of that
> >> Wikipedia page on category theory where the authors give the actual
> >> definition of a category:
>
> >>http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C...
>
> > Again, at this link, we have:
>
> > "Category theory emphasizes the morphisms - the structure-preserving
> > mappings - between these object."
>
> If you can't tell the difference between plain English commentary and
> actual definitions, then it's no wonder you're having trouble
> formalizing the definition.

I don't think I am having all that much trouble. It seems that my
definitions of Category and Arrow predicates will require at most
minor tweaking on one issue from what I have seen here.

>
> Honestly, Dan, you have to really be trying to misunderstand.  You claim
> that the term "category" aren't clearly defined,

There doesn't seem to be any widely accepted, formal definitions.
Doesn't that bother you?

> but you refuse to read
> the definition when it's on the same fucking web page as something else
> you quote.

The definitions given are not formal ones. That's my main problem with
them. And they are vague enough that a formal definition wasn't
immediately apparent.

>
> What is wrong with you?  Rotwang even pointed you to the definition, but
> instead you leap straight back to the loose, suggestive language of
> informal discussion.
>
> Don't you want to know what a category is?
>

I want a formal definition -- one incorporating the notion of
structure-preserving functions. If that is too much too ask, the
phrases "abstract nonsense" and "bullshit" comes to mind.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 14, 5:43 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > On Oct 14, 2:59 pm, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
>> >> On 15/10/2012 03:55, Dan Christensen wrote:

[...]

>>
>> > Then what meaning are we to attach to a structure-preserving function?
>>
>> How many times must I say this?
>>
>> Don't worry about "structure-preserving functions".
>
> Otherwise, we are only talking about a very uninteresting binary
> relation.

You are confusing the definition of "category" and the categories of
interest.

Your approach is analogous to the following:

I want to know about real-valued functions. I don't know the
definition of real-valued function, but I hear that the only good such
functions are infinitely differentiable. I don't know what that
means, but people say it means "smooth".

What nonsense! "Smooth" is such a vague word! Smooth like a baby's
butt, or smooth like a cue ball? The theory of real-valued functions
is just full of ambiguity!

Start with the definition of category. Have some interesting categories
in mind, but stop pretending that "structure-preserving functions" is
part of the *definition* of category or even concrete category. It
isn't.

>> Start with a simple
>> understanding of what a category is by the *definition* of category.
>> The phrase "structure-preserving function" is nothing but a bit of
>> suggestive language, not part of the definition.
>>
>
> On the contrary, it seems integral to the notion of a concrete
> category. (See links in my reply to Rotwang.)

You need to start with the definition of category, and functor, and
faithful, and *then* you can worry about what concrete category means.
Hint: the definition of concrete category does *not* include the phrase
structure-preserving.


[...]

>>
>> > Given a class with the a certain structure common to each element, how
>> > are we to formally determine whether this constitutes a category?
>>
>> Your question is utterly meaningless.
>>
>
> Not at all.

Yes, it is.

>> A category consists of a collection of objects, a collection of arrows
>> (with domain and codomain), a partial composition function on the
>> arrows, and an identity arrow for each object, where these things
>> satisfy the various conditions that I won't repeat here.
>>
>
> You mean a binary relation that is reflexive and transitive. <Yawn!>

No, I don't. It's clear that you need to go back to the definition,
because you don't get it yet.

>> The word "structure" does not occur in the definition of category.
>>
>
> Not in abstract categories where the objects in question are just
> atomic individuals, but it most certainly does on concrete categories.

No, it does not.

Definition: A concrete category is a category C with a faithful functor
C -> Set.

This is the definition. Show me where the word structure appears.

>> You keep looking at an informal description of concrete category,
>> pretending that it is a definition and complaining that it is vague.
>>
>
> Unfortunately, the only available descriptions of concrete category
> seem to be quite informal. If I want a formal one, it seems I will
> have to make it up.

Definition: A concrete category is a category C with a faithful functor
C -> Set.

>> That is just not a fruitful endeavor.
>
> I am quite happy with the latest versions of my definition of Category
> and Arrow predicates. The only question seems to be: surjections or
> not surjections?

Whatever you're doing, it isn't category theory.

As far as your silliness over whether it should be surjections or not,
ask yourself this: is every group homomorphism a surjection? Is every
continuous function a surjection? What about every monotone map between
partial orders?

But even these questions aren't correct. Your starting questions should
be: what is a category and what is a faithful functor? Because that's
the only way you'll learn what a concrete category is.

> Does anyone know of some widely accepted, fundamental results for
> concrete categories in general that I could use to test alternative
> formulations?

To be honest, no, because concrete categories just never appeared
particularly interesting in my studies. Oh, examples of such categories
are very useful, indeed the most common sort of examples, but as far as
special properties that concrete categories have, I know of none.

But I'd wager that any such properties would be too technical to explain
to you at this point anyway. After all, you still don't know the
fundamental definitions of category theory.

[...]

>> > We are talking about so-called concrete categories that involve
>> > morphisms that are structure-preserving functions. (Abstract
>> > categories seem to be nothing more than a reflexive and transitive
>> > relation defined on some class -- not very useful or interesting, I
>> > would think.)
>>
>> It includes all the concrete categories.
>>
>> I once wanted to study geometric figures, but that includes all these
>> weird shapes that aren't interesting or useful, so I decided to define
>> polygon so that only right triangles are polygons.  That seemed like a
>> great place to start.
>>
>> Dammit, Dan, the easiest place to start studying a field is in the basic
>> definitions.
>
> That is precisely what I have been doing.

No, you haven't.

>
>> In category theory, the most basic definition is
>> *category*.  Why do you insist on looking at concrete categories, when
>> you have no idea what a functor is?
>
> One thing at a time, Jesse. Let's formally nail down what a category
> is. I am more interested in concrete, less so in the abstract
> variety.

It doesn't matter that you're interested in concrete categories more
than the general definition. You have to start with the general
definition.

The concrete categories are a special case.

>
> "Most of the examples of categories that one encounters in the
> literature encode mathematical structures: the objects will be
> examples of this mathematical structure and the morphisms will be the
> structure-preserving maps between these. This kind of categories is
> usually referred to as concrete categories....
>
> "Since the key structural data of a category is its composition,
> emphasis is given to the structure preserving maps rather than the
> structures themselves."
>
> Source: "Categories for the practising physicist"
> http://www.cs.ox.ac.uk/people/bob.coecke/ctfwp1_final.pdf

Quote all you want, but until you start quoting actual definitions,
you're wasting your time.
--
Jesse F. Hughes

"A factor is simply something that multiplies against another factor
to produce a 'product'." -- James Harris offers a definition.

[Dan Christensen]

On Oct 14, 11:23 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

[snipping repetitious parts]

> > I am quite happy with the latest versions of my definition of Category
> > and Arrow predicates. The only question seems to be: surjections or
> > not surjections?
>
> Whatever you're doing, it isn't category theory.
>

So you have said.

> As far as your silliness over whether it should be surjections or not,
> ask yourself this: is every group homomorphism a surjection?

[snip]

I see structure-preserving morphisms like homomorphisms as
transformations of one set to another with no "leftovers" -- by
definition, a surjection onto an image set. If this is not part of a
workable formal definition, I remain open minded on this and any other
issue.

