Toward a Formal, Machine-Parsable Definition of a Category

[Dan Christensen]

Notation:

ob = class of objects
mor = class of morphisms
dom(f) = domain of f
cod(f) = codomain of f
comp(g,f) = the composition operator (for functions f and g, comp(g,f)
= g o f)
hom(x,y) = the class of morphisms with domain of x and codomain of y

@ = epsilon (indicates class membership)
& = AND-operator
=> = IMPLIES-operator
<=> = IFF-operator


The ordered sextuple (ob, mor, dom, cod, comp, hom) comprises a
category iff the following axioms hold:

Axioms 1-3 define dom and cod. Each morphisms has a unique domain and
codomain. Note that morphisms are seen here only as are mappings from
the ELEMENTS of one object to the elements of another object. If, for
example, you want to define to define a category on the set of natural
numbers, you would have to use the set of singletons of natural
numbers for ob = {{1}, {2}, {3}, ...}. This should not be an onerous
requirement.

1 ALL(f):[f @ mor => dom(f) @ ob]

2 ALL(f):[f @ mor => cod(f) @ ob]

3 ALL(f):[f @ mor => ALL(a):ALL(b):[(a,b) @ f => a @ dom(f) & b @
cod(f)]]

Identity morphisms exist for all objects.

4 ALL(a):[a @ ob => EXIST(i):[i @ mor & ALL(b):[b @ a => (b,b) @ i]]]

The composition operator comp

5 ALL(f):ALL(g):[f @ mor & g @ mor => [cod(f)=dom(g) => comp(g,f) @
mor
& dom(comp(g,f))=dom(f)
& cod(comp(g,f))=cod(g)
& ALL(a):ALL(b):[(a,b) @ comp(g,f) <=> EXIST(c):[(a,c) @ f & (c,b) @
g]]]]

Associativity of composition

6 ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [cod(f)=dom(g)
& cod(g)=dom(h)
=> comp(comp(h,g),f)=comp(h,comp(g,f))]]

hom classes exist for each ordered pair of objects.

7 ALL(a):ALL(b):[a @ ob & b @ ob => ALL(f):[f @ mor => [f @ hom(a,b)
<=> dom(f)=a & cod(f)=b]]]

Your comments are welcomed.

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

[Jesse F. Hughes]

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

> Notation:
>
> ob = class of objects
> mor = class of morphisms
> dom(f) = domain of f
> cod(f) = codomain of f
> comp(g,f) = the composition operator (for functions f and g, comp(g,f)
> = g o f)
> hom(x,y) = the class of morphisms with domain of x and codomain of y
>
> @ = epsilon (indicates class membership)
> & = AND-operator
> => = IMPLIES-operator
> <=> = IFF-operator

Well, you aren't doing what you claim in the subject line.

You aren't defining Category, per se, but really just a narrow view of
category: one in which the objects are sets and the arrows f are subsets
of dom(f) x cod(f).

So, no, this isn't a formal definition of a category (such definitions
can be tossed off in about four minutes, by the way, so why you are
failing after so many weeks is a bit of a mystery).

--
But in our enthusiasm, we could not resist a radical overhaul of the
system, in which all of its major weaknesses have been exposed,
analyzed, and replaced with new weaknesses.
-- Bruce Leverett (presumably with apologies to Ambrose Bierce)

[Dan Christensen]

On Nov 1, 4:53 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > Notation:
>
> > ob          = class of objects
> > mor        = class of morphisms
> > dom(f)     = domain of f
> > cod(f)      = codomain of f
> > comp(g,f) = the composition operator (for functions f and g, comp(g,f)
> > = g o f)
> > hom(x,y)  = the class of morphisms with domain of x and codomain of y
>
> > @    = epsilon (indicates class membership)
> > &     = AND-operator
> > =>   = IMPLIES-operator
> > <=> = IFF-operator
>
> Well, you aren't doing what you claim in the subject line.
>
> You aren't defining Category, per se, but really just a narrow view of
> category: one in which the objects are sets and the arrows f are subsets
> of dom(f) x cod(f).
>

Can you give a simple example of a category for which this is not the
case?

BTW, I am not claiming that all of the objects in question are sets,
just classes.

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

[Jesse F. Hughes]

Let M be any monoid (group, if you prefer). Then M defines a category
as follows: let Obj = {*} (that is, there is only one object of M, which
I denoted *). For each element m of the monoid M, there is an arrow
m:* -> * and let composition be given by the monoid multiplication.

All first chapter stuff.

To be sure, this is a concrete category and you can represent M as a set
with the functions set-based functions, but all of that is fairly
unnatural from a categorical perspective. Best to think of M as a
simple object, and the arrows and their composition as simply given.

There are, of course, other examples, but they need not be as simple.

But who cares whether the examples are simple or not? The fact is that
you're imposing additional restrictions on the notion of Category *FOR
NO DISCERNIBLE REASON*. Why do that?

--
Jesse F. Hughes
"We need to counter the shockwave of the evildoer by having individual
rate cuts accelerated and by thinking about tax rebates."
-- George W. Bush, Oct. 4, 2001

[Dan Christensen]

You are talking about the trivial monoid on a singleton? If so, why
not {{*}}?

> (that is, there is only one object of M, which
> I denoted *).  For each element m of the monoid M, there is an arrow
> m:* -> *

This would become have m: {x} -> {x}

m = {(x,x)}

> and let composition be given by the monoid multiplication.
>
> All first chapter stuff.
>
> To be sure, this is a concrete category and you can represent M as a set
> with the functions set-based functions, but all of that is fairly
> unnatural from a categorical perspective.  Best to think of M as a
> simple object, and the arrows and their composition as simply given.
>
> There are, of course, other examples, but they need not be as simple.
>
> But who cares whether the examples are simple or not?  The fact is that
> you're imposing additional restrictions on the notion of Category *FOR
> NO DISCERNIBLE REASON*.  Why do that?
>

To be able to use a truly formal (i.e. machine parsable) defintion --
one simple definition that should that can be used in every case, with
a bit of "massaging."

[Jesse F. Hughes]

NO.

I'm talking about an arbitrary monoid.

For instance, consider the category consisting of a single object * and
a morphism n:* -> * for each n in N. The rule of composition is given
by:

n o m = n + m.

This is a perfectly good category.

>
>> (that is, there is only one object of M, which
>> I denoted *).  For each element m of the monoid M, there is an arrow
>> m:* -> *
>
> This would become have m: {x} -> {x}
>
> m = {(x,x)}

Tain't so. There is a distinct map for every element m of M.

>
>> and let composition be given by the monoid multiplication.
>>
>> All first chapter stuff.
>>
>> To be sure, this is a concrete category and you can represent M as a set
>> with the functions set-based functions, but all of that is fairly
>> unnatural from a categorical perspective.  Best to think of M as a
>> simple object, and the arrows and their composition as simply given.
>>
>> There are, of course, other examples, but they need not be as simple.
>>
>> But who cares whether the examples are simple or not?  The fact is that
>> you're imposing additional restrictions on the notion of Category *FOR
>> NO DISCERNIBLE REASON*.  Why do that?
>>
>
> To be able to use a truly formal (i.e. machine parsable) defintion --
> one simple definition that should that can be used in every case, with
> a bit of "massaging."

Really, very silly. You're adding unnecessary assumptions and
needlessly complicating definitions. None of that is necessary to make
a formal definition (which is *truly* a trivial task, one which I
completed at your request in about ten minutes time weeks ago).

--
"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

[Jesse F. Hughes]

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

>> You aren't defining Category, per se, but really just a narrow view of
>> category: one in which the objects are sets and the arrows f are subsets
>> of dom(f) x cod(f).
>>
>
> Can you give a simple example of a category for which this is not the
> case?

Another simple example.

Let Prop be the following category:

An object P for each formula P of propositional logic

An arrow P -> Q for each proof of "P implies Q" in propositional logic.

Composition is given by concatenation of proofs. The identity "arrow"
is the canonical proof of "P implies P".

--
"There are people [...] who think it's socially acceptable to level
accusations of mental illness in insulting exchanges to make
points[...] [They] are rather sick [them]selves, and in reality, are
sociopathic." --- James Harris, evidently a self-described sociopath

[Jesse F. Hughes]

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

> Dan Christensen <Dan_Chr...@sympatico.ca> writes:
>
>>> You aren't defining Category, per se, but really just a narrow view of
>>> category: one in which the objects are sets and the arrows f are subsets
>>> of dom(f) x cod(f).
>>>
>>
>> Can you give a simple example of a category for which this is not the
>> case?
>
> Another simple example.
>
> Let Prop be the following category:
>
> An object P for each formula P of propositional logic
>
> An arrow P -> Q for each proof of "P implies Q" in propositional logic.
>
> Composition is given by concatenation of proofs. The identity "arrow"
> is the canonical proof of "P implies P".

My mistake. This doesn't quite work as stated, since concatenating the
canonical proof of "P implies P" with another proof of "P implies Q" is
not idempotent, but I'm sure it's not too difficult to twiddle around
with this general idea and fix things, perhaps by using a canonical cut
elimination of the concatenated proofs.

Another example of a category that Dan should keep in mind is the
category Rel, in which the objects are sets and the "arrows" are
arbitrary relations (not necessarily functional). Composition is
relation composition.

Of course, there are categories of functors as well, but Dan doesn't
know what a functor is.


--
"To solve this problem, we define a security flag, known as the 'evil'
bit, in the IPv4 [RFC791] header. Benign packets have this bit set to
0; those that are used for an attack will have the bit set to 1."
-- RFC 3514

[leland....@gmail.com]

On Thursday, November 1, 2012 8:23:04 PM UTC-4, Jesse F. Hughes wrote:
> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>
>
> Of course, there are categories of functors as well, but Dan doesn't
>
> know what a functor is.

Don't trouble him with functors; he'll have to deal with the collection of all functors from one category to another and since his categories are already classes his whole universe will explode. Certainly don't mention natural transformations (yes, I know Mac Lane has said that defining natural transformations was the whole motivation for him and Eilenberg to define categories, but missing the point seems to be Dan's goal). On and categories of sheaves are definite no-no, which pretty much rules out topos theory. Also stacks; stacks are right out. We wouldn't want to be able to do any useful category theory.

[William Hale]

In article <877gq4q...@phiwumbda.org>,

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

> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>
> > Dan Christensen <Dan_Chr...@sympatico.ca> writes:
> >
> >>> You aren't defining Category, per se, but really just a narrow view of
> >>> category: one in which the objects are sets and the arrows f are subsets
> >>> of dom(f) x cod(f).
> >>>
> >>
> >> Can you give a simple example of a category for which this is not the
> >> case?
> >
> > Another simple example.
> >
> > Let Prop be the following category:
> >
> > An object P for each formula P of propositional logic
> >
> > An arrow P -> Q for each proof of "P implies Q" in propositional logic.
> >
> > Composition is given by concatenation of proofs. The identity "arrow"
> > is the canonical proof of "P implies P".
>
> My mistake. This doesn't quite work as stated, since concatenating the
> canonical proof of "P implies P" with another proof of "P implies Q" is
> not idempotent, but I'm sure it's not too difficult to twiddle around
> with this general idea and fix things, perhaps by using a canonical cut
> elimination of the concatenated proofs.

How about the following:

ob: all formulas P of propositional logic

mor: P ->> Q [provided Q follows from P]

dom(P ->> Q) is P
cod(P ->> Q) is Q

comp(Q ->> R, P ->> Q) = P ->> R

hom(P,Q) is either empty or else the singleton P ->> Q

The identity is P ->> P.

