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)
 A: 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.
