"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."
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.
> "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."
> 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.
> > "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."
> > 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.
> 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."
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?
>> 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."
> 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.
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?
> 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)]]]
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.
> > 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)]]]
> 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.
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.
> > 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.
> 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.
go for the hat trick and add in symmetric and you get a partition!
>> 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.
> 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.
> 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.
> > 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>
> >> 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.
> > 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?
> > 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.
> On Oct 13, 6:32 am, Alan Eaton <alan.dennis.ea...@gmail.com> wrote:
> > 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.
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.
Dan Christensen <Dan_Christen...@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
> 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.
>> 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.
> > 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 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.
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).
-- 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 <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.
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.
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.
> 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.
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:
> > 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.
IMO group theory and category theory look like AI reruns of Search
Operands and Semantic Networks.
> 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?"
Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> 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.
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?"
> 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 <Dan_Christen...@sympatico.ca> writes:
> > 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.
> 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?
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.