Some concrete entities:

a) "p & q ->> p" is a morphism with domain "p & q" and codomain "p".
Here "p & q" and "p" are two objects.

b) hom(p v q, p & q & r) is empty

c) hom(p & q, p) = {p & q ->> p}

[Jack Campin]

>>> You aren't defining Category, per se, but really just a narrow view of
>>> category: one in which the objects are sets and the arrows f are subsets
>>> of dom(f) x cod(f).
>> Can you give a simple example of a category for which this is not the
>> case?
> Another simple example.
> Let Prop be the following category:
> An object P for each formula P of propositional logic
> An arrow P -> Q for each proof of "P implies Q" in propositional logic.
> Composition is given by concatenation of proofs. The identity "arrow"
> is the canonical proof of "P implies P".

Or while we're doing syntax: any formal grammar determines a category,
with the objects being words and the morphisms being sequences of rule
applications. Some of the standard concepts of category theory translate
directly into standard concepts of formal language theory.

-----------------------------------------------------------------------------
e m a i l : j a c k @ c a m p i n . m e . u k
Jack Campin, 11 Third Street, Newtongrange, Midlothian EH22 4PU, Scotland
mobile 07800 739 557 <http://www.campin.me.uk> Twitter: JackCampin

[William Hale]

In article
<f888316a-93ef-4914...@h9g2000yqd.googlegroups.com>,

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

> Notation:
>
> ob = class of objects
> mor = class of morphisms
> dom(f) = domain of f
> cod(f) = codomain of f
> comp(g,f) = the composition operator (for functions f and g, comp(g,f)
> = g o f)
> hom(x,y) = the class of morphisms with domain of x and codomain of y
>
> @ = epsilon (indicates class membership)
> & = AND-operator
> => = IMPLIES-operator
> <=> = IFF-operator
>
>
> The ordered sextuple (ob, mor, dom, cod, comp, hom) comprises a
> category iff the following axioms hold:
>
> Axioms 1-3 define dom and cod. Each morphisms has a unique domain and
> codomain. Note that morphisms are seen here only as are mappings from
> the ELEMENTS of one object to the elements of another object.

Are the morphisms functions? In particular, if each morphism contains no
elements of the form (a,b), then axiom 3 below is vacuously satisfied.
Or, do you intend each morphism to be a subset of dom(f) x cod(f)?

[Dan Christensen]

I misunderstood. I thought you meant there is only one object denoted
*.

> For instance, consider the category consisting of a single object *

If you have a class of several objects, there is no harm in putting
each one of them in its own little box (a singleton), is there?

> and
> a morphism n:* -> * for each n in N.  The rule of composition is given
> by:
>
>   n o m = n + m.
>
> This is a perfectly good category.
>

Sure.

>
>
> >> (that is, there is only one object of M, which
> >> I denoted *).  For each element m of the monoid M, there is an arrow
> >> m:* -> *
>
> > This would become have m: {x} -> {x}
>

Ooops. I meant m: {*} -> {*}

> m = {(x,x)}
>

Likewise m = {(*,*)}

> Tain't so.  There is a distinct map for every element m of M.
>

Didn't you say there is only one object? Is this another example?

Anyway, for the trivial monoid, we have:

ob={{*}}

m={(*,*)}

mor={m}

dom(m) = {*}

cod(m) = {*}

comp(m,m) = m

hom({*},{*}) = {m}

A very simple category.

> >> and let composition be given by the monoid multiplication.
>
> >> All first chapter stuff.
>
> >> To be sure, this is a concrete category and you can represent M as a set
> >> with the functions set-based functions, but all of that is fairly
> >> unnatural from a categorical perspective.  Best to think of M as a
> >> simple object, and the arrows and their composition as simply given.
>
> >> There are, of course, other examples, but they need not be as simple.
>
> >> But who cares whether the examples are simple or not?  The fact is that
> >> you're imposing additional restrictions on the notion of Category *FOR
> >> NO DISCERNIBLE REASON*.  Why do that?
>
> > To be able to use a truly formal (i.e. machine parsable) defintion --
> > one simple definition that should that can be used in every case, with
> > a bit of "massaging."
>
> Really, very silly.  You're adding unnecessary assumptions and
> needlessly complicating definitions.

You are making the definition needlessly complicated. If we are to
settle on one formal, machine-parsable definition, you can't have
morphisms that map the class ob to itself AND map the elements of each
object to the elements of another. Maybe we could have simple and
compound categories? Would that make you happy? My preference is for
one definition.

> None of that is necessary to make
> a formal definition (which is *truly* a trivial task, one which I
> completed at your request in about ten minutes time weeks ago).
>

Perhaps you could repost it again here? I have forgotten all about it.

[Dan Christensen]

On Nov 1, 9:30 pm, William Hale <bill...@yahoo.com> wrote:
[snip]

> How about the following:
>
> ob: all formulas P of propositional logic
>

[snip]

Why not dress up each one of them with a nice pair of {}'s? Make them
all singletons, et voila!

[Dan Christensen]

On Nov 1, 11:19 pm, William Hale <bill...@yahoo.com> wrote:

[snip]

> > Axioms 1-3 define dom and cod. Each morphisms has a unique domain and
> > codomain. Note that morphisms are seen here only as are mappings from
> > the ELEMENTS of one object to the elements of another object.
>
> Are the morphisms functions?

They sets or classes of ordered pairs of elements of objects.

> In particular, if each morphism contains no
> elements of the form (a,b), then axiom 3 below is vacuously satisfied.

The so-called empty category?


> Or, do you intend each morphism [f] to be a subset of dom(f) x cod(f)?
>
[snip]

Informally speaking, yes.

Dan

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

[Jesse F. Hughes]

William Hale <bil...@yahoo.com> writes:

> In article <877gq4q...@phiwumbda.org>,
> "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>>
>> > Dan Christensen <Dan_Chr...@sympatico.ca> writes:
>> >
>> >>> You aren't defining Category, per se, but really just a narrow view of
>> >>> category: one in which the objects are sets and the arrows f are subsets
>> >>> of dom(f) x cod(f).
>> >>>
>> >>
>> >> Can you give a simple example of a category for which this is not the
>> >> case?
>> >
>> > Another simple example.
>> >
>> > Let Prop be the following category:
>> >
>> > An object P for each formula P of propositional logic
>> >
>> > An arrow P -> Q for each proof of "P implies Q" in propositional logic.
>> >
>> > Composition is given by concatenation of proofs. The identity "arrow"
>> > is the canonical proof of "P implies P".
>>
>> My mistake. This doesn't quite work as stated, since concatenating the
>> canonical proof of "P implies P" with another proof of "P implies Q" is
>> not idempotent, but I'm sure it's not too difficult to twiddle around
>> with this general idea and fix things, perhaps by using a canonical cut
>> elimination of the concatenated proofs.
>
> How about the following:
>
> ob: all formulas P of propositional logic
>
> mor: P ->> Q [provided Q follows from P]

Yes, but this is duller than I had hoped, since it's just the entailment
partial order (as you point out below). I was hoping for a category
with multiple arrows.

I'm sure a little fiddling would do it.

> dom(P ->> Q) is P
> cod(P ->> Q) is Q
>
> comp(Q ->> R, P ->> Q) = P ->> R
>
> hom(P,Q) is either empty or else the singleton P ->> Q
>
> The identity is P ->> P.
>
> Some concrete entities:
>
> a) "p & q ->> p" is a morphism with domain "p & q" and codomain "p".
> Here "p & q" and "p" are two objects.
>
> b) hom(p v q, p & q & r) is empty
>
> c) hom(p & q, p) = {p & q ->> p}

--
"So why talk [about my factoring method] out on Usenet? Because it's a
highly public place so I'm unlikely to disappear[...] You people are
my protection. [...] You may be what's keeping me free and walking out
in the open air." -- James S. Harris, theory guy on the edge.

[William Hale]

In article
<3c40cf3f-b74d-4aed...@r6g2000yqd.googlegroups.com>,

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

> On Nov 1, 11:19 pm, William Hale <bill...@yahoo.com> wrote:
>
> [snip]
>
> > > Axioms 1-3 define dom and cod. Each morphisms has a unique domain and
> > > codomain. Note that morphisms are seen here only as are mappings from
> > > the ELEMENTS of one object to the elements of another object.
> >
> > Are the morphisms functions?
>
>
> They sets or classes of ordered pairs of elements of objects.
>
>
> > In particular, if each morphism contains no
> > elements of the form (a,b), then axiom 3 below is vacuously satisfied.
>
> The so-called empty category?

No.

Let the objects be sets. For each set A, let f = {(1,a,a) | a in A} be
the one and only morphism with domain A and codomain A. There are no
other morphisms. The composition of two morphisms f and g is only
defined when f and g are the same, in which case the composition is f.

For example, if A is {1, 2}, then the only morphism with domain A is
{(1,1,1), (1,2,2)}. If B is {2,3,5}, then the only morphism with domain
B is {(1,2,2), (1,3,3), (1,5,5)}.

For each set there is a morphism. There are many objects and many
morphisms. This category is not empty. But the morphisms are not a set
of ordered pairs (since I intentionally made them not to be ordered
pairs).

It seems that my category meets your axioms. It seems that you need an
axiom to say that each morphism needs to be a subset of dom(f)xcod(f).

[Jesse F. Hughes]

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

> On Nov 1, 9:30 pm, William Hale <bill...@yahoo.com> wrote:
> [snip]
>
>> How about the following:
>>
>> ob: all formulas P of propositional logic
>>
>
> [snip]
>
> Why not dress up each one of them with a nice pair of {}'s? Make them
> all singletons, et voila!

Why do that? What good comes from it? (Hint: none at all.)

You act as if we should naturally want the objects of a category to be
sets, but that is simply not true. There are plenty of interesting (but
not so easy to explain to someone who won't learn category theory)
categories in which the objects are not sets. Leland McInnes named a
few.

--
Jesse F. Hughes
"I think the burden is on those people who think he didn't have
weapons of mass destruction to tell the world where they are."
-- White House spokesman Ari Fleischer

[Jesse F. Hughes]

Yes, there's only one object. You are still misunderstanding.

The monoid N as a category has ONE object, denoted *, and one arrow for
each natural number. I can't draw it worth a damn in ASCII, but think
of a single object (ONE, count it, ONE object). That object has one
arrow called 0, looping from the object to itself. It has another
arrow, called 1, again looping from * to *. And another arrow labeled
2. And so on and so on.

How do we compose arrows? The result of composing arrow n with arrow m
is arrow n + m.

>> For instance, consider the category consisting of a single object *
>
> If you have a class of several objects, there is no harm in putting
> each one of them in its own little box (a singleton), is there?

There's only one object.

>
>> and
>> a morphism n:* -> * for each n in N.  The rule of composition is given
>> by:
>>
>>   n o m = n + m.
>>
>> This is a perfectly good category.
>>
>
> Sure.
>
>
>>
>>
>> >> (that is, there is only one object of M, which
>> >> I denoted *).  For each element m of the monoid M, there is an arrow
>> >> m:* -> *
>>
>> > This would become have m: {x} -> {x}
>>
>
> Ooops. I meant m: {*} -> {*}
>
>> m = {(x,x)}
>>
>
> Likewise m = {(*,*)}

No, you're just confused.

>
>
>
>> Tain't so.  There is a distinct map for every element m of M.
>>
>
> Didn't you say there is only one object? Is this another example?

Yes, there's only one object. There are |M| morphisms (which I called
maps, but "arrows" or "morphisms" are better terms).

> Anyway, for the trivial monoid, we have:
>
> ob={{*}}
>
> m={(*,*)}

No.

