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Ki Song
Posts:
221
Registered:
9/19/09
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Re: Toward a Formal, Machine-Parsable Definition of a Category
Posted:
Nov 6, 2012 1:04 AM
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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:
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> >
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> > > 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
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> > >>> whether g o f = h1 or h2 -- I should just pick one. If I chose g o f =
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> > >>> h1, then, even though h2 is, in a sense, also a composition of f and
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> > >>> g, we would have g o f =/= h2.
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> >
>
> > >> In what sense is h2 also a composition of f and g?
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> >
>
> > > In the sense that dom(h2)=dom(f) and cod(h2)=cod(g).
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> >
>
> > But that isn't what "composition of f and g" means.
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> >
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>
>
> That's all it means in CT -- in effect, just putting 2 arrows head-to-
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> tail.
You are not getting it.
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>
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>
>
> > >> All you know is
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> > >> that h2 has suitable domain and codomain.
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> >
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> > > That should be enough.
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> >
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> > > "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."
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> >
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> > > Source: "Category Theory," WikiBooks,http://en.wikibooks.org/wiki/Category_Theory/Categories#Definition
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> >
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> > > Doesn't that suggest that ANY morphism mapping A to C (in the
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> > > defintion here) would do for the composite of f and g?
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> >
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> > No, it doesn't.
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> >
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> [snip]
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>
>
>
>
> 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.
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>
>
> > So when you write, for example,
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> >
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> > How can g o f can be the one and only composition of f followed by g
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> > (assuming cod(f)=dom(g)) in a category with multiple distinct
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> > morphisms from dom(f) to cod(g)? It seems to me that the only way that
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> > 'o' can be a function in such a case is for g o f to be simply one
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> > possible composition. If so, then g o f should not be seen as The
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> > Composition, but simply as one POSSIBLE composition.
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> >
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> > this becomes (denoting a group product by *)
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> >
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> > How can a*b can be the one and only product of b and a in a group
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> > with multiple distinct elements?
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>
>
> [snip]
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>
>
> As I have said elsewhere, if all your morphisms are functions (as in
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> your group examples), their composition is always uniquely defined.
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> But morphisms are not always functions in other categories.
>
That doesn't imply what you are saying.
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>
> > >> Whether it's the composition
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> > >> of f and g depends on the f and g in question, and the relevant notion
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> > >> of composition given by the category at issue.
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> >
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> > > I'm just going by the above definition, and it suggests that the
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> > > domain and codomain are the ONLY relevant criteria. A problem arises,
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> > > of course, when there are multiple, distinct morphisms from A to C --
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> > > which one to pick? It's beginning to look to me like the choice is
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> > > completely arbitrary.
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> >
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> > > Can you cite any authoritative source to contrary? Online would be
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> > > nice.
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> >
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> > Look at your own link, in particular the section about groups. Look at
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> > the sentence that says "We take as composition the group
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> > multiplication."
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>
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> This is only a particular case -- the category of groups where all
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> morphisms are functions. Our definitions must also cover the case
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> where morphisms are not functions.
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>
The definition does. You are misunderstanding the definition.
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>
>
> > > It seems to me that would it work perfectly fine to arbitrary select
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> > > as the composite any morphism with the required domain and codoemain.
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> > > This would at least be consistent with all the definitions we have
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> > > looked at.
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> >
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> > No it wouldn't, since many such arbitrary choices would fail to satisfy
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> > associativity.
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> >
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>
>
> How so?
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>
>
> Dan
>
> Download my DC Proof 2.0 software at http://www.dcproof.com
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