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I want to ask a stupid question. Let $I$ be an infinite set and suppose $i$ belongs to $I$. I wonder whether following morphisms exist in general:

Hom($A$,colim $B_i) \to$ lim Hom($A,B_i$) and

Colim Hom($A,B_i) \to$ Hom($A$,colim $B_i$)

What I know is: if we replace lim by infinite product and colim by infinite coproduct, it exists. But I am not sure in this general case above.

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I would suggest that you reread what you have written and correct the misprints, for starters. The first would-be morphism does not make much sense due to some mixup between lim and colim, apparently. In the second one, one "Hom" is missing and there is a misprint "B1". – Leonid Positselski Dec 12 '09 at 12:44
I am sorry for the misprints,I have corrected – Shizhuo Zhang Dec 12 '09 at 12:48
No, he is asking about morphisms, not isomorphisms. So the first formula is indeed wrong, but the morphism in the second formula does actually exist. Neither it matters whether the colimit in the second formula is finite or infinite. The morphism exists in both cases, and it is not an isomorphism, in general, in both cases. – Leonid Positselski Dec 12 '09 at 12:55
The existence of that morphism is trivial and I'm pretty sure that's not what he's looking for. – Harry Gindi Dec 12 '09 at 12:59
@Shizhou, if you intend these objecs to be abelian groups, you should say so in the question. (cf your comment below.) – Scott Morrison Dec 12 '09 at 16:40
up vote 9 down vote accepted

For any diagram $B_i$ and an object $A$ in a category, there are natural maps of sets:

  1. colim Hom($A,B_i) \to$ Hom($A$, colim $B_i$)
  2. colim Hom($B_i,A) \to$ Hom(lim $B_i, A$)

These maps need not be isomorphisms, in general (neither even when the diagram is filtered, nor when it is finite). Nor are they isomorphisms for infinite products and coproducts, in general (for finite products and coproducts in an additive category they are isomorphisms, though).

Besides, for any diagram $B_i$ and an object $A$ there are natural isomorphisms of sets:

  1. Hom($A$, lim $B_i$) = lim Hom($A,B_i$)
  2. Hom(colim $B_i, A$) = lim Hom($B_i,A$)

These isomorphisms hold for any diagram (it does not have to be filtered, nor does it have to be finite). Actually, they hold by the definition of lim and colim.

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In general, your category has to admit small limits for that to even start to begin to make any sense at all. Also, as I noted in my question, I'm fairly sure that you've got it backwards. It should be:

Hom(colim(F(-)),X) is isomorpic to lim Hom(F(-),X), and Hom(Y,lim(F(-))) is isomorphic to lim Hom(Y,F(-)), where we're limiting and colimiting over the domain of F, where F is a functor into our category from some other category (Diagrams for example.)

I don't know if this is what you actually wanted, but if I read your question the way you typed it out, the first one doesn't make sense, since covariant hom is covariant. The second one might be true provided that the limit has certain restrictions on it or if covariant hom has an appropriate adjoint. There might be other cases, but it's not true in general. If you're just looking for the existence of a map in the second one, then it's trivial.

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actually, the original problem is whether these two morphism exist for infinite products and infinite coproduts. And A,Bi are abelian groups – Shizhuo Zhang Dec 12 '09 at 12:54
Covariant hom is covariant, so the first is just false. – Harry Gindi Dec 12 '09 at 12:56
Thank you. In fact, I did not have much motivation to ask this question. It is just from some of my homework.In my homework,the question is whether following two morphism exists 1. Hom(A, infinite coproduct Bi)--->infinite product Hom(A,Bi) 2. infinite coproduct Hom(A,Bi)---->Hom(A, infinite coproduct Bi) where A,Bi are all abelian groups. I just wonder know if I use co(limit)instead of co(product),whether they still exist. – Shizhuo Zhang Dec 12 '09 at 13:15

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