OBJ = {*}

MOR = {m: * -> * | m in M}

>
> mor={m}
>
> dom(m) = {*}
>
> cod(m) = {*}
>
> comp(m,m) = m
>
> hom({*},{*}) = {m}
>
> A very simple category.

You've misunderstood. The category N above is an example of what I'm
discussing.

[...]

>> > To be able to use a truly formal (i.e. machine parsable) defintion --
>> > one simple definition that should that can be used in every case, with
>> > a bit of "massaging."
>>
>> Really, very silly.  You're adding unnecessary assumptions and
>> needlessly complicating definitions.
>
>
> You are making the definition needlessly complicated. If we are to
> settle on one formal, machine-parsable definition, you can't have
> morphisms that map the class ob to itself AND map the elements of each
> object to the elements of another. Maybe we could have simple and
> compound categories? Would that make you happy? My preference is for
> one definition.

We're doing category theory. There's no requirement that the arrows map
"elements" of the objects to a darned thing, since there's no
requirement that the objects have elements.

You are needlessly adding requirements to the definition of category.
You aren't simplifying it. You're making it more complicated and more
restrictive.

Not every category involves sets as objects.

>> None of that is necessary to make
>> a formal definition (which is *truly* a trivial task, one which I
>> completed at your request in about ten minutes time weeks ago).
>>
>
> Perhaps you could repost it again here? I have forgotten all about it.

No. You can go find it. We discussed it briefly, but as usual, you got
distracted and I won't do the legwork to find the post again.

It's extraordinarily rude of you to request that I find the definition
I've already posted for you. It's in one of the threads from the past
month. I took the time to post it once and you cared so little that
you've forgotten it. Why should I spend any more time on it?


--
"You can do 'math' until you drop dead, but if you're wrong, you're
still wrong, and the world will keep on turning regardless. Heaven
will not open up and God shout at you to stop, you'll just live in
error until one day you die." -- A James S. Harris self-reflection

[Dan Christensen]

On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

[snip]

> The monoid N as a category has ONE object, denoted *, and one arrow for
> each natural number.  I can't draw it worth a damn in ASCII, but think
> of a single object (ONE, count it, ONE object).  That object has one
> arrow called 0, looping from the object to itself.  It has another
> arrow, called 1, again looping from * to *.  And another arrow labeled
> 2.  And so on and so on.
>

So, for all n in N, your nth object would be {(*,*)}??? They are all
the same. So you really only have one object, namely {(*,*)} not {*}
as I originally thought. Weird, but OK.

> How do we compose arrows?  The result of composing arrow n with arrow m
> is arrow n + m.
>

[snip]

Let x={(*,*)}, a singleton. Then we would have:

ob = {x}

Let f = {(x,x)}. Then:

mor = {f}

dom(f) = x

cod(f) = x

comp(f,f)=f

hom(x,x) = {f}

> >> > To be able to use a truly formal (i.e. machine parsable) defintion --
> >> > one simple definition that should that can be used in every case, with
> >> > a bit of "massaging."
>
> >> Really, very silly.  You're adding unnecessary assumptions and
> >> needlessly complicating definitions.
>
> > You are making the definition needlessly complicated. If we are to
> > settle on one formal, machine-parsable definition, you can't have
> > morphisms that map the class ob to itself AND map the elements of each
> > object to the elements of another. Maybe we could have simple and
> > compound categories? Would that make you happy? My preference is for
> > one definition.
>
> We're doing category theory.  There's no requirement that the arrows map
> "elements" of the objects to a darned thing, since there's no
> requirement that the objects have elements.
>

Can you express that in a list of simple, machine-parsable axioms?
I'll bet you can't.

> You are needlessly adding requirements to the definition of category.
> You aren't simplifying it.  You're making it more complicated and more
> restrictive.
>

I am trying to clarify the notion of a category before I move on. Do
away with all the hand-waving.

> Not every category involves sets as objects.
>

For those "categories", you can easily reformulate a set of objects
e.g. {a,b,c} as {{a}, {b}, {c}}. Then you only need one definition for
a category.

> >> None of that is necessary to make
> >> a formal definition (which is *truly* a trivial task, one which I
> >> completed at your request in about ten minutes time weeks ago).
>
> > Perhaps you could repost it again here? I have forgotten all about it.
>
> No.  You can go find it.

I'm awful, I know. But I did try to find it. Perhaps you deleted it by
mistake? It would be interesting to look at again.

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

[Rupert]

On Nov 2, 2:32 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

Is there any problem with this definition?

http://en.wikipedia.org/wiki/Category_%28mathematics%29#Definition

[Dan Christensen]

On Nov 1, 11:58 pm, William Hale <bill...@yahoo.com> wrote:
> In article
> <3c40cf3f-b74d-4aed-ab3b-efb3ae32b...@r6g2000yqd.googlegroups.com>,

This precisely what my 3rd axiom states. Again:

ALL(f):[f @ mor => ALL(a):ALL(b):[(a,b) @ f => a @ dom(f) & b @
cod(f)]]

where @ = epsilon (indicates set or class membership)

[Aatu Koskensilta]

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

> On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> The monoid N as a category has ONE object, denoted *, and one arrow
>> for each natural number.  I can't draw it worth a damn in ASCII, but
>> think of a single object (ONE, count it, ONE object).  That object
>> has one arrow called 0, looping from the object to itself.  It has
>> another arrow, called 1, again looping from * to *.  And another
>> arrow labeled 2.  And so on and so on.
>
> So, for all n in N, your nth object would be {(*,*)}???

No, that's not what Jesse said. He's describing a category N that
consists of a single object * and has an arrow an : * --> * for every
natural n. The arrows a0, a1, a2, ... are all distinct. To get past this
stumbling block you need to forget all about arrows as sets of ordered
pairs etc, and simply recall the abstract characterization of
category. In particular, rid yourself of the notions that objects have
elements and that an arrow from one object to another maps elements of
the first to those of the second -- looking up the definition of
category you will find no mention of such things.

> I am trying to clarify the notion of a category before I move on. Do
> away with all the hand-waving.

Why not read an introductory category theory text?

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[Dan Christensen]

Yeah, too many words. It could never be machine-parsable.

It is also not clear exactly what a morphism is. Is it just a set (or
class) of ordered pairs of elements of ob(C) or what? It think I have
sharpened the definition somewhat, although I am open to suggestions
for improvement.

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

[leland....@gmail.com]

A morphism is a morphism; it's a primitive in the definition. It is not "made up of" anything. It's is an opaque primitive object, and urelement. Objects of a category are, likewise, urelements.

[Dan Christensen]

On Nov 2, 9:49 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> >> The monoid N as a category has ONE object, denoted *, and one arrow
> >> for each natural number. I can't draw it worth a damn in ASCII, but
> >> think of a single object (ONE, count it, ONE object). That object
> >> has one arrow called 0, looping from the object to itself. It has
> >> another arrow, called 1, again looping from * to *. And another
> >> arrow labeled 2. And so on and so on.
>
> > So, for all n in N, your nth object would be {(*,*)}???
>
>   No, that's not what Jesse said. He's describing a category N that
> consists of a single object * and has an arrow an : * --> * for every
> natural n. The arrows a0, a1, a2, ... are all distinct.

How can they be distinct? They are all the SAME morphism.

> To get past this
> stumbling block you need to forget all about arrows as sets of ordered
> pairs etc, and simply recall the abstract characterization of
> category. In particular, rid yourself of the notions that objects have
> elements and that an arrow from one object to another maps elements of
> the first to those of the second -- looking up the definition of
> category you will find no mention of such things.

That's the problem.

>
> > I am trying to clarify the notion of a category before I move on. Do
> > away with all the hand-waving.
>
>   Why not read an introductory category theory text?
>

I find the definitions given in most introductory notes to be too
vague for my liking. Exactly what is morphism, for example? Since I
now have the time I didn't as a student, I have tried to pull these
definitions apart and, with a few admittedly false starts, tried to
sharpen the definition by coming up with a truly formal, machine-
parsable version. Do you see any problem with my current version here?
Does it exclude any structures that ought to be included? Does it
include ones that it shouldn't?

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

[billh04]

On Nov 2, 8:42 am, Dan Christensen <Dan_Christen...@sympatico.ca>

Does my f = {(1,1,1), (1,2,2)} satisfy your 3rd axiom? I say yes. Is
that what you want?

[Rupert]

On Nov 2, 2:55 pm, Dan Christensen <Dan_Christen...@sympatico.ca>

You just have a domain of discourse consisting of things called
morphisms, that's all. You don't define the notion of a morphism in
terms of other things.

I don't see any reason why you couldn't put the definition I gave you
into machine-parsable form.

[Aatu Koskensilta]

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

> It is also not clear exactly what a morphism is. Is it just a set (or
> class) of ordered pairs of elements of ob(C) or what?

The definition of group is similarly problematic. What are the
elements of a group? Are they numbers? Matrices? Functions? What?!?!
Categories are an abstraction of a recurring pattern in our mathematical
experience. Anything we meet in mathematics that satisfies the laws for
categories can be thought of as a category. In this categories are no
different from groups, vector spaces, fields, lattices, etc. etc.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"

[Frederick Williams]

Dan Christensen wrote:
>
> [...] I am open to suggestions
> for improvement.

(1) Don't bother.
Or (2) Use an actual definition of category instead of a made-up one.
Or (3) Persist with your own definition but don't call the thing defined
a category, call it a Danegory.

(1) is best.

--
On this front, eventually I'll demonstrate that the "transposition" [],
or "replace" [], between the truth/falsehood requirement at the
beginning of Godel's paper, and the "syntactical" OC (Omega Consistency)
is UN-attainable, due to this language-structure-to-undecidability
UN-inferenceability barrier.

Nam Nguyen appeals to the UN

[Dan Christensen]

On Nov 2, 10:23 am, Rupert <rupertmccal...@yahoo.com> wrote:

[snip]

> > > Is there any problem with this definition?
>
> > >http://en.wikipedia.org/wiki/Category_%28mathematics%29#Definition
>
> > Yeah, too many words. It could never be machine-parsable.
>
> > It is also not clear exactly what a morphism is. Is it just a set (or
> > class) of ordered pairs of elements of ob(C) or what? It think I have
> > sharpened the definition somewhat, although I am open to suggestions
> > for improvement.
>
> You just have a domain of discourse consisting of things called
> morphisms, that's all. You don't define the notion of a morphism in
> terms of other things.
>
> I don't see any reason why you couldn't put the definition I gave you
> into machine-parsable form.

Just try it yourself. Write every part of the definition in the
notation of FOL or set theory with no words -- something like my 7
axioms in my original posting.

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

[Frederick Williams]

Dan Christensen wrote:

>
> How can they be distinct? They are all the SAME morphism.

Quoting Jesse: For instance, consider the category consisting of a
single object * and
a morphism n:* -> * for each n in N....

Not all elements of N (the natural numbers) are equal. You don't
believe that.

>
> Exactly what is morphism, for example?

If you don't know what a morphism is, how can you claim that they
(Jesse's) are all the same?

A morphism is anything that satisfies the definition.

I refer to Jesse's morphisms but that category is an example you'll find
in textbooks. You should read one.

[Aatu Koskensilta]

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

> I find the definitions given in most introductory notes to be too
> vague for my liking. Exactly what is morphism, for example?

Anything you want.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"

[Dan Christensen]

On Nov 2, 10:25 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > It is also not clear exactly what a morphism is. Is it just a set (or
> > class) of ordered pairs of elements of ob(C) or what?
>
>   The definition of group is similarly problematic. What are the
> elements of a group?