[William Hale]

In article
<c91d37f1-2883-4c93...@l18g2000vbv.googlegroups.com>,

I don't think "structure-preserving" function precedes the notion of
"category": rather, "category" allows one to say that two categories
have the "same" structure.

[William Hale]

In article
<df71b188-30bb-4332...@s14g2000vba.googlegroups.com>,

Homomorphisms need not be surjections.

[Alan Eaton]

On 15/10/2012 13:41, Dan Christensen wrote:

<snip>

> I see structure-preserving morphisms like homomorphisms as
> transformations of one set to another with no "leftovers" -- by
> definition, a surjection onto an image set. If this is not part of a
> workable formal definition, I remain open minded on this and any other
> issue.

If that is what you see then you need to get your eyes fixed.

I think your notion of structure might be a bit off. See:

http://en.wikipedia.org/wiki/Structure_(mathematical_logic)

for a general definition of structures and their homomorphisms.

[Alan Eaton]

On 15/10/2012 13:00, Dan Christensen wrote:
> On Oct 14, 5:43 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

<snip>

>> A category consists of a collection of objects, a collection of arrows
>> (with domain and codomain), a partial composition function on the
>> arrows, and an identity arrow for each object, where these things
>> satisfy the various conditions that I won't repeat here.
>>
>
> You mean a binary relation that is reflexive and transitive. <Yawn!>

No he doesn't.
You keep making this kind of statement. Stop it! It is wrong.

A category can only be viewed in such a way if for each pair of
objects x and y there is at most 1 morphism from x to y.

The following is an example category(and it *IS* a category) that
shows your statement to be false.

You might find this category of interest since its morphisms are
functions but they are not functions from their domains to their
codomains.


Objects
The points of V=R^42

Morphisms
The continuous functions f:[0,a]->V where [0,a] c R and 0 <= a


We will use the following helper function in further definitions:

length:Morphisms->R
(length(f) = a) == (f:[0,a]->V)


dom:Morphisms->Objects
dom(f) = f(0)

cod:Morphisms->Objects
cod(f) = f(length(f))

o:(Morphisms x Morphisms)->Morphisms
o is a partial function.
(dom(f) = cod(g)) -> (
(f o g):[0,length(f)+length(g)]->V
& (t e [0,length(g)] -> (f o g)(t) = g(t))
& (t e [length(g),length(f)+length(g)] -> (f o g)(t) = f(t-length(g)))
)

from the above:
length(f o g) = length(f) + length(g)

dom(f o g) = dom(g)

cod(f o g) = cod(f)

f o (g o h) = (f o g) o h

hom:(Objects x Objects)->pow(Morphisms)
hom(x,y) = {f e Morphisms| dom(f)=x & cod(f)=y}

id:Objects->Morphisms
id(x):[0,0]->V
id(x)(0) = x


Note that for any pair of objects x and y, hom(x,y) has uncountably
many elements.

This category is most certainly NOT a yawn-worthy binary relation.
(and its morphisms are not surjections)

[Jesse F. Hughes]

Alan Eaton <alan.den...@gmail.com> writes:

> On 15/10/2012 13:00, Dan Christensen wrote:
>> On Oct 14, 5:43 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> <snip>
>
>>> A category consists of a collection of objects, a collection of arrows
>>> (with domain and codomain), a partial composition function on the
>>> arrows, and an identity arrow for each object, where these things
>>> satisfy the various conditions that I won't repeat here.
>>>
>>
>> You mean a binary relation that is reflexive and transitive. <Yawn!>
>
> No he doesn't.
> You keep making this kind of statement. Stop it! It is wrong.
>
> A category can only be viewed in such a way if for each pair of
> objects x and y there is at most 1 morphism from x to y.

In other words, if the category is a pre-order.

> The following is an example category(and it *IS* a category) that
> shows your statement to be false.
>
> You might find this category of interest since its morphisms are
> functions but they are not functions from their domains to their
> codomains.

[Snip example]

Yet another example which Dan will ignore:

Let M be a monoid. We define a category as follows:

Obj: * That is, there is only one object, which we denote *.
Mor: for each m in M, an arrow m:* -> *.
Composition: Every pair of arrows is compatible, and m o m' is defined
to be the arrow m . m' (where . is the monoid multiplication)
Identity: The identity element e in M is obviously the identity
morphism.

--
"The job of the Justices is to make long lists of benefit versus harm
[...] if the benefits outweigh the harm, then the Health Care Bill is
constitutional. As simple as that."
--Archimedes Plutonium, on legal utilitarianism.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 14, 10:28 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > Oh, I don't know....
>>
>> > "Category theory emphasizes the morphisms - the structure-preserving
>> > mappings - between these objects."
>> >http://en.wikipedia.org/wiki/Category_theory
>>
>> > "Morphisms are structure-preserving functions."
>> >http://en.wikipedia.org/wiki/Morphism
>>
>> Those aren't mathematical definitions.  They're informal language that
>> suggests the point of category theory.
>>
>> Look at the definitions.
>>
>
> The ones I have seen online aren't very formal or mathematical either.
> Hence, my quest here.

Stop quoting informal, motivational discussion and start quoting
definitions.

Tell me what's wrong with the definition found on the WP page.

[...]

> There doesn't seem to be any widely accepted, formal definitions.
> Doesn't that bother you?

Most mathematicians don't do purely formal reasoning, but anyone with a
modicum of basic logic can certainly transform the WP definition (which
is, despite your claim about "widespread acceptance", the absolutely
standard definition found in any text) into a formal definition. I've
done so in this thread. You ignored it.

Here's the simple formalization. For an explanation, see the original
post.

Cat(O,M,c) <=> dom:M -> O & cod:M -> O & c subset M x M x M & id:O -> M &
(A o in O)(dom(id(o)) = O & cod(id(o)) = O) &
(A m,m' in M)(E m'')(A m''')( <m,m',m'''> in c ->
m''' = m'' ) &
(A m,m' in M)((E m'')( <m,m',m''> in c ) <->
cod(m') = dom(m)) &
(A m,m',m'' in M)( <m,m',m''> in c -> (*)
dom(m'') = dom(m') &
cod(m'') = cod(m) ) &
(A m in M)( <m, id(dom(m)), m> in c & (**)
<id(cod(m)), m, m> in c ) &
Assoc(c)

>> but you refuse to read
>> the definition when it's on the same fucking web page as something else
>> you quote.
>
> The definitions given are not formal ones. That's my main problem with
> them. And they are vague enough that a formal definition wasn't
> immediately apparent.

They are not at all vague. They are as precise as any standard
mathematical definition. They are just as precise as the definition of
partially ordered set, of group and so on.

>
>
>>
>> What is wrong with you?  Rotwang even pointed you to the definition, but
>> instead you leap straight back to the loose, suggestive language of
>> informal discussion.
>>
>> Don't you want to know what a category is?
>>
>
> I want a formal definition -- one incorporating the notion of
> structure-preserving functions. If that is too much too ask, the
> phrases "abstract nonsense" and "bullshit" comes to mind.

There is no clear definition of "structure-preserving functions" in
category theory or elsewhere[1], because that phrase is informal and
suggestive. It is not a defined mathematical term at all.

How many times must I say that?

Show me *any* definition (know what a definition is?) of "(concrete)
category" in which that term appears. It doesn't.

Footnotes:
[1] At least not that I've seen.

