Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Ki Song
Posts:
549
Registered:
9/19/09


Re: Toward a Formal, MachineParsable Definition of a Category
Posted:
Nov 6, 2012 1:04 AM


On Tuesday, November 6, 2012 12:27:34 AM UTC5, 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 headto > > 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, > > > > there is a law of composition: If f:A>B and g:B>C, then the > > > > 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 online.
> > > > > 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. > > > > > > > Can you cite any authoritative source to contrary? Online would be > > > > nice. > > > > > > Look at your own link, in particular the section about groups. Look at > > > 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 > > Download my DC Proof 2.0 software at http://www.dcproof.com