[snip]

It is not problematic. They are elements of an underlying set G. Then
a binary operator *, is defined on G. In my definition of a category,
the underlying set (or class) is ob. Then I define various operators
on ob.

[Dan Christensen]

On Nov 2, 10:40 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > I find the definitions given in most introductory notes to be too
> > vague for my liking. Exactly what is morphism, for example?
>
>   Anything you want.
>

I use the set (or class) ob as my starting point and define morphisms
and everything else in terms of it. Do you see a problem with that?

[Dan Christensen]

Sorry, I didn't look closely at your example. Informally, I have
morphisms as sets of ordered pairs. They map the elements of one
object to the elements of another (possibly the same) object -- these
"objects" being elements of some set (or class) ob.

[leland....@gmail.com]

On Friday, November 2, 2012 10:50:52 AM UTC-4, Dan Christensen wrote:
> On Nov 2, 10:40 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>
> > Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>
> > > I find the definitions given in most introductory notes to be too
>
> > > vague for my liking. Exactly what is morphism, for example?
>
> >
>
> >   Anything you want.
>
> >
>
>
>
> I use the set (or class) ob as my starting point and define morphisms
>
> and everything else in terms of it. Do you see a problem with that?
>

Yes, there is a problem with that. morphisms are primitives like objects, so if you want to go about nit this way you need to define a set (or class) ob *and* a set (or class) mor, both of which are primitive and not defined in terms of anything else, and go from there.

[Aatu Koskensilta]

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

> On Nov 2, 10:25�am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>
>> The definition of group is similarly problematic. What are the
>> elements of a group?
>

> It is not problematic.

It's exactly as problematic.

> They are elements of an underlying set G. Then a binary operator *, is
> defined on G. In my definition of a category, the underlying set (or
> class) is ob. Then I define various operators on ob.

The morphisms of a category are not determined by its objects. To make
a category we need a set (or class) Ob together with a set (or class)
Hom. Then we define a way to associate with every f in Hom two elements,
rn(f) and dm(f), in Ob. Next we define a (partial) binary operation o on
Hom, such that g o f is defined whenever dm(g) = rn(f). We require that
o is associative, that for every object there is an identity morphisms,
and so on.

[Aatu Koskensilta]

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

> I use the set (or class) ob as my starting point and define morphisms
> and everything else in terms of it. Do you see a problem with that?

Yes. The morphisms of a category are not determined by its objects.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen"

[Frederick Williams]

"Jesse F. Hughes" wrote:

>
> I'm sure a little fiddling would do it.

Lambek and Scott's _Introduction to higher-order categorical logic_ is
about such stuff, there's also a chapter in Troelstra and
Schwichtenberg's _Basic proof theory_.

[Dan Christensen]

On Nov 2, 11:03 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 2, 10:25 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>
> >>   The definition of group is similarly problematic. What are the
> >> elements of a group?
>
> > It is not problematic.
>
>   It's exactly as problematic.
>
> > They are elements of an underlying set G. Then a binary operator *, is
> > defined on G. In my definition of a category, the underlying set (or
> > class) is ob. Then I define various operators on ob.
>
>   The morphisms of a category are not determined by its objects. To make
> a category we need a set (or class) Ob together with a set (or class)
> Hom.

Or you can define the hom operator in terms of a set (or class) in
terms of a set (or class) of morphisms -- much simpler to work with, I
would think. I'm not exactly sure that it is even necessary to define
a hom operator, but it created such an uproar when I left it out that
I put it in as an afterthought. Maybe it is used in the theorems of
CT, but there is no reference to it in my other 6 axioms.

(How about another ale, Aatu? On me.)

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

[Dan Christensen]

On Nov 2, 11:04 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > I use the set (or class) ob as my starting point and define morphisms
> > and everything else in terms of it. Do you see a problem with that?
>
>   Yes. The morphisms of a category are not determined by its objects.
>

See my previous reply to you, just now.

[Aatu Koskensilta]

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

> Or you can define the hom operator in terms of a set (or class) in
> terms of a set (or class) of morphisms -- much simpler to work with, I
> would think.

I'm afraid you've lost me. The salient point here is that a category
does not consists of objects alone. Rather, a category consists of a
class of objects (Ob) together with a class of morphisms (Hom). As in
case of groups, the nature of objects and morphisms is entirely
immaterial as long as they (together with o, rn and dm) satisfy the
axioms for categories.

> (How about another ale, Aatu? On me.)

I'm always game for another ale.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen"

[Dan Christensen]

On Nov 2, 11:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > Or you can define the hom operator in terms of a set (or class) in
> > terms of a set (or class) of morphisms -- much simpler to work with, I
> > would think.
>
>   I'm afraid you've lost me.

It's just alternative formulation. I define the class of all morphisms
in the category instead of the class of all morphisms for each pair of
objects in it. Actually, I do define the latter as well, but I don't
actually refer to it in the other axioms. (A minor concession to the
readers here.)

> The salient point here is that a category
> does not consists of objects alone.

Just like a group is not just a set. As I said in the original
posting, a category can be seen as an ordered sextuple (ob, mor, dom,
cod, comp, hom).

Not sure about hom, but it's there if you need to use it.

[Jesse F. Hughes]

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

> On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> [snip]
>
>> The monoid N as a category has ONE object, denoted *, and one arrow for
>> each natural number.  I can't draw it worth a damn in ASCII, but think
>> of a single object (ONE, count it, ONE object).  That object has one
>> arrow called 0, looping from the object to itself.  It has another
>> arrow, called 1, again looping from * to *.  And another arrow labeled
>> 2.  And so on and so on.
>>
>
> So, for all n in N, your nth object would be {(*,*)}??? They are all
> the same. So you really only have one object, namely {(*,*)} not {*}
> as I originally thought. Weird, but OK.

No. I have one object, *.

The arrows are not the same. They're different. Each arrow m is
distinct from n when n != m.

The arrows are *NOT* sets of ordered pairs. They are primitives.

I'm snipping the rest, since there's no point in going on until you
realize this fact.

--
Jesse F. Hughes
"Mathematicians don't fit in with a consistent view, unless you accept
that to a strangely large extent they are acting under the influence
of something very powerful, dark, and negative." -- James S. Harris

[Jesse F. Hughes]

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

> I'm awful, I know. But I did try to find it. Perhaps you deleted it by
> mistake? It would be interesting to look at again.

See
https://groups.google.com/forum/?fromgroups=#!search/sci.math$20jesse$20hughes$20associativity$20category$20formal/sci.logic/vY14ySDh0ow/B01x6KruabYJ

Nothing at all to it. It is an utterly trivial task to write a formal
axiomatization for category theory.

Real category theory, mind you. Not the borked version you insist is an
improvement.

--
"Now I'm informing all of you that the people arguing against me are EVIL,
yes they are real, live EVIL people as mathematics is that important, so
it's important enough for Evil itself to send minions like them."
-- James Harris on Evil's interest in Algebraic Number Theory

[Jesse F. Hughes]

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

> On Nov 2, 9:49 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>
>> >> The monoid N as a category has ONE object, denoted *, and one arrow
>> >> for each natural number. I can't draw it worth a damn in ASCII, but
>> >> think of a single object (ONE, count it, ONE object). That object
>> >> has one arrow called 0, looping from the object to itself. It has
>> >> another arrow, called 1, again looping from * to *. And another
>> >> arrow labeled 2. And so on and so on.
>>
>> > So, for all n in N, your nth object would be {(*,*)}???
>>
>>   No, that's not what Jesse said. He's describing a category N that
>> consists of a single object * and has an arrow an : * --> * for every
>> natural n. The arrows a0, a1, a2, ... are all distinct.
>
> How can they be distinct? They are all the SAME morphism.

No, they're not. I know they're not, because I defined the category so
that they are distinct.

These arrows are *NOT* subsets of * x *. They are primitives.

>> To get past this
>> stumbling block you need to forget all about arrows as sets of ordered
>> pairs etc, and simply recall the abstract characterization of
>> category. In particular, rid yourself of the notions that objects have
>> elements and that an arrow from one object to another maps elements of
>> the first to those of the second -- looking up the definition of
>> category you will find no mention of such things.
>
> That's the problem.

No, that's not the problem. It is a virtue of the theory. We focus on
the arrows themselves, not some particular notion of function.

>> > I am trying to clarify the notion of a category before I move on. Do
>> > away with all the hand-waving.
>>
>>   Why not read an introductory category theory text?
>>
>
> I find the definitions given in most introductory notes to be too
> vague for my liking. Exactly what is morphism, for example? Since I
> now have the time I didn't as a student, I have tried to pull these
> definitions apart and, with a few admittedly false starts, tried to
> sharpen the definition by coming up with a truly formal, machine-
> parsable version. Do you see any problem with my current version here?
> Does it exclude any structures that ought to be included? Does it
> include ones that it shouldn't?

Morphism is a primitive notion, not defined in terms of objects.

Your current version obviously fails to represent the category N we're
discussing here.

--
"...you are around so that I have something else to do when I'm not
figuring something important out. I was especially intrigued on this
iteration by cursing, which I think I'll continue at some later date
as it's so amusing." --- James S. Harris

[Jesse F. Hughes]

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

> Dan Christensen <Dan_Chr...@sympatico.ca> writes:
>
>> On Nov 2, 9:49 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>>> > On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>>
>>> >> The monoid N as a category has ONE object, denoted *, and one arrow
>>> >> for each natural number. I can't draw it worth a damn in ASCII, but
>>> >> think of a single object (ONE, count it, ONE object). That object
>>> >> has one arrow called 0, looping from the object to itself. It has
>>> >> another arrow, called 1, again looping from * to *. And another
>>> >> arrow labeled 2. And so on and so on.
>>>
>>> > So, for all n in N, your nth object would be {(*,*)}???
>>>
>>>   No, that's not what Jesse said. He's describing a category N that
>>> consists of a single object * and has an arrow an : * --> * for every
>>> natural n. The arrows a0, a1, a2, ... are all distinct.
>>
>> How can they be distinct? They are all the SAME morphism.
>
> No, they're not. I know they're not, because I defined the category so
> that they are distinct.
>
> These arrows are *NOT* subsets of * x *. They are primitives.

Pardon me, but that notation is very hard to read! What I mean is that
the arrows are not subsets of the Cartesian product of * with itself.

Which is, of course, sensible, since * is not a set, but an object in a
category in which there are no products (in either the set-theoretic or
categorical senses).

--
"[The author of the article] refuses to comment further. '/Nonlinear
Analysis/ editors have evaluated the paper, they accepted it for
publication and they have the copyright of its contents - and thus
they are responsible for its correctness,' she [said]."

[Dan Christensen]

On Nov 2, 12:28 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 2, 12:18 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > [snip]
>
> >> The monoid N as a category has ONE object, denoted *, and one arrow for
> >> each natural number.  I can't draw it worth a damn in ASCII, but think
> >> of a single object (ONE, count it, ONE object).  That object has one
> >> arrow called 0, looping from the object to itself.  It has another
> >> arrow, called 1, again looping from * to *.  And another arrow labeled
> >> 2.  And so on and so on.
>
> > So, for all n in N, your nth object would be {(*,*)}??? They are all
> > the same. So you really only have one object, namely {(*,*)} not {*}
> > as I originally thought. Weird, but OK.
>
> No.  I have one object, *.
>

I despair.

> The arrows are not the same.  They're different.  Each arrow m is
> distinct from n when n != m.
>
> The arrows are *NOT* sets of ordered pairs.  They are primitives.
>

Ummmmm.... OK.