--
"They are anti-mathematicians, evil incarnate, dedicated to undermining
intellectual development in this area. If you never thought such
people could actually exist, outside of myths or legends, welcome to
the real world." --James S Harris on evil incarnate's Usenet presence

[Jesse F. Hughes]

Well, you think that because you have apparently no familiarity with the
intuitions behind terms like "structure" and "structure-preserving". In
terms of topology, for instance, the structure-preserving morphisms are
the continuous functions. Do you think that every continuous function
is a surjection?

But none of this matters. Everyone has to learn these intuitions at
some point. Want to know a good way to get an informal idea of what
structure means?

Start with the basic definition of category. Look at many examples of
categories (all of the examples I think of off the top of my head are
concrete). Work with them.

Try the Barr and Wells book "Category Theory for Computing Science".
It's an excellent introductory text.

--
"I am the barbarian at the gates, raw creative force, willpower, and
the will to fight for the truth no matter what, no matter who stands
against me, no matter how many of you band [...] together in your
weakness to fight against the math." -- James S. Harris

[Shmuel Metz]

In <87pq4kp...@phiwumbda.org>, on 10/14/2012

at 05:40 PM, "Jesse F. Hughes" <je...@phiwumbda.org> said:

>Yes, but that's not the category Set, of course.

Nor did I claim that it was, or even that it was as useful as the
category Set, just that I could attach meaning to "preserve the
structure".

>The boy insists on misunderstanding everything he touches.

Well, he certainly has trouble distinguishing definitions, examples
and motivation from each other. However, even a stopped (12-hour)
clock is right twice a day. Whether he will draw the correct
inferences after fortuitously being right about something is a
separate question, but I suspect that quantifier dyslexia will
interfere.

Wait until you start explaining the difference between a small
category and a large category ;-)

[Shmuel Metz]

In <87y5j9q...@phiwumbda.org>, on 10/14/2012
at 09:58 AM, "Jesse F. Hughes" <je...@phiwumbda.org> said:

>Alan has given one formal definition. I'll give another,

It needs some work.

>the pair <O,M,d,cod,c,id>

That's not a pair, and if you're going to use ordered tuples then you
need to give some axioms for them.

>Let Delta(O) be the set

You need axioms for Set Theory. A modern formulation of Category
Theory avoids sets entirely[1], although its application includes such
things as Local Set Theory.

[1] That's necessary if you want to define sets in terms of
categories et al.

[Jesse F. Hughes]

Shmuel (Seymour J.) Metz <spam...@library.lspace.org.invalid> writes:

> In <87pq4kp...@phiwumbda.org>, on 10/14/2012
> at 05:40 PM, "Jesse F. Hughes" <je...@phiwumbda.org> said:
>
>>Yes, but that's not the category Set, of course.
>
> Nor did I claim that it was, or even that it was as useful as the
> category Set, just that I could attach meaning to "preserve the
> structure".

Right. And there's other structure we could use as well. For instance,
we could be interested in functions f:X -> Y where, for each x in X,

|x| <= |f(x)|,

or functions such that f({}) = {}, and so on.

So, perhaps what I said was misleading, but what I meant is that no
structure is preserved in the category Set itself.

>>The boy insists on misunderstanding everything he touches.
>
> Well, he certainly has trouble distinguishing definitions, examples
> and motivation from each other. However, even a stopped (12-hour)
> clock is right twice a day. Whether he will draw the correct
> inferences after fortuitously being right about something is a
> separate question, but I suspect that quantifier dyslexia will
> interfere.

I haven't noticed quantifier dyslexia among Dan's many shortcomings so
far.

> Wait until you start explaining the difference between a small
> category and a large category ;-)

Yeah, that may turn out to be an issue, since many of the categories
he's interested in will be large. But so far, that's a distant concern.

--
Jesse F. Hughes
Did you lay down in heaven? Did you wake up in hell?
I bet you never guessed that it would be so hard to tell.
-- The Flatlanders, /Judgment Day/

[Jesse F. Hughes]

Shmuel (Seymour J.) Metz <spam...@library.lspace.org.invalid> writes:

> In <87y5j9q...@phiwumbda.org>, on 10/14/2012
> at 09:58 AM, "Jesse F. Hughes" <je...@phiwumbda.org> said:
>
>>Alan has given one formal definition. I'll give another,
>
> It needs some work.
>
>>the pair <O,M,d,cod,c,id>
>
> That's not a pair, and if you're going to use ordered tuples then you
> need to give some axioms for them.

I meant tuple, not pair, and I presumed we have axioms for them. After
all, a tuple can easily be implemented as nested pairs, so if we won't
balk at ordered pairs, we shouldn't balk at tuples either.

>>Let Delta(O) be the set
>
> You need axioms for Set Theory. A modern formulation of Category
> Theory avoids sets entirely[1], although its application includes such
> things as Local Set Theory.
>
> [1] That's necessary if you want to define sets in terms of
> categories et al.

Yes, I agree that for my definition, I am assuming the axioms of set
theory. And, yes, of course modern formulations of category theory do
not presume set theory as a background, but here I was trying to give
Dan something that he might like to work with.

He thinks of set theory as the foundation for mathematics and has
developed some theory of sets. So, in order to make the definition easy
for him to understand, I made it a definition in the context of set
theory.

I really don't think that Dan is ready for category theory in all its
generality yet.
--
Jesse F. Hughes
"Well, you know as soon as you have a new number I will be happy to
add it to the list. Don't try those childish tit-for-tat games with
me." -- Ross Finlayson on Cantor's theorem.

[Dan Christensen]

On Oct 15, 7:08 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Oct 14, 10:28 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> >> > Oh, I don't know....
>
> >> > "Category theory emphasizes the morphisms - the structure-preserving
> >> > mappings - between these objects."
> >> >http://en.wikipedia.org/wiki/Category_theory
>
> >> > "Morphisms are structure-preserving functions."
> >> >http://en.wikipedia.org/wiki/Morphism
>
> >> Those aren't mathematical definitions.  They're informal language that
> >> suggests the point of category theory.
>
> >> Look at the definitions.
>
> > The ones I have seen online aren't very formal or mathematical either.
> > Hence, my quest here.
>
> Stop quoting informal, motivational discussion and start quoting
> definitions.
>
> Tell me what's wrong with the definition found on the WP page.
>

See my comments on your formalism below.

> [...]
>
> > There doesn't seem to be any widely accepted, formal definitions.
> > Doesn't that bother you?
>
> Most mathematicians don't do purely formal reasoning, but anyone with a
> modicum of basic logic can certainly transform the WP definition (which
> is, despite your claim about "widespread acceptance", the absolutely
> standard definition found in any text) into a formal definition.  I've
> done so in this thread.  You ignored it.
>

I was getting around to it. I thought I should deal with Alan's
formulation first.

> Here's the simple formalization.  For an explanation, see the original
> post.
>
> Cat(O,M,c) <=> dom:M -> O & cod:M -> O
> & c subset M x M x M
> & id:O -> M

Are you saying that id maps the class O to the class M? Isn't id an
element of M?

>              & (A o in O)(dom(id(o)) = O & cod(id(o)) = O)

Shouldn't you be quantifying over M?

>              & (A m,m' in M)(E m'')(A m''')( <m,m',m'''> in c ->
>                                               m''' = m'' )

c is a binary function on M.

