# Natural Isomorphism $\mbox{Hom}(\oplus_\alpha A_\alpha,G) \simeq \prod_\alpha \mbox{Hom}(A_\alpha,G)$

In his algebraic topology book, Hatcher makes the comment that there is a natural isomorphism $\mbox{Hom}(\oplus_\alpha A_\alpha,G) \simeq \prod_\alpha \mbox{Hom}(A_\alpha,G)$

It is not immediately clear why this is true. I can take a morphism $f:\oplus_\alpha A_\alpha \to G$, but I can't see how to construct an isomorphism ($G$ is an abelian group)

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The direct sum of abelian groups $\bigoplus_\alpha A_\alpha$ satisfies a universal property: for any choice of homomorphisms $f_\alpha:A_\alpha\rightarrow G$ for each $\alpha$, there is a unique homomorphism $f:\bigoplus_\alpha A_\alpha\rightarrow G$ such that $f_\alpha=f\circ i_\alpha$, where $i_\alpha:A_\alpha\rightarrow\bigoplus_\alpha A_\alpha$ is the canonical inclusion map. In other words, the direct sum of abelian groups gives the coproduct in the category of abelian groups (this is a statement that has to be proven).

Given any $f:\bigoplus_\alpha A_\alpha\rightarrow G$, I can get a collection of $f_\alpha:A_\alpha\rightarrow G$'s by simply setting $f_\alpha=f\circ i_\alpha$. Conversely, given any choice of $f_\alpha:A_\alpha\rightarrow G$ for each $\alpha$, I can create an $f:\bigoplus_\alpha A_\alpha\rightarrow G$ that agrees with all of the $f_\alpha$'s on each factor, by setting $f((x_\alpha)_\alpha)=\prod_\alpha f_\alpha(x_\alpha)$ - the order that this product is taken in $G$ doesn't matter since $G$ is abelian, and this is well defined because the direct sum consists of precisely those tuples of elements from each $A_\alpha$ where all but finitely many are the identity (this is required for well-definedness because the product of infinitely many non-identity elements of a group is not well-defined).

Thus, the natural isomorphism $$\text{Hom}\left(\bigoplus_\alpha A_\alpha,G\right)\xrightarrow{\,\,\,\,\,\phi\,\,\,\,\,}\prod_\alpha\text{Hom}(A_\alpha,G)$$ is defined by $\phi(f)=(f\circ i_\alpha)_\alpha$, where the $i_\alpha$ are the natural inclusions.

This argument is actually completely general. This is what the statement would look like in an arbitrary category.

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thanks. Excellent explanation - I am used to a coproduct of two objects in a category, so I hadn't seen the connection to this – Juan S Apr 28 '11 at 7:55
@Qwirk, no problem, glad to help. As my explanation hopefully makes clear, this is why the direct sum of abelian groups $\bigoplus_\alpha A_\alpha$ (i.e. the coproduct) and and the direct product of abelian groups $\prod_\alpha A_\alpha$ (i.e. the product) only agree when the indexing set is finite. – Zev Chonoles Apr 28 '11 at 8:13
@Qwirk: That's precicely what happens in this case as well, but with different words in it. By definition, a morphism from the coproduct is the same thing as a morphism from each of its factors. – Eivind Dahl Apr 28 '11 at 8:16
@Eivind: Exactly! I think Zev Chonoles was spot on in explaining that – Juan S Apr 29 '11 at 1:58