So, in some categories, do these "arrows" of yours go for one object
(as a whole) to to another object (as a whole)? And in other
categories, from the ELEMENTS of one object to the ELEMENTS of
another? You can't have both ways -- not without a lot of vigorous
hand-waving.

[Dan Christensen]

On Nov 2, 12:33 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > I'm awful, I know. But I did try to find it. Perhaps you deleted it by
> > mistake? It would be interesting to look at again.
>

> Seehttps://groups.google.com/forum/?fromgroups=#!search/sci.math$20jesse...

>
> Nothing at all to it.  It is an utterly trivial task to write a formal
> axiomatization for category theory.
>

Suit yourself, Jesse.

[Jesse F. Hughes]

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

> On Nov 2, 12:33 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>> > I'm awful, I know. But I did try to find it. Perhaps you deleted it by
>> > mistake? It would be interesting to look at again.
>>
>> Seehttps://groups.google.com/forum/?fromgroups=#!search/sci.math$20jesse...
>>
>> Nothing at all to it.  It is an utterly trivial task to write a formal
>> axiomatization for category theory.
>>
>
> Suit yourself, Jesse.

Does this mean that you agree that it's an utterly trivial task which I
(and Rupert) completed weeks ago, while you are still incapable of
giving the properly general definition?

If so, congrats on the progress.

--
"...[W]hatever gifts I have, they are mine. And I do fully intend on NOT
doing more research, NOT teaching, and NOT doing any number of things
that other people may feel they have a right to tell me I should do, as
when you had the chance with me, you crapped out." --James S. Harris

[Jesse F. Hughes]

No arrows ever "go from" elements of an object to elements of another
object.

The arrows go from one object to another. In some categories, the
arrows represent functions, which map elements of the domain to its
codomain. In other categories, arrows are not defined as functions, but
are simply primitives.

There's no handwaving to be had here. Aatu's analogy with group theory
is apt. In some groups, the elements of the group are numbers. In
other groups, matrices, and in others, more exotic beasties. But none
of this matters, so long as the set and its multiplication and identity
element do what they are supposed to do.

In exactly the same way, any structure of objects and morphisms which
satisfy the definition of "category" is a category. The objects *DO NOT
NEED TO BE SETS* and the morphisms *DO NOT NEED TO BE FUNCTIONS*.

It is truly stunning that, after so many weeks, you don't get this quite
simple fact.

--
Jesse F. Hughes
"Women aren't that unpredictable."
"Well, I can't guess what you're getting at, honey."
-- Hitchcock's _Rear Window_

[Rupert]

On 2 Nov., 15:34, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

To be honest I really can't be bothered. I have absolutely no doubt
that it can be done, but I have a vibe that making the effort to do so
would be unproductive, especially given that Jesse Hughes already
posted a definition of category for you before which you ignored.

I know that the theory of categories is a recursively axiomatizable
first-order theory. It's obvious.

[Frederick Williams]

"Jesse F. Hughes" wrote:
>
> [...] Aatu's analogy with group theory

> is apt. In some groups, the elements of the group are numbers. In
> other groups, matrices, and in others, more exotic beasties.

Is that a good analogy? I'm thinking of the Cayley representation
theorem.

[Jesse F. Hughes]

Frederick Williams <freddyw...@btinternet.com> writes:

> "Jesse F. Hughes" wrote:
>>
>> [...] Aatu's analogy with group theory
>> is apt. In some groups, the elements of the group are numbers. In
>> other groups, matrices, and in others, more exotic beasties.
>
> Is that a good analogy? I'm thinking of the Cayley representation
> theorem.

It's certainly a start at an analogy, though I see your point.

--
Jesse F. Hughes
"All Chinese are Confucianists when successful, and Taoists when they
are failures."
-- Lin Yutang, /My Country and My People/

[William Hale]

In article
<1677f59c-37b9-4766...@y8g2000yqy.googlegroups.com>,

Here's my attempt.

A) Consider some primitive things. These things will be of two kinds:
objects and morphisms.

B) There is a primitive predicate obj of arity one. If x is a thing, I
will write "x is an object" instead of "obj(x)".

C) There is a primitive predicate mor of arity one. If x is a thing, I
will write "x is a morphism" instead of "mor(x)".

D) There is a primitive function dom of arity one from morphisms to
objects.

E) There is a primitive function cod of arity one from morphisms to
objects.

F) There is partial function comp of arity two from two morphisms to a
morphism.

Def1: Morphism f is composable with morphism g iff comp can be applied
to (f, g).

Axiom1: Morphism f is composable with morphism g iff cod(f) = dom(g).

Axiom2: If morphism f is composable with morphism g, then dom(comp(f,g))
= dom(f) and cod(comp(f,g)) = cod(g).

Theorem1: If morphism f is composable with morphism g and g is
composable with morphism h, then f is composable with comp(g,h) and
comp(f, g) is composable with h.

Axiom3: If morphism f is composable with morphism g and g is composable
with morphism h, then comp(f, comp(g,h)) = comp(comp(f,g), h).

Axiom4: For every object A, there is a morphism f such that i) dom(f) =
cod(f) = A; ii) for every morphism g with dom(g) = A, then comp(f,g) =
g; and iii) for every morpism h with cod(h) = A, then comp(h,f) = h.

Def2: A morphism f satisfying Axiom4 is called an identity morphism on
the object A.

[William Hale]

In article
<20b34937-3703-4fac...@h15g2000yqe.googlegroups.com>,

I don't understand your "informally".

Does my f = {(1,1,1), (1,2,2)} satisfy your 3rd axiom?

Don't you see that f satisfies your 3rd axiom vacuously since there are
no (a,b) belonging to f?

What I am leading up to is the following. I suspect that you really want
your morphism to be just a function from its domain to its codomain. If
that is so, then your axiom 6 is not an axiom since it follows from set
theory. If that is not so, then your axiom 3 could have examples that
satisfy its condition vacuously.

[Dan Christensen]

I have just found an enlightening, introductory article from
Wikiversity, "Introduction to Category Theory/Monoids" at
http://en.wikiversity.org/wiki/Introduction_to_Category_Theory/Monoids#From_Binary_Operators_to_Arrows

I think that, after digesting this, I will be tweaking my axioms a
bit. Thanks again for all your help and, ahem... your not always
deserved patience.

[Dan Christensen]

A category is really just as it says and NOTHING MORE in the
definition at Wiki http://en.wikipedia.org/wiki/Category_(mathematics)#Definition

The terminology, borrowed from set theory, is misleading in way. It's
really more like graph theory: objects = nodes, morphisms = directed
edges (arrows) with the possibility of any (possibly uncountable)
number of distinct arrows from one node to another.

Here we have:

ob = a class of objects (the nodes)
mor(a,b) = a class of morphisms (the arrows)
dom = the domain operator (gives the source node for any given arrow)
cod = codomain operator (gives the target node for any given arrow)
hom(a,b) = the class of morphisms (arrows) from a to b
comp = composition operator
id = the identity operator (giving the identity arrow for any node)

Having tweaked my original axioms here considerably, the ordered
septuple (ob,mor,dom,cod,hom,comp,id) is said to comprise a category
iff the following axioms are met:

Define the dom and cod operators:

1 ALL(f):[f @ mor => dom(f) @ ob]

2 ALL(f):[f @ mor => cod(f) @ ob]

Define the hom operator (not used by other axioms):

3 ALL(a):ALL(b):[a @ ob & b @ ob => ALL(f):[f @ mor => [f @ hom(a,b)
<=> dom(f)=a & cod(f)=b]]]

Define the comp operator:

4 ALL(f):ALL(g):[f @ mor & g @ mor => [cod(f)=dom(g) => comp(g,f) @
mor]]

5 ALL(f):ALL(g):[f @ mor & g @ mor
=> [cod(f)=dom(g) => dom(comp(g,f))=dom(f) &
cod(comp(g,f))=cod(g)]]

6 ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [cod(f)=dom(g)
& cod(g)=dom(h)
=> comp(comp(h,g),f)=comp(h,comp(g,f))]]

Define the id operator:

7 ALL(a):[a @ ob => id(a) @ mor]

8 ALL(a):[a @ ob => dom(id(a))=a & cod(id(a))=a]

9 ALL(f):[f @ mor => ALL(a):[a @ ob => [dom(f)=a =>
comp(f,id(a))=f]]]

10 ALL(f):[f @ mor => ALL(a):[a @ ob => [cod(f)=a =>
comp(id(a),f)=f]]]

[Bill Taylor]

On Nov 3, 4:03 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> To make a category we need a set (or class) Ob together with a set (or class)
> Hom. Then we define a way to associate with every f in Hom two elements,
> rn(f) and dm(f), in Ob.

It's a totally trivial point, given that you can call the bits
whatever you like,
but I would suggest it's better to refer to that 2nd-last thingy as
cd(f) rather than rn(f) .

You're welcome...

-- Tendentious Taylor

[Rotwang]

On 03/11/2012 05:02, Dan Christensen wrote:
> A category is really just as it says and NOTHING MORE in the
> definition at Wiki http://en.wikipedia.org/wiki/Category_(mathematics)#Definition
>
> The terminology, borrowed from set theory, is misleading in way. It's
> really more like graph theory: objects = nodes, morphisms = directed
> edges (arrows) with the possibility of any (possibly uncountable)
> number of distinct arrows from one node to another.

That's one way of thinking about it, though it doesn't emphasise the
important fact that certain arrows are composable. Here's another way to
think about the definition of a category: category is to monoid as
groupoid is to group. Hence a category could instead be called a
"monoidoid".

That looks correct to me, apart from a minor niggle: the axiom defining
the hom operator doesn't exclude the possibility that hom(a, b) has
members that aren't in mor.


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

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

[Bill Taylor]

On Nov 3, 6:28 am, Frederick Williams <freddywilli...@btinternet.com>

wrote:

> > [...]  Aatu's analogy with group theory is apt.

As an idle comment, I recall that one of my colleagues
once called group theory the study of how to put an object
back in its box. (Thinking of the box as being a lovely
snug-fitting container exactly the right shape for it.)

I thought it was quite cool.

I have similarly seen category theory described as
the calculus of silly diagrams, though with less justice
I would imagine...

-- Woolly Willy

** Creation science - one of the flat earth sciences?

[Frederick Williams]

Dan Christensen wrote:
>
> A category is really just as it says and NOTHING MORE in the
> definition at Wiki http://en.wikipedia.org/wiki/Category_(mathematics)#Definition
>
> The terminology, borrowed from set theory, is misleading in way. It's
> really more like graph theory: objects = nodes, morphisms = directed
> edges (arrows) with the possibility of any (possibly uncountable)
> number of distinct arrows from one node to another.

You're getting there. And if a node x in a graph was being discussed,
you wouldn't insist that x was a set (or class) would you? And if your
interlocutor said that a node was just a "thing" subject only to the
definition of graph, you wouldn't insist, would you, on making it a set
by writing {x} where everyone else would write x? And if there were two
directed edges each leaving x and each joining x, you wouldn't insist,
would you, that those edges were identical?

[A graph with such a node is called a quiver, but no matter a quiver is
still a graph.]

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[LudovicoVan]

"Frederick Williams" <freddyw...@btinternet.com> wrote in message
news:509555F1...@btinternet.com...