>              & (A m,m' in M)((E m'')( <m,m',m''> in c ) <->
>                              cod(m') = dom(m))

Shouldn't that be dom(m')=img(m)?

>             &  (A m,m',m'' in M)( <m,m',m''> in c ->                    (*)
>                                    dom(m'') = dom(m')  &

Shouldn't that be dom(m'')=dom(m)?

>                                    cod(m'') = cod(m) )

Shouldn't that be cod(m'') = cod(m')?

>              & (A m in M)( <m, id(dom(m)), m> in c &                   (**)
>                            <id(cod(m)), m, m> in c )

Identity property.

>             &  Assoc(c)
>

Shouldn't composition also be transitive? I'm not sure that you have
fully characterized composition of morphisms here.

This definition could not be more vague on morphisms as structure-
preserving functions. There is indeed NOTHING at all about here about
them! As the subject heading says, however, I am interested in a
definition of CONCRETE categories -- categories with morphisms that
ARE structure-preserving functions, the main emphasis in applications
of category theory IIUC. If concrete categories are what you are
interested in, I think my relatively compact list definitions may be
all you need.

> >> but you refuse to read
> >> the definition when it's on the same fucking web page as something else
> >> you quote.
>
> > The definitions given are not formal ones. That's my main problem with
> > them. And they are vague enough that a formal definition wasn't
> > immediately apparent.
>
> They are not at all vague.  They are as precise as any standard
> mathematical definition.  They are just as precise as the definition of
> partially ordered set, of group and so on.
>
>
>
> >> What is wrong with you?  Rotwang even pointed you to the definition, but
> >> instead you leap straight back to the loose, suggestive language of
> >> informal discussion.
>
> >> Don't you want to know what a category is?
>
> > I want a formal definition -- one incorporating the notion of
> > structure-preserving functions. If that is too much too ask, the
> > phrases "abstract nonsense" and "bullshit" comes to mind.
>
> There is no clear definition of "structure-preserving functions" in
> category theory or elsewhere[1], because that phrase is informal and
> suggestive.  It is not a defined mathematical term at all.
>

How can that be? Isn't a class not uniquely determined by some
property common to all of its elements? If the elements of class X are
themselves classes, you can think of this property as a "structure" on
each of these elements. This structure will be preserved by any
surjection from one element of X to another.

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 15, 7:08 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> > There doesn't seem to be any widely accepted, formal definitions.
>> > Doesn't that bother you?
>>
>> Most mathematicians don't do purely formal reasoning, but anyone with a
>> modicum of basic logic can certainly transform the WP definition (which
>> is, despite your claim about "widespread acceptance", the absolutely
>> standard definition found in any text) into a formal definition.  I've
>> done so in this thread.  You ignored it.
>>
>
> I was getting around to it. I thought I should deal with Alan's
> formulation first.

No problem there.

>
>> Here's the simple formalization.  For an explanation, see the original
>> post.
>>
>> Cat(O,M,c) <=> dom:M -> O & cod:M -> O
>> & c subset M x M x M
>> & id:O -> M
>
> Are you saying that id maps the class O to the class M? Isn't id an
> element of M?

For each object o in O, there is an identity morphism id(o) (usually
written id_o) in M. id_o is a morphism o -> o.

Thus, id maps objects to morphisms, by

o |-> id_o:o -> o.

>
>>              & (A o in O)(dom(id(o)) = O & cod(id(o)) = O)

OOPS! That should read:
>>              & (A o in O)(dom(id(o)) = o & cod(id(o)) = o)
^ ^

(Note lowercase o's.)

> Shouldn't you be quantifying over M?

No, for reasons given above. Here, all I'm saying is that for each
object o in O, id_o is a morphism with domain and codomain o.

>
>>              & (A m,m' in M)(E m'')(A m''')( <m,m',m'''> in c ->
>>                                               m''' = m'' )
>
> c is a binary function on M.

A partial binary function on M. I don't require

(A m,m' in M)(E m'')(<m,m',m''> in c).

>>              & (A m,m' in M)((E m'')( <m,m',m''> in c ) <->
>>                              cod(m') = dom(m))
>
> Shouldn't that be dom(m')=img(m)?

First, it shouldn't be img, but cod. The codomain is typically larger
than the image.

Second, I wrote it in the traditional order for composition of
functions. That is, f o g is defined whenever dom(f) = cod(g).

>
>>             &  (A m,m',m'' in M)( <m,m',m''> in c ->                    (*)
>>                                    dom(m'') = dom(m')  &
>
> Shouldn't that be dom(m'')=dom(m)?

See above.

>
>>                                    cod(m'') = cod(m) )
>
> Shouldn't that be cod(m'') = cod(m')?

Ditto.

>
>>              & (A m in M)( <m, id(dom(m)), m> in c &                   (**)
>>                            <id(cod(m)), m, m> in c )
>
> Identity property.
>
>>             &  Assoc(c)
>>
>
> Shouldn't composition also be transitive? I'm not sure that you have
> fully characterized composition of morphisms here.

I don't know what it means for a partial function to be transitive. It
is associative.

> This definition could not be more vague on morphisms as structure-
> preserving functions. There is indeed NOTHING at all about here about
> them! As the subject heading says, however, I am interested in a
> definition of CONCRETE categories -- categories with morphisms that
> ARE structure-preserving functions, the main emphasis in applications
> of category theory IIUC. If concrete categories are what you are
> interested in, I think my relatively compact list definitions may be
> all you need.

OF COURSE there's nothing here about structure-preserving. This is what
I've been saying all along. The term "structure-preserving" is just an
informal description of how we think about concrete categories. But
that term *DOES NOT* appear in the definition of concrete category.

Defn: A concrete category is a category C with a faithful functor
C -> Set.

That is the definition. It is intended to formally capture the informal
notion of "structure-preserving" morphisms, but you are not yet ready to
understand how, because you are not yet ready to understand the
definition.

>> >> but you refuse to read
>> >> the definition when it's on the same fucking web page as something else
>> >> you quote.
>>
>> > The definitions given are not formal ones. That's my main problem with
>> > them. And they are vague enough that a formal definition wasn't
>> > immediately apparent.
>>
>> They are not at all vague.  They are as precise as any standard
>> mathematical definition.  They are just as precise as the definition of
>> partially ordered set, of group and so on.
>>
>>
>>
>> >> What is wrong with you?  Rotwang even pointed you to the definition, but
>> >> instead you leap straight back to the loose, suggestive language of
>> >> informal discussion.
>>
>> >> Don't you want to know what a category is?
>>
>> > I want a formal definition -- one incorporating the notion of
>> > structure-preserving functions. If that is too much too ask, the
>> > phrases "abstract nonsense" and "bullshit" comes to mind.
>>
>> There is no clear definition of "structure-preserving functions" in
>> category theory or elsewhere[1], because that phrase is informal and
>> suggestive.  It is not a defined mathematical term at all.
>>
>
> How can that be? Isn't a class not uniquely determined by some
> property common to all of its elements? If the elements of class X are
> themselves classes, you can think of this property as a "structure" on
> each of these elements. This structure will be preserved by any
> surjection from one element of X to another.

I've no idea what you think a structure is, but your guess is not
well-informed.

