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Topic: Toward a Formal, Machine-Parsable Definition of a Category
Replies: 8   Last Post: Nov 6, 2012 8:39 AM

 Messages: [ Previous | Next ]
 Ki Song Posts: 549 Registered: 9/19/09
Re: Toward a Formal, Machine-Parsable Definition of a Category
Posted: Nov 6, 2012 1:04 AM

On Tuesday, November 6, 2012 12:27:34 AM UTC-5, Dan Christensen wrote:
> On Nov 5, 11:57 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>

> > On 06/11/2012 03:46, Dan Christensen wrote:
>
> >
>
> > > On Nov 5, 7:03 pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>
> > >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
>
> > >>> 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.
>
> >
>
> > >>    In what sense is h2 also a composition of f and g?
>
> >
>
> > > In the sense that dom(h2)=dom(f) and cod(h2)=cod(g).
>
> >
>
> > But that isn't what "composition of f and g" means.
>
> >
>
>
>
> That's all it means in CT -- in effect, just putting 2 arrows head-to-
>
> tail.

You are not getting it.

>
>
>
>
>

> > >> All you know is
>
> > >> that h2 has suitable domain and codomain.
>
> >
>
> > > That should be enough.
>
> >
>
> > > "For each ordered triple of objects A, B, C in category [curly] C,
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> > > there is a law of composition: If f:A->B and g:B->C, then the
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> > > composite of f and g is a morphism gf:A->C."
>
> >
>
> > > Source: "Category Theory," WikiBooks,http://en.wikibooks.org/wiki/Category_Theory/Categories#Definition
>
> >
>
> > > Doesn't that suggest that ANY morphism mapping A to C (in the
>
> > > defintion here) would do for the composite of f and g?
>
> >
>
> > No, it doesn't.
>
> >
>
> [snip]
>
>
>
>
>
> I disagree.

Then you are being stubborn. Look, why don't you look at some explicit examples of categories. There are a lot of different examples available on-line.

>
>
>

> > So when you write, for example,
>
> >
>
> >    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.
>
> >
>
> > this becomes (denoting a group product by *)
>
> >
>
> >    How can a*b can be the one and only product of b and a in a group
>
> >    with multiple distinct elements?
>
>
>
> [snip]
>
>
>
> As I have said elsewhere, if all your morphisms are functions (as in
>
> your group examples), their composition is always uniquely defined.
>
> But morphisms are not always functions in other categories.
>

That doesn't imply what you are saying.

>
>

> > >> Whether it's the composition
>
> > >> of f and g depends on the f and g in question, and the relevant notion
>
> > >> of composition given by the category at issue.
>
> >
>
> > > I'm just going by the above definition, and it suggests that the
>
> > > domain and codomain are the ONLY relevant criteria. A problem arises,
>
> > > of course, when there are multiple, distinct morphisms from A to C --
>
> > > which one to pick? It's beginning to look to me like the choice is
>
> > > completely arbitrary.
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> >
>
> > > Can you cite any authoritative source to contrary? Online would be
>
> > > nice.
>
> >
>
>
> > the sentence that says "We take as composition the group
>
> > multiplication."
>
>
>
> This is only a particular case -- the category of groups where all
>
> morphisms are functions. Our definitions must also cover the case
>
> where morphisms are not functions.
>
>

The definition does. You are misunderstanding the definition.

>
>
>

> > > It seems to me that would it work perfectly fine to arbitrary select
>
> > > as the composite any morphism with the required domain and codoemain.
>
> > > This would at least be consistent with all the definitions we have
>
> > > looked at.
>
> >
>
> > No it wouldn't, since many such arbitrary choices would fail to satisfy
>
> > associativity.
>
> >
>
>
>
> How so?
>
>
>
> Dan
>

Date Subject Author
11/5/12 Aatu Koskensilta
11/5/12 Dan Christensen
11/5/12 Rotwang
11/6/12 Dan Christensen
11/6/12 Rotwang
11/6/12 Ki Song
11/6/12 Frederick Williams
11/6/12 Ki Song