> Dan Christensen wrote:
>>
>> A category is really just as it says and NOTHING MORE in the
>> definition at Wiki
>> http://en.wikipedia.org/wiki/Category_(mathematics)#Definition
>>
>> The terminology, borrowed from set theory, is misleading in way. It's
>> really more like graph theory: objects = nodes, morphisms = directed
>> edges (arrows) with the possibility of any (possibly uncountable)
>> number of distinct arrows from one node to another.
>
> You're getting there. And if a node x in a graph was being discussed,
> you wouldn't insist that x was a set (or class) would you? And if your
> interlocutor said that a node was just a "thing" subject only to the
> definition of graph, you wouldn't insist, would you, on making it a set
> by writing {x} where everyone else would write x? And if there were two
> directed edges each leaving x and each joining x, you wouldn't insist,
> would you, that those edges were identical?
>
> [A graph with such a node is called a quiver, but no matter a quiver is
> still a graph.]

And this is a falsification method too, though of another kind.

-LV

[George Greene]

There was a whole thread 9 years ago.
Unfortunately it started with a feud/flame-war between myself and a
(more) mentally-ill amateur before settling down into trying to
axiomatize category theory.

http://groups.google.com/group/sci.logic/tree/browse_frm/thread/52e0b4186e86870e/59f29a0fa5bfdf32?rnum=21
That attempts to start at the 21st message of the thread.
The first 10 messages are the feud and the 11th-20th are the
transition.

[George Greene]

On Nov 1, 4:41 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

> 1 ALL(f):[f @ mor => dom(f) @ ob]

> 2 ALL(f):[f @ mor => cod(f) @ ob]

> 3 ALL(f):[f @ mor => ALL(a):ALL(b):[(a,b) @ f => a @ dom(f) & b @ cod(f)]]

> Identity morphisms exist for all objects.

> 4 ALL(a):[a @ ob => EXIST(i):[i @ mor & ALL(b):[b @ a => (b,b) @ i]]]

> The composition operator comp

> 5 ALL(f):ALL(g):[f @ mor & g @ mor => [cod(f)=dom(g) => comp(g,f) @
> mor
> & dom(comp(g,f))=dom(f)
> & cod(comp(g,f))=cod(g)
> & ALL(a):ALL(b):[(a,b) @ comp(g,f) <=> EXIST(c):[(a,c) @ f & (c,b) @ g]]]]

> Associativity of composition

> 6 ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [cod(f)=dom(g)
> & cod(g)=dom(h)
> => comp(comp(h,g),f)=comp(h,comp(g,f))]]

> hom classes exist for each ordered pair of objects.

> 7 ALL(a):ALL(b):[a @ ob & b @ ob => ALL(f):[f @ mor => [f @ hom(a,b)
> <=> dom(f)=a & cod(f)=b]]]

> Your comments are welcomed.

Here is a 6-axiom treatment.
I think this is close to equivalent to yours in that it doesn't have a
hom-sets-exist axiom but the others seem to match.
It is better than yours in lacking all the @ clutter and the separate
sorts, though. Speaking of something that is
"not an onerous requirement", there is usually little-or-nothing to be
lost by simply tossing the object in favor of its
identity-arrow, which already MUST exist in ALL treatments ANYway.
Composition, since it is not always defined between every 2 arrows, is
represented as a ternary
predicate that behaves functionally when the arrows connect and is
false when they don't.
domain(x) is d(x); codomain(x) is c(x). K(f,g,h)<=df=>gof=h.

> C.1. Ax1[ d(c(x1)) = c(x1) & c(d(x1)) = d(x1) ]
> The domain of the codomain of x1 is the codomain of x1,
> and the codomain of the domain of x1 is the domain of x1.

> C.2. Ax1x2x3x4 [ K(x1,x2,x3) & K(x1,x2,x4) --> x3=x4
> The composition of x1 with x2 is unique when it is defined.

> C.3. Ax1x2[ Ex3[ K(x1,x2,x3) <-> c(x1)=d(x2) ] ]
> The composition of x1 with x2 is defined if and only
> if the codomain of x1 is the domain of x2.

> C.4. Ax1x2x3[ K(x1,x2,x3) --> d(x3)=d(x1) & c(x3)=c(x2) ]
> If x3 is the composition of x1 with x2 then the
> domain of x3 is the domain of x1 and the codomain of x3 is the codomain of x2.

> C.5. Ax1[ K(d(x1),x1,x1) & K(x1,c(x1),x1)
> For any x1, the domain of x1 is a left identity
> for x1 under composition and the codomain is a right identity.

> C.6. Ax1x2x3x4x5x6x7 [ ( K(x1,x2,x4) & K(x2,x3,x5)
> & K(x1,x5,x6) & K(x4,x3,x7) ) --> x6=x7 ]
> Composition is associative when it is defined.

Phrased in more natural language:
C1. Every arrow is from an object to an object.
C2. Composition is a binary operator on arrows.
C3. The binary composition operator is partial, and defined only when
the argument-arrows chain.
C4. what you would expect about composing arrows that chain, except
for order-of-arguments; that may need restating;
K(f,g,h) as written here is gof=h, or g(f(x)) = h(x) ; you have to
apply f first.
C5. Every object is an identity arrow.
C6. Just as it already says, composition is associative (but partial).

[Shmuel Metz]

In <877gq4l...@phiwumbda.org>, on 11/02/2012

at 12:41 PM, "Jesse F. Hughes" <je...@phiwumbda.org> said:

>Pardon me, but that notation is very hard to read! What I mean is
>that the arrows are not subsets of the Cartesian product of * with
>itself.

And that remains true even when we define categories in terms of sets;
the objects and arrow are arbitrary and only the relations among them
determine whether a class is a category.

--
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

[Dan Christensen]

On Nov 3, 12:02 am, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

> A category is really just as it says and NOTHING MORE in the

> definition at Wikihttp://en.wikipedia.org/wiki/Category_(mathematics)#Definition
>
> The terminology, borrowed from set theory, is misleading in a way. It's

> really more like graph theory: objects = nodes, morphisms = directed
> edges (arrows) with the possibility of any (possibly uncountable)
> number of distinct arrows from one node to another.
>

In particular, the beginner (me, for example) can be mislead by the
use of suggestive set-theoretic notation like f: A -> B and g o f and
(h o g) o f = h o (g o f) (specifically the used of symbols '->', 'o'
and '='). I will elaborate at a later date.

[Jesse F. Hughes]

You were misled because you refused to listen to advice for literally
weeks. We told you from the beginning that, in category theory,
morphisms are primitives.

You cannot fault anyone else for your own willful stupidity, Dan.

(And set theory doesn't own the notation for morphisms, either.)

--
"Your knowledge is the power that promote good thought, how then can you have
good thought without powerful knowledge or how can you have powerful knowledge
without learning or how can you learn without a teacher and how can a teacher
teach if he or she has not learned the subject." --CA Alternative High School

[Dan Christensen]

On Nov 4, 4:18 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 3, 12:02 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> > wrote:
> >> A category is really just as it says and NOTHING MORE in the
> >> definition at Wikihttp://en.wikipedia.org/wiki/Category_(mathematics)#Definition
>
> >> The terminology, borrowed from set theory, is misleading in a way. It's
> >> really more like graph theory: objects = nodes, morphisms = directed
> >> edges (arrows) with the possibility of any (possibly uncountable)
> >> number of distinct arrows from one node to another.
>
> > In particular, the beginner (me, for example) can be mislead by the
> > use of suggestive set-theoretic notation like f: A -> B  and g o f and
> > (h o g) o f = h o (g o f) (specifically the used of symbols '->', 'o'
> > and '='). I will elaborate at a later date.
>
> You were misled because you refused to listen to advice for literally
> weeks.  We told you from the beginning that, in category theory,
> morphisms are primitives.
>

Sorry, Jesse, but, since you brought it up, your powers of explanation
do leave a bit to be desired.

[Jesse F. Hughes]

Sure, Dan. Whatever you say.

You wouldn't come off as such a moron if you would actually take the
time to learn a thing or two, rather than going off half-cocked on your
little misunderstandings. Just read over the threads from this month
and count the ways that others tried to correct your mistaken
impressions about category theory.

And then ask yourself whether or not it would have been better to do
what was suggested: make an honest effort to go through a chapter or two
of a textbook rather than jumping right in without any basic
understanding at all.

But if it makes you feel better to instead complain about my powers of
explanation, well, then do that.

--
"If your community has been lying about my research hoping I'd never
find a way to prove that with some super dramatic discovery that's
almost yanked out of the clear blue because I am a great discoverer
then yeah, maybe you should worry."--James S. Harris: great discoverer

[Dan Christensen]

Suffice it to say, I don't recall anyone here pulling it all together
the way that Wikiversity article did. It was a real find!

[FredJeffries]

On Nov 4, 7:04 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:
>

> Sorry, Jesse, but, since you brought it up, your powers of explanation
> do leave a bit to be desired.

You are an ungrateful pig

[Dan Christensen]

I should be "grateful" for Jesse's childish name-calling and insults?
Rather than saying, yes, that it is correct, he can only lash out like
some insecure, school-yard bully. Really rather pathetic.

[Dan Christensen]

On Nov 3, 1:07 am, Rotwang <sg...@hotmail.co.uk> wrote:
> On 03/11/2012 05:02, Dan Christensen wrote:
>
> > A category is really just as it says and NOTHING MORE in the

> > definition at Wikihttp://en.wikipedia.org/wiki/Category_(mathematics)#Definition

Could you elaborate? I haven't come across anything like that?

I also think there may be a larger problem with the functionality of
composition. Suppose, for example, that f is morphism from object A to
object B, that g is morphism from B to C, and that h1 and h2 are
distinct morphisms from A to C. Then comp(g,f) as defined here could
be either h1 or h2, could it not? Should we define some equivalence
relation on mor. Should comp be seen as a non-functional mapping from
mor x mor to mor?

[Dan Christensen]

On Nov 5, 1:11 am, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

How about something like:

1 ALL(f):ALL(g):ALL(h):[(f,g,h) @ comp => f @ mor & g @ mor & h @
mor]

2 ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [(f,g,h) @
comp
<=> cod(f)=dom(g) & dom(f)=dom(h) & cod(h)=cod(g)]]

3 ALL(f):ALL(g):[f @ mor & g @ mor => [cod(f)=dom(g)
=> EXIST(h):[h @ mor & dom(h)=dom(f) & cod(h)=cod(g) & (f,g,h)@
comp]]]

[Frederick Williams]

Dan Christensen wrote:

> >
>
> Sorry, Jesse, but, since you brought it up, your powers of explanation
> do leave a bit to be desired.

You've got a cheek. You were told repeatedly by Jesse and others that
objects are not (necessarily) sets and morphisms are not (necessarily)
functions.

I am sympathetic (but see my btw) with your:

> > In particular, the beginner (me, for example) can be mislead by the
> > use of suggestive set-theoretic notation like f: A -> B and g o f and
> > (h o g) o f = h o (g o f) (specifically the used of symbols '->', 'o'
> > and '='). I will elaborate at a later date.

But why would anyone but a pig headed idiot continue to be mislead when
people such as Jesse have told you the facts and furnished you with
examples to illustrate those facts?

Btw '=' in '(h o g) o f = h o (g o f)' means what '=' means throughout
mathematics.

[Frederick Williams]

Dan Christensen wrote:

> Suffice it to say, I don't recall anyone here pulling it all together
> the way that Wikiversity article did. It was a real find!

Is a Usenet newsgroup the right place to make posts like Wikiversity
articles? Should someone who posts to a Usenet newsgroup expect replies
other than in newsgroup style? (Especially a person who is familiar
with at least two Usenet newsgroups.)

What I recall is a number of people telling you that objects weren't
(necessarily) sets and morphisms weren't (necessarily) functions. And I
recall examples being offered. And I recall you insisting on ignoring
them.