I've given you about a dozen examples of concrete categories. In the
bulk of them, the morphisms are not surjections. How much plainer can
this be?
--
"I AM serious about this being a short route to a Ph.d for some of
you, but just remember, I'm the guy who proved Fermat's Last Theorem
in just a bit over 6 years [...] My standards are kind of high."
--James Harris, founding a new mathematical school

[Alan Smaill]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> How can that be? Isn't a class not uniquely determined by some
> property common to all of its elements? If the elements of class X are
> themselves classes, you can think of this property as a "structure" on
> each of these elements. This structure will be preserved by any
> surjection from one element of X to another.

Typically the "structure" is not something that holds
of individual elements, but of relationships between the elements.

A good example is that of groups with morphisms just the usual
group homomorphisms. What is preserved is the group
structure - eg composition of elements.

Collectively the image of a group under a homomorphism is a group wrt
the target group operations. There is no need for these homomorphisms
to be surjective, though -- subgroup inclusion is
a perfectly good morphism.

> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com
>
>
>

--
Alan Smaill

[Dan Christensen]

On Oct 15, 11:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

[snip]

> > This definition could not be more vague on morphisms as structure-
> > preserving functions. There is indeed NOTHING at all about here about
> > them! As the subject heading says, however, I am interested in a
> > definition of CONCRETE categories -- categories with morphisms that
> > ARE structure-preserving functions, the main emphasis in applications
> > of category theory IIUC. If concrete categories are what you are
> > interested in, I think my relatively compact list definitions may be
> > all you need.
>
> OF COURSE there's nothing here about structure-preserving.  This is what
> I've been saying all along.  The term "structure-preserving" is just an
> informal description of how we think about concrete categories.

[snip]

I think I have successfully formalized this notion of structure-
preserving. Here it is again:

Define: Category predicate

1 ALL(a):[Category(a)
<=> ALL(b):[b @ a => EXIST(f):ALL(c):[c @ b => f(c)=c]]]

Informally, a class is a category iff each element of that class is
itself a class that has an identity function defined on it. (Note that
an identity function can be constructed on any set.)

Define: Arrow predicate

2 ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob

& EXIST(f):[ALL(c):[c @ a => f(c) @ b]
& ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]

Informally, the arrow relation indicates the existence of a morphism
-- a transformation or surjection -- mapping elements of one class
onto another. And since the image and pre-image are both in the same
class, both will exhibit the property (or structure) that defines
membership in that class.

Morphisms here are just functions (surjections) mapping one element of
a class to another. Therefore, the Arrow relation is reflexive and
transitive.

The composition of morphisms here is just the ordinary composition of
functions. So, the composition of morphisms must also be associative.

[Dan Christensen]

On Oct 15, 12:05 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > How can that be? Isn't a class not uniquely determined by some
> > property common to all of its elements? If the elements of class X are
> > themselves classes, you can think of this property as a "structure" on
> > each of these elements. This structure will be preserved by any
> > surjection from one element of X to another.
>
> Typically the "structure" is not something that holds
> of individual elements, but of relationships between the elements.
>
> A good example is that of groups with morphisms just the usual
> group homomorphisms. What is preserved is the group
> structure - eg composition of elements.
>
> Collectively the image of a group under a homomorphism is a group wrt
> the target group operations.  There is no need for these homomorphisms
> to be surjective, though -- subgroup inclusion is
> a perfectly good morphism.

I guess it depends on what you want to do with these "morphisms" in
your category theory. But it seems an odd notion that we necessarily
want to consider a function mapping an infinite space to single point
to be a structure-preserving morphism. What am I missing here?

[Dan Christensen]

On Oct 15, 3:16 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

In abstract algebra, we cannot have a ring with only one element. So,
if map ring X into a single point in X, how can we say that this
mapping is a structure-preserving morphism?

[Jesse F. Hughes]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 15, 11:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> [snip]
>
>> > This definition could not be more vague on morphisms as structure-
>> > preserving functions. There is indeed NOTHING at all about here about
>> > them! As the subject heading says, however, I am interested in a
>> > definition of CONCRETE categories -- categories with morphisms that
>> > ARE structure-preserving functions, the main emphasis in applications
>> > of category theory IIUC. If concrete categories are what you are
>> > interested in, I think my relatively compact list definitions may be
>> > all you need.
>>
>> OF COURSE there's nothing here about structure-preserving.  This is what
>> I've been saying all along.  The term "structure-preserving" is just an
>> informal description of how we think about concrete categories.
>
> [snip]
>
> I think I have successfully formalized this notion of structure-
> preserving. Here it is again:

Thanks, but I won't bother reading this.

It's clear that you don't know the intuition behind the informal
language of "structure-preserving morphisms", because I don't think you
actually have familiarity with many such structures.

Whatever you're formalizing here, it isn't "concrete category", because
that term is already well-defined and it has to do with faithful
functors into set.

>
> Define: Category predicate
>
> 1 ALL(a):[Category(a)
> <=> ALL(b):[b @ a => EXIST(f):ALL(c):[c @ b => f(c)=c]]]
>
> Informally, a class is a category iff each element of that class is
> itself a class that has an identity function defined on it. (Note that
> an identity function can be constructed on any set.)
>
> Define: Arrow predicate
>
> 2 ALL(ob):ALL(a):ALL(b):[Arrow(ob,a,b) <=> a @ ob & b @ ob
> & EXIST(f):[ALL(c):[c @ a => f(c) @ b]
> & ALL(c):[c @ b => EXIST(d):[d @ a & f(d)=c]]]]
>
> Informally, the arrow relation indicates the existence of a morphism
> -- a transformation or surjection -- mapping elements of one class
> onto another. And since the image and pre-image are both in the same
> class, both will exhibit the property (or structure) that defines
> membership in that class.
>
> Morphisms here are just functions (surjections) mapping one element of
> a class to another. Therefore, the Arrow relation is reflexive and
> transitive.

Continuous maps need not be surjections.
Algebraic homomorphisms need not be surjections.
And so on and so on.

But never mind. You just go on and guess what "structure-preserving"
means and don't worry about actual examples of structures and their
morphisms.

> The composition of morphisms here is just the ordinary composition of
> functions. So, the composition of morphisms must also be associative.

--
"It's one of the easiest tickets to true fame--not this silly stuff
where people cheer you for a few years and then forget about you--but
the kind of fame where school kids have to read your biography and do
reports on you." -- Another reason to support James S. Harris.

[Jesse F. Hughes]

Any curiosity regarding how others use mathematical terms, for starters.

Just work through one chapter of Category Theory for Computer Science.
See examples of categories. Why try to re-invent the wheel, when all
you know are triangles?

--
"Even if a man has good parts, still, if he carries philosophy into
later life, he is necessarily ignorant of all those things which a
gentleman and a person of honor ought to know."
--Callicles, in "Gorgias"

[Jesse F. Hughes]

Perhaps you're laboring under a delusion that every map is either a
surjection or a constant map. It isn't so.

But let's talk about rings. Know what a ring homomorphism is? It's
*any* map f:R -> S (R and S are rings) satisfying

f(a + b) = f(a) + f(b) for all a and b in R
f(ab) = f(a) f(b) for all a and b in R

For instance, the inclusion map Z -> Q is a ring homomorphism (Z is the
ring of integers, Q the ring of rationals).

Is this a surjection?

No?

