Where Am I Going Wrong in My Category Theory Solution?

[nateHI]

I'm trying to learn Category Theory; this isn't homework or anything. I've attached a problem from the text "Basic Homological Algebra" by Osborne and I show my attempt at a solution. My solution doesn't seem exactly correct and I state why in the attachment as well. Can someone take a look and let me know where I'm going wrong?

Thanks!

 

[fresh_42]

I'm trying to learn Category Theory; this isn't homework or anything. I've attached a problem from the text "Basic Homological Algebra" by Osborne and I show my attempt at a solution. My solution doesn't seem exactly correct and I state why in the attachment as well. Can someone take a look and let me know where I'm going wrong?

Thanks!

I think your problems start with the definition of the $S_i$ where you hide ℤ somehow. What you really want to show is that for each $x ∈ S$ there is a copy of ℤ in $A$. To "redefine" ℤ via $|S_i|$ seems artificial to me. What should be the elements of $S_i$? Consider the free groups generated by a single $x∈S$ instead. The only thing that makes it all a little abstract is that you don't know how big $S$ is. You could try to prove it for finite $S$ first.

 

[mathwonk]

To me category theory is the art of phrasing everything in terms of mapping properties and avoiding use of individual elements. So I would do this problem this way:

Lemma (Yoneda): An object in a category is determined by the Hom functor it defines, i.e. two objects A,B are isomorphic in any category if and only if the functors Hom(A, ), and Hom(B, ) are equivalent.

So if Frab(S) is the free abelian group on the set S, and if Copr(Z;S) is the coproduct of the family of copies of the integers Z indexed by the set S, we want to show the Hom functors Hom(F(S), ) and Hom(Copr(Z;S), ) are equivalent.

By definition of coproduct, we have Hom(Copr(Z;S), ) ≈ Prod(Hom(Z, );S) ≈
(*) Prod( ;S), where Prod is the product functor and Hom is homomorphisms in ab.

But by definition of the free abelian group, also Hom(F(S), ) ≈ Map(S, ) ≈ Prod( ;S), where Map denotes set functions, i.e. morphisms in the category of sets.

Since we have shown Hom(F(S), ) and Hom(Copr(Z;S), ) are equivalent, thus F(S) and Copr(Z;S), are isomorphic.

(*) Now in this step, i.e. the equivalence of Hom(Z, ) with the identity functor, we do need to use elements to prove e.g. for every abelian group G, that Hom(Z,G) ≈ G, because Z has been given to us as a concrete group, not in terms of its mapping properties.

 

[mathwonk]

to answer your question about your solution, isn't your map phi:S-->sigma(A) the map they speak of from S to "A".

 

[nateHI]

Thanks mathwonk and fresh_42 for the great comments. I've read them and will rework this problem shortly. Seems like it needs a lot of work though so it might take a day or two.

 

[nateHI]

So I cleaned my solution up a little based on the comments but I also worked through mathwonks solution which is much nicer. It used some mathematics that were a little beyond where I'm at currently but I managed to get a handle on those mechanics by "reading ahead."

Anyway, if anyone is reading this I propose you follow mathwonks solution and ignore my clunky attempt as his seems to be the more elegant (and correct) method.

 

[mathwonk]

thank you for your kind comments, it is so much more satisfying to answer a question when the OP reads the answer and responds as graciously as you did about the benefit he/she derived. this beautiful approach to the topic was explained to us by the great maurice auslander in his first year graduate algebra class at brandeis in 1965.