[Jesse F. Hughes]

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

> On Nov 4, 11:14 pm, FredJeffries <fredjeffr...@gmail.com> wrote:
>> On Nov 4, 7:04 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
>> wrote:
>>
>>
>>
>> > Sorry, Jesse, but, since you brought it up, your powers of explanation
>> > do leave a bit to be desired.
>>
>> You are an ungrateful pig
>
> I should be "grateful" for Jesse's childish name-calling and insults?
> Rather than saying, yes, that it is correct, he can only lash out like
> some insecure, school-yard bully. Really rather pathetic.

For weeks, people have tried to correct your errors, but you've
stubbornly clung to them.

For weeks, people have suggested that you read an actual text. You
resisted the suggestion, because you're far to clever to have to read
something.

You are a deliberate moron, Dan. You take an active role in your own
stupidity. If it's bullying for me to tell you so, then I reckon I'm a
bully.

--
Jesse F. Hughes

"It's much better to live with my parents than with a wife."
-- Quincy P. Hughes, age 4 1/2

[Frederick Williams]

Dan Christensen wrote:

>
> On Nov 2, 10:40 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> > Dan Christensen <Dan_Christen...@sympatico.ca> writes:

> > > I find the definitions given in most introductory notes to be too
> > > vague for my liking. Exactly what is morphism, for example?
> >
> > Anything you want.
> >
>
> I use the set (or class) ob as my starting point and define morphisms
> and everything else in terms of it. Do you see a problem with that?

You can start with morphisms and define the objects in terms of them if
you wish. As to what the morphisms are, they are just anything that
satisfies the definition.

[Jesse F. Hughes]

Goodness, no.

But congrats on finding a gross new misconception.

Composition is a partial function. Please, please read an entire
chapter of an introductory text.

[Dan Christensen]

On Nov 5, 9:53 am, Jesse F. Hughes <je...@phiwumbda.org> wrote:
> On Sun, 4 Nov 2012 22:35:19 -0800 (PST), Dan Christensen
>

> <Dan_Christen...@sympatico.ca> wrote:
> > On Nov 5, 1:11 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> > wrote:
> > 2  ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [(f,g,h) @
> > comp
> >    <=> cod(f)=dom(g) & dom(f)=dom(h) & cod(h)=cod(g)]]
>
> Goodness, no.
>
> But congrats on finding a gross new misconception.
>

You are a real piece of work, Jesse.

> Composition is a partial function.

Even partial functions map elements of one set to a UNIQUE element of
another. Unless there is some kind of selection mechanism (e.g.
invoking AC), I don't see how composition of morphisms (arrows) can be
a function. Please note, here I am questioning my own definition of a
category.

[FredJeffries]

On Nov 5, 7:39 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:
> Dan Christensen wrote:
>
> > [...] Do we invoke the Axiom of
> > Choice?
>
> What is the axiom of choice in category theoretic terms?  I know what it
> is for toposes but not for categories generally.

Every epimorphism in C splits.

http://ncatlab.org/nlab/show/axiom+of+choice

[Frederick Williams]

Thank you.

[Rotwang]

On 05/11/2012 06:11, Dan Christensen wrote:
> On Nov 3, 1:07 am, Rotwang <sg...@hotmail.co.uk> wrote:
>> On 03/11/2012 05:02, Dan Christensen wrote:
>>

>>> [...]

What I mean is that your axiom 3 says that for any morphism f, f is in
hom(a, b) iff dom(f) = a and cod(f) = b, but it doesn't say anything
about whether some entity f is in hom(a, b) when f is not a morphism.
That is, if our category were the category of sets, for example, then
hom(a, b) could include as members all functions a to b, but could also
include the number 2, and would still satisfy your axiom 3. That would
not be the usual definition of hom(a, b).

> I also think there may be a larger problem with the functionality of
> composition. Suppose, for example, that f is morphism from object A to
> object B, that g is morphism from B to C, and that h1 and h2 are
> distinct morphisms from A to C. Then comp(g,f) as defined here could
> be either h1 or h2, could it not?

If I tell you the answer to your question, will you believe me? Or will
you insist that I'm wrong and then complain about my lack of explanatory
skills when you realise a month from now that I'm correct?

Anyway, here's the answer: in the usual definition of a category,
composition is a partial function - that is, if cod(f) = dom(g) then
there is exactly one h such that h = comp(g, f). I had assumed that this
was implicit in your axioms (since the notation comp(g, f) doesn't
really make any sense if comp is not functional), but if it isn't then
you should include an additional axiom that says that comp is functional
and specifies its domain; something like

3 and a half: ALL(x):[x @ comp => EXISTS(h)EXISTS(g)EXISTS(f)[h @ mor &
g @ mor & f @ mor & dom(g) = cod(f) & x = ((g, f), h)]]

3 and two thirds: ALL(h)All(h'):[EXISTS(f)EXISTS(g):[((g, f), h) @ mor &
((g, f), h') @ mor] => h = h']

Incidentally, if you're not allowed to assume from the axioms you gave
earlier that comp is a partial function, then you probably shouldn't be
assuming that cod and dom are partial functions either - you should add
axioms that say so.

[Dan Christensen]

On Nov 5, 6:43 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 4, 11:14 pm, FredJeffries <fredjeffr...@gmail.com> wrote:
> >> On Nov 4, 7:04 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
> >> wrote:
>
> >> > Sorry, Jesse, but, since you brought it up, your powers of explanation
> >> > do leave a bit to be desired.
>
> >> You are an ungrateful pig
>
> > I should be "grateful" for Jesse's childish name-calling and insults?
> > Rather than saying, yes, that it is correct, he can only lash out like
> > some insecure, school-yard bully. Really rather pathetic.
>
> For weeks, people have tried to correct your errors, but you've
> stubbornly clung to them.
>
> For weeks, people have suggested that you read an actual text.  You
> resisted the suggestion, because you're far to clever to have to read
> something.
>
> You are a deliberate moron, Dan.  You take an active role in your own
> stupidity.  If it's bullying for me to tell you so, then I reckon I'm a
> bully.
>

Your insecurity is showing again, Jesse. Why can't you just admit that
I may finally I have got it right? (Still outstanding is the question
of functionality of my composition operator -- a question that I
myself have subsequently raised.)

[Jesse F. Hughes]

A category is a structure which includes a partial function of
composition. It's that simple.

Your complaint is no more sensible than the following: There is no four
element group. Because, if {1,a,b,c} is a group, what is a * b? It
might be c. It might be 1. I don't know which it is, so * cannot be a
function.

A category is a tuple consisting of objects, morphisms, identity
morphisms and composition. Among other things, we require that
composition is a partial function.

You will notice that, in every example of a category I've given to you
over the past few weeks, composition is functional. If this were not
the case, then the example would not be a category. The axioms of
category theory don't tell you *which* arrow is the result of composing
f with g, but *one* arrow must be the result.

Honestly, Dan, you're confusing yourself again.

--
Jesse F. Hughes

"I have written many words to sci.math, some of them are not even
meaningless." --Ross Finlayson

[Jesse F. Hughes]

What you've gotten right (ignoring, for a moment, your brand new
misconception) was a triviality that took you literally weeks. I
offered you a perfectly good formalization of the definition of category
some time ago, but never mind.

Better to do it on your own, going for some time through the
misconception that all morphisms are surjections, then hang on for a
while on the requirement that all morphisms are functions between sets
and now the new stupidity that composition isn't a partial function.

I can see why you're so darned proud of your accomplishment. It's much
more challenging to do things without understanding than with, and so in
that respect, your definition is a much better accomplishment than mine.

Aside from your new misunderstanding, of course.

--
Jesse F. Hughes
"Well, I'm a pragmatist. I've been wrong MANY TIMES and it seems to
me that it would be simpler to be wrong with this paper."
--James S. Harris explains his latest paper

[Dan Christensen]

Good point! Axiom 3 should be:

ALL(a):ALL(b):[a @ ob & b @ ob => ALL(f):[f @ hom(a,b)
<=> f @ mor & dom(f)=a & cod(f)=b]]

I think this will address your concern.

> That is, if our category were the category of sets, for example, then
> hom(a, b) could include as members all functions a to b, but could also
> include the number 2, and would still satisfy your axiom 3. That would
> not be the usual definition of hom(a, b).
>
> > I also think there may be a larger problem with the functionality of
> > composition. Suppose, for example, that f is morphism from object A to
> > object B, that g is morphism from B to C, and that h1 and h2 are
> > distinct morphisms from A to C. Then comp(g,f) as defined here could
> > be either h1 or h2, could it not?
>
> If I tell you the answer to your question, will you believe me? Or will
> you insist that I'm wrong and then complain about my lack of explanatory
> skills when you realise a month from now that I'm correct?
>
> Anyway, here's the answer: in the usual definition of a category,
> composition is a partial function - that is, if cod(f) = dom(g) then
> there is exactly one h such that h = comp(g, f). I had assumed that this
> was implicit in your axioms (since the notation comp(g, f) doesn't
> really make any sense if comp is not functional),

[snip]

Yes, I subsequently began to question whether or not this notation was
justified. My definition can be "rescued" if comp(g,f) is not The
Composition, but only a possible composition.

> Incidentally, if you're not allowed to assume from the axioms you gave
> earlier that comp is a partial function, then you probably shouldn't be
> assuming that cod and dom are partial functions either - you should add
> axioms that say so.
>

As I have defined them, dom and cod are full functions. dom and cod
are defined for EVERY element of the class of morphisms (mor). If x is
ANY morphism, than dom(x) is an element of ob. Likewise, if y is ANY
morphism, than cod(y) is also an element of ob.

[Rotwang]

Yes.

>> That is, if our category were the category of sets, for example, then
>> hom(a, b) could include as members all functions a to b, but could also
>> include the number 2, and would still satisfy your axiom 3. That would
>> not be the usual definition of hom(a, b).
>>
>>> I also think there may be a larger problem with the functionality of
>>> composition. Suppose, for example, that f is morphism from object A to
>>> object B, that g is morphism from B to C, and that h1 and h2 are
>>> distinct morphisms from A to C. Then comp(g,f) as defined here could
>>> be either h1 or h2, could it not?
>>
>> If I tell you the answer to your question, will you believe me? Or will
>> you insist that I'm wrong and then complain about my lack of explanatory
>> skills when you realise a month from now that I'm correct?
>>
>> Anyway, here's the answer: in the usual definition of a category,
>> composition is a partial function - that is, if cod(f) = dom(g) then
>> there is exactly one h such that h = comp(g, f). I had assumed that this
>> was implicit in your axioms (since the notation comp(g, f) doesn't
>> really make any sense if comp is not functional),
>
> [snip]
>
> Yes, I subsequently began to question whether or not this notation was
> justified. My definition can be "rescued" if comp(g,f) is not The
> Composition, but only a possible composition.

But in that case your axioms say nothing about the actual composition
that is part of the definition of a category, surely?

>> Incidentally, if you're not allowed to assume from the axioms you gave
>> earlier that comp is a partial function, then you probably shouldn't be
>> assuming that cod and dom are partial functions either - you should add
>> axioms that say so.
>>
>
> As I have defined them, dom and cod are full functions. dom and cod
> are defined for EVERY element of the class of morphisms (mor). If x is
> ANY morphism, than dom(x) is an element of ob. Likewise, if y is ANY
> morphism, than cod(y) is also an element of ob.

That's exactly as it should be (though I don't see anywhere that says so
in the earlier material I quoted). But it's not clear from your reply
whether you accept that composition is supposed to be a partial
function. If your axioms don't require that it is then they are wrong.