Does it map everything in Z to a single point in Q?

No?

Well, how about that. A morphism which is neither surjective nor
constant.

--
Jesse F. Hughes

"I may have invented it, but Bill [Gates] made it famous."
-- David Bradley, inventor of the Ctrl-Alt-Del reboot sequence

[Alan Smaill]

What's odd?

The properties (structure) preserved by morphism f:

f(x.y) = f(x).f(y)
f(i(x)) = i(f(x))

(with i as inverse, . as composition in the respective groups),
and this works fine if everything maps to the identity element
in the target group.

I suspect the name "morphism" comes from this notion of algebraic
homomorphism. Anyone know?

If you want to preserve more, then particular sorts of
morphism will do that.

> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com

--
Alan Smaill

[Alan Smaill]

If you're working in the category of Rings (with a unit)
and ring homomorphisms, no, since the map you describe is
not a ring homomorphism (whatever you take the target ring to be).

> Dan
> Download my DC Proof 2.0 software athttp://www.dcproof.com
>

--
Alan Smaill

[Shmuel Metz]

In <87ipabo...@phiwumbda.org>, on 10/15/2012

at 09:10 AM, "Jesse F. Hughes" <je...@phiwumbda.org> said:

>He thinks of set theory as the foundation for mathematics

So do I, but I recognize that the modern use of categories as
fundamental is equally legitimate and that my preference for sets is
mostly habit. I also reject the description "abstract nonsense".

>I really don't think that Dan is ready for category theory in all
>its generality yet.

Assuming that what you quoted is representative, that's an
understatement. In particular, I found "I see structure-preserving

morphisms like homomorphisms as transformations of one set to another
with no "leftovers" -- by definition, a surjection onto an image set.

" rather astonishing.

[Dan Christensen]

Perhaps you can answer my question, Jesse. Is such a mapping a
structure-preserving morphism or not?

[Dan Christensen]

So, morphisms in a concrete category MUST be surjections. Agreed?

[Jesse F. Hughes]

Look, that does not follow!

There are maps which are neither surjections *nor* constant maps.

Furthermore, there are categories in which every constant map is a
morphism. In the category Top of topological spaces[1], every constant
map is a morphism (i.e. continuous). This is a concrete category.

Here are the two most obvious ways in which your reasoning has run off
the rails in that single sentence.

(1) You think that if a constant map is not a morphism, then all
morphisms are surjections.

(2) You think that if in a single category, constant maps are not
morphisms, then they're not morphisms in any category.

All this is in addition to the more fundamental problem. You think you
can formalize concrete category without a single, fucking clue what it
means.

Footnotes:
[1] No offense, but you don't know what those are, do you?

--
Jesse F. Hughes
"Contrariwise," continued Tweedledee, "if it was so, it might be, and
if it were so, it would be; but as it isn't, it ain't. That's logic!"
-- Lewis Carroll

[Alan Smaill]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 15, 4:00 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > On Oct 15, 3:16 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
>> > wrote:
>> >> On Oct 15, 12:05 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>>

...

>> >> > Typically the "structure" is not something that holds
>> >> > of individual elements, but of relationships between the elements.
>>
>> >> > A good example is that of groups with morphisms just the usual
>> >> > group homomorphisms. What is preserved is the group
>> >> > structure - eg composition of elements.
>>
>> >> > Collectively the image of a group under a homomorphism is a group wrt
>> >> > the target group operations.  There is no need for these homomorphisms
>> >> > to be surjective, though -- subgroup inclusion is
>> >> > a perfectly good morphism.
>>
>> >> I guess it depends on what you want to do with these "morphisms" in
>> >> your category theory. But it seems an odd notion that we necessarily
>> >> want to consider a function mapping an infinite space to single point
>> >> to be a structure-preserving morphism. What am I missing here?
>>
>> > In abstract algebra, we cannot have a ring with only one element. So,
>> > if map ring X into a single point in X, how can we say that this
>> > mapping is a structure-preserving morphism?
>>
>> If you're working in the category of Rings (with a unit)
>> and ring homomorphisms, no, since the map you describe is
>> not a ring homomorphism (whatever you take the target ring to be).
>>
>
> So, morphisms in a concrete category MUST be surjections. Agreed?

No.

I have no idea where you draw that conclusion from.

Here's an example from group theory, in the category
of groups with group homomorphisms:

let G1 be the unique group wih 1 element.
let G2 be the unique group with 2 elements.

The homomorphism f: G1 -> G2 that maps the identity element
in G1 to the identity element in G2 is a morphism in this category.
This function is not surjective.

> Dan
> Download my DC Proof 2.0 software athttp://www.dcproof.com

--
Alan Smaill

[Jesse F. Hughes]

Honestly, I haven't any idea why he keeps asking this question when he's
been given example after example of non-surjective morphisms in concrete
categories.

He really ought to take the time to work through a chapter or two of an
introductory category theory text. Though, I'm beginning to wonder
whether he has the background for that.

--
"It is my opinion that since neither Spight nor Hughes can see or
understand their moral trespass [namely, quoting AP in .sigs], that
their degrees from whatever university they earned their degree should
be annulled." -- Archimedes Plutonium (12/1/09)

[Dan Christensen]

Yeah, it does.

> There are maps which are neither surjections *nor* constant maps.
>
> Furthermore, there are categories in which every constant map is a
> morphism.

In group theory, a mapping of all points to the identity element is a
homomorphism. It depends on whether the image set is an object in the
category. That's why you must consider only surjections.

> In the category Top of topological spaces[1], every constant
> map is a morphism (i.e. continuous).  This is a concrete category.
>
> Here are the two most obvious ways in which your reasoning has run off
> the rails in that single sentence.
>
> (1) You think that if a constant map is not a morphism, then all
> morphisms are surjections.
>

All morphisms in a concrete category anyway.

> (2) You think that if in a single category, constant maps are not
> morphisms, then they're not morphisms in any category.
>

Didn't say that.

> All this is in addition to the more fundamental problem. You think

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com

PS: Just admit you were wrong, Jesse. (Hee, hee!)

[Dan Christensen]

See my reply to Jesse.

[Jesse F. Hughes]

No, it doesn't, for reasons which I gave below.

>> There are maps which are neither surjections *nor* constant maps.
>>
>> Furthermore, there are categories in which every constant map is a
>> morphism.
>
>
> In group theory, a mapping of all points to the identity element is a
> homomorphism. It depends on whether the image set is an object in the
> category. That's why you must consider only surjections.

No, you're mistaken.

The codomain is not the same as the image. The inclusion map N -> Q is
a ring homomorphism which is not a surjection.

You are confusing codomain and image. (By the way, it is not
necessarily the case that the "image" of a morphism is an object in the
category. I don't know whether it is always the case for concrete
categories.)

>
>> In the category Top of topological spaces[1], every constant
>> map is a morphism (i.e. continuous).  This is a concrete category.
>>
>> Here are the two most obvious ways in which your reasoning has run off
>> the rails in that single sentence.
>>
>> (1) You think that if a constant map is not a morphism, then all
>> morphisms are surjections.
>>
>
>
> All morphisms in a concrete category anyway.

Top is a concrete category. Can't you read?

I would prove to you that Top is a concrete category, by showing you
that the evident forgetful functor Top -> Set is faithful. Indeed, this
is trivial. But you don't know what a functor is. More amazingly, you
still don't understand what a category is, though you've been pointed to
the definition repeatedly.

>> (2) You think that if in a single category, constant maps are not
>> morphisms, then they're not morphisms in any category.
>>
>
>
> Didn't say that.

You did. Your (addled) reasoning was this:

(1) No constant map r |-> 0 is a ring homomorphism.
(2) So every morphism is a surjection (in every category).

>
>> All this is in addition to the more fundamental problem. You think
>
>
> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com
>
> PS: Just admit you were wrong, Jesse. (Hee, hee!)

Son, I wrote a doctoral thesis in category theory in a pure and applied
logic program. I don't say that to appeal to authority here -- after
all, I've not done any category theory in some time -- but just to
indicate that I'm not the least bit concerned that you magically
understand the field better than I do.

I am sure that there are others here who understand it much better than
I. You just aren't one of them.

Hell, I'll make you a wager. I wager I can name two dozen concrete
categories in which not every morphism is a surjection. (I'm pretty
sure it's a triviality to name infinitely many such categories, but I'll
just go for two dozen for starters[1].) We'll pick a neutral judge (since
you're incompetent and you may reasonably accuse me of bias). If I win,
you give a quarter to the girl scouts. If you win, I publicly admit
that Dan Christensen is an amazing genius who intuits more category
theory than I could ever learn -- and I'll give fifty cents to the girl
scouts.


Footnotes:
[1] Every monoid is a concrete category and I'm pretty sure that every
morphism in this category is epic iff the map m |-> x * m is onto for
every x in m. Consequently, all we have to do is find infinitely many
monoids for which this is not the case, and I can't imagine that this is
in the least bit difficult. I'll wager that playing around with the
monoid N^k will do it in two shakes.

But I'll find two dozen concrete categories that are actually different
in flavor in order to win the wager.

Anyway, the above monoid stuff has not been checked.

--
Jesse F. Hughes
"I don't know if you noticed but I had a tremendous drop in confidence
concomittant [sic] with a dramatic grip of existential crisis."
--- James S. Harris even has better diseases than you

[Alan Smaill]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> On Oct 15, 4:47 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > On Oct 15, 4:00 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> >> > On Oct 15, 3:16 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
>> >> > wrote:

...

>>
>> >> >> I guess it depends on what you want to do with these "morphisms" in
>> >> >> your category theory. But it seems an odd notion that we necessarily
>> >> >> want to consider a function mapping an infinite space to single point
>> >> >> to be a structure-preserving morphism. What am I missing here?
>>
>> >> > In abstract algebra, we cannot have a ring with only one element. So,
>> >> > if map ring X into a single point in X, how can we say that this
>> >> > mapping is a structure-preserving morphism?
>>
>> >> If you're working in the category of Rings (with a unit)
>> >> and ring homomorphisms, no, since the map you describe is
>> >> not a ring homomorphism (whatever you take the target ring to be).
>>
>> > So, morphisms in a concrete category MUST be surjections. Agreed?
>>
>> Look, that does not follow!
>>
>
>
> Yeah, it does.

No.

>> There are maps which are neither surjections *nor* constant maps.
>>
>> Furthermore, there are categories in which every constant map is a
>> morphism.
>
>
> In group theory, a mapping of all points to the identity element is a
> homomorphism. It depends on whether the image set is an object in the
> category. That's why you must consider only surjections.

Even if the image set (with appropriate structure) is an object,
nothing says that the *image* set is the same as the *co-domain*
of the function. What you need in the (concrete) category
is that the image set is a *subset* of the co-domain.
That's why you need to consider more than surjections.

These really are examples right at the start of any introductory text
on category theory.

> Dan
> Download my DC Proof 2.0 software at http://www.dcproof.com
>
> PS: Just admit you were wrong, Jesse. (Hee, hee!)

Hee, hee, indeed.

--
Alan Smaill

[Jesse F. Hughes]

I thought I had answered this previously, but I don't see my post.

No, the function you describe is not a morphism in the category Ring,
but this fact does not entail that every morphism is onto.

Why, if only I could give you an example of a morphism which isn't
onto. Maybe an example involving rings.

Oh, wait. I did give you that example. You snipped it in this post.

Here it is again: a ring homomorphism which is not a surjection. (Nor
is it constant, of course.)

Know what a ring homomorphism is? It's *any* map f:R -> S (R and S are
rings) satisfying

f(a + b) = f(a) + f(b) for all a and b in R
f(ab) = f(a) f(b) for all a and b in R

For instance, the inclusion map Z -> Q is a ring homomorphism (Z is the
ring of integers, Q the ring of rationals).

Is this a surjection?

No?

Does it map everything in Z to a single point in Q?

No?

Well, how about that. A morphism which is neither surjective nor
constant.
--

"Phil & Larry, you guys sounded excellent in the show. You came across
very well, and sounded like you were doing a professional job
regarding investigation, and did not sound in any way crazy."
-- Conspiracy nut Phil Jayhan gets a positive review

[Dan Christensen]

Where or where can it be?

> No, the function you describe is not a morphism in the category Ring,
> but this fact does not entail that every morphism is onto.
>

[snip]


By definition, however, every function is a surjection onto its image
set. And it is a function's image set that determines whether it is a
morphism in a concrete category. So, you only want to be looking for
surjections.

[Dan Christensen]

On Oct 15, 5:23 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

[snip]

> > In group theory, a mapping of all points to the identity element is a
> > homomorphism. It depends on whether the image set is an object in the
> > category. That's why you must consider only surjections.
>
> No, you're mistaken.
>
> The codomain is not the same as the image.

Never said it was.

> The inclusion map N -> Q is
> a ring homomorphism which is not a surjection.
>
> You are confusing codomain and image.

I don't think so, Jesse.

> (By the way, it is not
> necessarily the case that the "image" of a morphism is an object in the
> category.

Those must be some weird categories!

> I don't know whether it is always the case for concrete
> categories.)
>
>
>
> >> In the category Top of topological spaces[1], every constant
> >> map is a morphism (i.e. continuous).  This is a concrete category.
>
> >> Here are the two most obvious ways in which your reasoning has run off
> >> the rails in that single sentence.
>
> >> (1) You think that if a constant map is not a morphism, then all
> >> morphisms are surjections.
>
> > All morphisms in a concrete category anyway.
>
> Top is a concrete category.  Can't you read?
>

Let me rephrase that. To determine if a function is a morphism in a
concrete category, you must look at that function's image set. By
definition, every function is a surjection onto its image set, so you
only need to look at functions that are surjections. Happy now?

> I would prove to you that Top is a concrete category, by showing you
> that the evident forgetful functor Top -> Set is faithful.  Indeed, this
> is trivial. But you don't know what a functor is.  More amazingly, you
> still don't understand what a category is, though you've been pointed to
> the definition repeatedly.
>

Now, now, Jesse. It was just that kind of attitude that got into this
mess in the first place.

> Son, I wrote a doctoral thesis in category theory in a pure and applied
> logic program.  I don't say that to appeal to authority here -- after
> all, I've not done any category theory in some time -- but just to
> indicate that I'm not the least bit concerned that you magically
> understand the field better than I do.
>

Gosh. I'm flattered.

> I am sure that there are others here who understand it much better than
> I.  You just aren't one of them.
>
> Hell, I'll make you a wager.  I wager I can name two dozen concrete
> categories in which not every morphism is a surjection.