[Dan Christensen]

On Nov 5, 11:48 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 5, 9:53 am, Jesse F. Hughes <je...@phiwumbda.org> wrote:
> >> On Sun, 4 Nov 2012 22:35:19 -0800 (PST), Dan Christensen
>
> >> <Dan_Christen...@sympatico.ca> wrote:
> >> > On Nov 5, 1:11 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> >> > wrote:
> >> > 2  ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [(f,g,h) @
> >> > comp
> >> >    <=> cod(f)=dom(g) & dom(f)=dom(h) & cod(h)=cod(g)]]
>
> >> Goodness, no.
>
> >> But congrats on finding a gross new misconception.
>
> > You are a real piece of work, Jesse.
>
> >> Composition is a partial function.
>
> > Even partial functions map elements of one set to a UNIQUE element of
> > another. Unless there is some kind of selection mechanism (e.g.
> > invoking AC), I don't see how composition of morphisms (arrows) can be
> > a function. Please note, here I am questioning my own definition of a
> > category.
>
> A category is a structure which includes a partial function of
> composition.  It's that simple.
>

[snip]

How can g o f can be the one and only composition of f followed by g
(assuming cod(f)=dom(g)) in a category with multiple distinct
morphisms from dom(f) to cod(g)? It seems to me that the only way that
'o' can be a function in such a case is for g o f to be simply one
possible composition. If so, then g o f should not be seen as The
Composition, but simply as one POSSIBLE composition. It could still be
a function.

[Rotwang]

On 05/11/2012 17:33, Dan Christensen wrote:
> On Nov 5, 11:48 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>

>> [...]

>>
>>> Even partial functions map elements of one set to a UNIQUE element of
>>> another. Unless there is some kind of selection mechanism (e.g.
>>> invoking AC), I don't see how composition of morphisms (arrows) can be
>>> a function. Please note, here I am questioning my own definition of a
>>> category.
>>
>> A category is a structure which includes a partial function of
>> composition. It's that simple.
>>
> [snip]
>
> How can g o f can be the one and only composition of f followed by g
> (assuming cod(f)=dom(g)) in a category with multiple distinct
> morphisms from dom(f) to cod(g)? It seems to me that the only way that
> 'o' can be a function in such a case is for g o f to be simply one
> possible composition. If so, then g o f should not be seen as The
> Composition, but simply as one POSSIBLE composition. It could still be
> a function.

You seem to be labouring under the misapprehension that the composition
in a category is defined in terms of the other properties. It isn't -
the composition is part of the definition of a category, and two
categories with the same objects, morphisms and dom and cod functions
may have different composition operators. Your question is akin to
saying that the group product in a group should not be seen as The
Product, but simply as one possible product. Different products give
rise to different groups, and a set of objects is not a group until a
product has been specified.

[Dan Christensen]

[snip]

> >> Anyway, here's the answer: in the usual definition of a category,
> >> composition is a partial function - that is, if cod(f) = dom(g) then
> >> there is exactly one h such that h = comp(g, f). I had assumed that this
> >> was implicit in your axioms (since the notation comp(g, f) doesn't
> >> really make any sense if comp is not functional),
>
> > [snip]
>
> > Yes, I subsequently began to question whether or not this notation was
> > justified. My definition can be "rescued" if comp(g,f) is not The
> > Composition, but only a possible composition.
>
> But in that case your axioms say nothing about the actual composition
> that is part of the definition of a category, surely?
>

I'm not sure what you are getting at.

Axiom 4 says the comp is a partial function mapping mor x mor to mor.

Axiom 5 says the domain and codomain of comp(g,f) are compatible with
those of f and g.

Axiom 6 says the comp is associative.

> >> Incidentally, if you're not allowed to assume from the axioms you gave
> >> earlier that comp is a partial function, then you probably shouldn't be
> >> assuming that cod and dom are partial functions either - you should add
> >> axioms that say so.
>
> > As I have defined them, dom and cod are full functions. dom and cod
> > are defined for EVERY element of the class of morphisms (mor). If x is
> > ANY morphism, than dom(x) is an element of ob. Likewise, if y is ANY
> > morphism, than cod(y) is also an element of ob.
>
> That's exactly as it should be (though I don't see anywhere that says so
> in the earlier material I quoted). But it's not clear from your reply
> whether you accept that composition is supposed to be a partial
> function.

That, in effect, is what Axiom 4 says. You don't agree?

[Dan Christensen]

Suppose some category C includes 4 distinct morphisms f, g, h1 and h2
such that

1. dom(f)=dom(h1)=dom(h1)

2. cod(f)=dom(g)

3. cod(g)=cod(h1)=cod(h2)

What is g o f? Is it h1 or h2 or both or possibly neither?

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

[Aatu Koskensilta]

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

> Suppose some category C includes 4 distinct morphisms f, g, h1 and h2
> such that
>
> 1. dom(f)=dom(h1)=dom(h1)
>
> 2. cod(f)=dom(g)
>
> 3. cod(g)=cod(h1)=cod(h2)
>
> What is g o f? Is it h1 or h2 or both or possibly neither?

Possibly neither, if there are morphisms apart from h1 and h2 with
suitable domain and codomain. What is it that's bothering you here? The
objects and morphisms do not determine the composition operation;
rather, the composition operation is part of the data making up the
category, just as with the operation * for groups etc.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[Dan Christensen]

On Nov 5, 1:43 pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > Suppose some category C includes 4 distinct morphisms f, g, h1 and h2
> > such that
>
> > 1.  dom(f)=dom(h1)=dom(h1)
>
> > 2.  cod(f)=dom(g)
>
> > 3.  cod(g)=cod(h1)=cod(h2)
>
> > What is g o f? Is it h1 or h2 or both or possibly neither?
>
>   Possibly neither, if there are morphisms apart from h1 and h2 with
> suitable domain and codomain.

So the choice is arbitrary? You just have to choose one? So, you COULD
have:

g o f = h1?

[Dan Christensen]

On Nov 3, 12:02 am, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

> A category is really just as it says and NOTHING MORE in the
> definition at Wikihttp://en.wikipedia.org/wiki/Category_(mathematics)#Definition
>
> The terminology, borrowed from set theory, is misleading in way. It's
> really more like graph theory: objects = nodes, morphisms = directed
> edges (arrows) with the possibility of any (possibly uncountable)
> number of distinct arrows from one node to another.
>

> Here we have:
>
> ob = a class of objects (the nodes)
> mor(a,b) = a class of morphisms (the arrows)
> dom = the domain operator  (gives the source node for any given arrow)
> cod = codomain operator    (gives the target node for any given arrow)
> hom(a,b) = the class of morphisms (arrows) from a to b
> comp = composition operator
> id = the identity operator (giving the identity arrow for any node)
>
> Having tweaked my original axioms here considerably, the ordered
> septuple (ob,mor,dom,cod,hom,comp,id) is said to comprise a category
> iff the following axioms are met:
>
> Define the dom and cod operators:
>
> 1  ALL(f):[f @ mor => dom(f) @ ob]
>
> 2  ALL(f):[f @ mor => cod(f) @ ob]
>
> Define the hom operator (not used by other axioms):
>
> 3  ALL(a):ALL(b):[a @ ob & b @ ob => ALL(f):[f @ mor => [f @ hom(a,b)
> <=> dom(f)=a & cod(f)=b]]]
>

Should be:

ALL(a):ALL(b):[a @ ob & b @ ob => ALL(f):[f @ hom(a,b)
<=> f @ mor & dom(f)=a & cod(f)=b]]

Thanks to Rotwang.

> Define the comp operator:
>
> 4  ALL(f):ALL(g):[f @ mor & g @ mor => [cod(f)=dom(g) => comp(g,f) @
> mor]]
>
> 5  ALL(f):ALL(g):[f @ mor & g @ mor
>    => [cod(f)=dom(g) => dom(comp(g,f))=dom(f) &
> cod(comp(g,f))=cod(g)]]
>
> 6  ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [cod(f)=dom(g)
> & cod(g)=dom(h)
>    => comp(comp(h,g),f)=comp(h,comp(g,f))]]
>
> Define the id operator:
>
> 7  ALL(a):[a @ ob => id(a) @ mor]
>
> 8  ALL(a):[a @ ob => dom(id(a))=a & cod(id(a))=a]
>
> 9  ALL(f):[f @ mor => ALL(a):[a @ ob => [dom(f)=a =>
> comp(f,id(a))=f]]]
>
> 10 ALL(f):[f @ mor => ALL(a):[a @ ob => [cod(f)=a =>
> comp(id(a),f)=f]]]
>

[Dan Christensen]

Correction

On Nov 5, 1:43 pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > Suppose some category C includes 4 distinct morphisms f, g, h1 and h2
> > such that
>
> > 1.  dom(f)=dom(h1)=dom(h1)
>
> > 2.  cod(f)=dom(g)
>
> > 3.  cod(g)=cod(h1)=cod(h2)
>
> > What is g o f? Is it h1 or h2 or both or possibly neither?
>
>   Possibly neither, if there are morphisms apart from h1 and h2 with
> suitable domain and codomain.

So the choice is arbitrary? You just have to choose one? So, you COULD
have

g o f = h1

Then, of course, we would also have:

g o f =/= h2

even though, in a sense, h2 is also a composition of f followed by g.

[Jesse F. Hughes]

How can a*b be 1 or c in a group with four elements?

The axioms of category theory require composition to be a partial
function. A particular structure <Ob,Mor,dom,cod,id,comp> is a category
only if comp actually *is* a partial function.

The composition function is part of the specification of a category,
just like multiplication is part of the specification of a group. And
just as multiplication in groups must be functional, so must composition
in a category.

Example:

Consider the category with one object M and two arrows, f_0 and f_1,
where

dom(f_0) = dom(f_1) = cod(f_0) = cod(f_1) = M
id_M = f_1

f_1 o f_0 = f_0 o f_1 = f_0 o f_0 = f_0.
f_1 o f_1 = f_1.

There's a category for you. It satisfies all of the axioms, including
the axiom that composition be functional, even though it has distinct
morphisms.

Here's another category. It has one object M' and two arrows g_0 and
g_1 and its composition is given by

dom(g_0) = dom(g_1) = cod(g_0) = cod(g_1) = M'
id_M' = g_1

g_1 o g_0 = g_0 o g_1 = g_0.
g_1 o g_1 = g_0 o g_0 = g_1.

Distinct categories. In the second category, for instance, every arrow
is an isomorphism, whereas this is not true in the first category.

I sincerely hope this clears up your misunderstanding. Category theory
places certain restrictions on what counts as composition, but it
doesn't *determine* the composition function for each category.

--
"If you go to my blog you may notice that I have a link to a Princeton
graduate student who seems to think I'm a great 'mathematican', and I
do wonder why he has that word wrongly spelled."
-- James S. Harris has mathematic supporters

[Dan Christensen]

That should be a PARTIAL function.

> in such a case is for g o f to be simply one
> > possible composition. If so, then g o f should not be seen as The
> > Composition, but simply as one POSSIBLE composition. It could still be
> > a function.
>

Again, a PARTIAL function.

> How can a*b be 1 or c in a group with four elements?
>
> The axioms of category theory require composition to be a partial
> function.

[snip]

Agreed. But leaving aside your own examples for now, in the example I
just posed to Rotwang and Aatu, you then agree that it doesn't matter
whether g o f = h1 or h2 -- I should just pick one. If I chose g o f =
h1, then, even though h2 is, in a sense, also a composition of f and
g, we would have g o f =/= h2. I can live with that. It just means
that, in some cases g o f is NOT the only composition of f and g.