[snip]

Still don't get it, do you, Jesse? Re-read my comments above. Slowly.

[Jesse F. Hughes]

You fail to understand that codomain is not the same as image.

In category theory, we can consider the function

{0} -> {0,1,2,3,...}.

This is distinct from the function

{0} -> {0},

even though *as sets*, they are equal, namely both functions literally
are the set {<0,0>}.

In some categories (like the category Set, for instance), every arrow
can be decomposed into an epimorphism (analogous to surjection) followed
by a monomorphism (analogous to injection). Nonetheless, we still
distinguish the two arrows above. Categorically, they are different,
because they have different codomain (though the same image).

I genuinely hope that helps. I have my doubts, though, since you've
shown no willingness to learn category theory.

--
"Memoirists like Frey and Augusten Burroughs belong to the long list of
those who should never have stopped using drugs. The drugs might have
made Frey more interesting, or they might have killed him. Either way,
American literature would have benefited." --John Dolan, www.exile.ru

[Jesse F. Hughes]

This is an utterly trivial fact about the category set. For every
category C with a functor U: C -> Set, and for every morphism f in C, we
have that Uf is onto its image in the category Set[1]. This has nothing
at all to do with "structure-preserving" and everything to do with a
basic fact about the category Set.

>
>> I would prove to you that Top is a concrete category, by showing you
>> that the evident forgetful functor Top -> Set is faithful.  Indeed, this
>> is trivial. But you don't know what a functor is.  More amazingly, you
>> still don't understand what a category is, though you've been pointed to
>> the definition repeatedly.
>>
>
>
> Now, now, Jesse. It was just that kind of attitude that got into this
> mess in the first place.

If you are admitting that your refusal to learn category theory coupled
with your insistence to formalize concrete categories got us into this
mess, then I couldn't agree more.

>> Son, I wrote a doctoral thesis in category theory in a pure and applied
>> logic program.  I don't say that to appeal to authority here -- after
>> all, I've not done any category theory in some time -- but just to
>> indicate that I'm not the least bit concerned that you magically
>> understand the field better than I do.
>>
>
> Gosh. I'm flattered.
>
>
>> I am sure that there are others here who understand it much better than
>> I.  You just aren't one of them.
>>
>> Hell, I'll make you a wager.  I wager I can name two dozen concrete
>> categories in which not every morphism is a surjection.
>
> [snip]
>
> Still don't get it, do you, Jesse? Re-read my comments above. Slowly.

You have finally spelled out that you're not claiming that morphisms are
surjections, but simply that their underlying set maps factor through an
image.

The interesting question, which we haven't answered and which I don't
know the answer to offhand, is this: Is it the case that in every
concrete category, a morphism can also be factored into an epi followed
by a mono. I probably knew the answer once, but I don't recall it at
present.

I suspect the answer is no.


Footnotes:
[1] More categorically stated, every morphism g in Set can be
"factored" into an epi followed by a mono, i.e., g = i o e, where e is
epi and i is mono.

--
Jesse F. Hughes
"I love Mathematics[...] She doesn't care about my feelings, or my
pride or how I so wish to get out and travel. Truth is truth, and
anything else is just plain wrong." -- James Harris, /A Love Story/

[Dan Christensen]

On Oct 15, 5:30 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

[snip]

> > In group theory, a mapping of all points to the identity element is a
> > homomorphism. It depends on whether the image set is an object in the
> > category. That's why you must consider only surjections.
>
> Even if the image set (with appropriate structure) is an object,
> nothing says that the *image* set is the same as the *co-domain*
> of the function.

Agreed.

> What you need in the (concrete) category
> is that the image set is a *subset* of the co-domain.
> That's why you need to consider more than surjections.
>

You can get ALL the morphisms by looking only at surjections. That's
the point of my definition of the Arrow predicate (see above).

[Jesse F. Hughes]

"Jesse F. Hughes" <je...@phiwumbda.org> writes:

> The interesting question, which we haven't answered and which I don't
> know the answer to offhand, is this: Is it the case that in every
> concrete category, a morphism can also be factored into an epi followed
> by a mono. I probably knew the answer once, but I don't recall it at
> present.

A brief sketch that the answer is no, to be filled in later. This is
not really at an appropriate level for Dan, to be sure, but I'll try to
fill in some details later.

Again, what I'm trying to show here is that, even though the underlying
set-functions factor into an epi followed by a mono, it is not the case
that this factorization has to be the image of a factorization in the
concrete category, C.

Idea: Every monoid is a concrete category, where the functor from the
monoid M (as a category) to C takes the unique object * to the set of
elements of M (written M), and each arrow m in M to the function
M -> M defined by left multiplication, i.e. m(m') = m * m'.

This is a faithful functor.

Now, consider the monoid consisting of {0,1,2,3} where multiplication is
integer multiplication mod 4. I claim that the arrow 2: * -> * cannot
be factored into a (set) surjection followed by an (set) inclusion in
the category M.

I haven't worked out the details, but I'll try to get to it tomorrow or
later. I'm not sure that I can make it apparent to Dan, but so be it.

In any case, this isn't a very weird category at all.
--
One these mornings gonna wake | Ain't nobody's doggone business how
up crazy, | my baby treats me,
Gonna grab my gun, kill my baby. | Nobody's business but mine.
Nobody's business but mine. | -- Mississippi John Hurt

[Dan Christensen]

On Oct 15, 11:08 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

[snip]

> > By definition, however, every function is a surjection onto its image
> > set. And it is a function's image set that determines whether it is a
> > morphism in a concrete category. So, you only want to be looking for
> > surjections.
>
> You fail to understand that codomain is not the same as image.
>

Wrong again.

> In category theory, we can consider the function
>
>   {0} -> {0,1,2,3,...}.
>
> This is distinct from the function
>
>   {0} -> {0},
>
> even though *as sets*, they are equal, namely both functions literally
> are the set {<0,0>}.

Not sure what you are getting at here.

>
> In some categories (like the category Set, for instance), every arrow
> can be decomposed into an epimorphism (analogous to surjection) followed
> by a monomorphism (analogous to injection).  Nonetheless, we still
> distinguish the two arrows above.  Categorically, they are different,
> because they have different codomain (though the same image).
>

Nice try, Jesse. Remember, a function is just a set of ordered n-
tuples. Arbitrarily expanding what you consider to be the codomain
will have no effect on that set.

[Jesse F. Hughes]

In category theory, those two functions are distinct. This is a
departure from the usual treatment in set theory.

This is fundamental to the nature of the study, Dan. You have to learn
to work with it, or you'll get nowhere. The inclusion function
{0} -> N is, categorically speaking, a different function than the
identity function {0} -> {0}.

--
Jesse F. Hughes
"But you probably aren't a person with the ability to make any kind of
checks for yourself. But you do talk a lot in posts on Usenet where
you probably live out some fantasy." --James S. Harris is funning, no?

[Dan Christensen]

On Oct 16, 12:13 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

Category theory needs to get its act together.

> This is a
> departure from the usual treatment in set theory.
>

Indeed. A rather silly departure. What possible advantage can there be
to see different functions when there is only the one underlying set?
It's like saying 1/2 and 2/4 are really two different numbers.

What was that about abstract nonsense?