## The interaction of Hom with direct sums and direct products

Let $R$ be a ring with 1. Let $A$ be a left unital $R$-module, and let $\{B_i\}_I$ be a nonempty family of left unital $R$-modules. Prove that, as abelian groups, $\mathsf{Hom}_R(\bigoplus_I B_i, A) \cong \prod_I \mathsf{Hom}_R(B_i,A)$ and $\mathsf{Hom}_R(A, \prod_I B_i) \cong \prod_I \mathsf{Hom}_R(A,B_i)$. Prove also that if $R$ is commutative, these pairs are $R$-module isomorphic.

First we show that $\mathsf{Hom}_R(\bigoplus_I B_i, A) \cong \prod_I \mathsf{Hom}_R(B_i,A)$ as abelian groups. Recall that for each $i$, we have the canonical injection $\iota_i : B_i \rightarrow \bigoplus_I B_i$. Define for each $i \in I$ the map $\varphi_i : \mathsf{Hom}_R( \bigoplus_I B_i, A) \rightarrow \mathsf{Hom}_R(B_i,A)$ by $\varphi_i(\alpha) = \alpha \circ \iota_i$; certainly each $\varphi_i$ is well defined. By the universal property of direct products of abelian groups, there exists a unique group homomorphism $\Phi : \mathsf{Hom}_R(\bigoplus_I B_i, A) \rightarrow \prod_I \mathsf{Hom}_R(B_i, A)$ such that $\pi_i \circ \Phi = \varphi_i$, where $\pi_i$ denotes the $i$th natural projection from a direct product. We claim that this $\Phi$ is a group isomorphism.

Suppose $\alpha \in \mathsf{ker}\ \Phi$, so $\Phi(\alpha) = 0$. Then $(\pi_i \circ \Phi)(\alpha) = 0$ for each $i$, so that $\varphi_i(\alpha) = 0$ for each $i$. Thus $\alpha \circ \iota_i = 0$ for all $i$. That is, $\alpha$ applied to the $i$th component of any element in $\bigoplus_I B_i$ is zero, for all $i$. So $\alpha = 0$, and thus $\mathsf{ker}\ \Phi = 0$. So $\Phi$ is injective.

Now suppose $\psi = (\psi_i) \in \prod_I \mathsf{Hom}_R(B_i,A)$. Define $\alpha_\psi : \bigoplus_I B_i \rightarrow A$ by $\alpha_\psi(b_i) = \sum \psi_i(b_i)$; this map is well defined since only finitely many terms of $(b_i)$ are nonzero. Moreover, it is clear that $\alpha_\psi$ is an $R$-module homomorphism, so in fact $\alpha_\psi \in \mathsf{Hom}_R(\bigoplus_I B_i, A)$. Note that for all $i \in I$, $\Phi(\alpha_\psi)_i(b) = \varphi_i(\alpha_\psi)(b)$ $= (\alpha_\psi \circ \iota_i)(b)$ $= \alpha_\psi(\iota_i(b))$ $= \psi_i(b)$. Thus $\Phi(\alpha_\psi)_i = \psi_i$ for all $i$, and so $\Phi(\alpha_\psi) = \psi$. Thus $\Phi$ is surjective.

So $\Phi$ is an isomorphism of abelian groups. Suppose now that $R$ is commutative, so that both $\mathsf{Hom}_R(\bigoplus_I B_i,A)$ and $\prod_I \mathsf{Hom}_R(B_i,A)$ are naturally left $R$-modules. For all $r \in R$ and all module homomorphisms $\alpha : \bigoplus_I B_i \rightarrow A$, we have $\Phi(ra) = ((r \alpha) \circ \iota_i)$ $= (r(\alpha \circ \iota_i))$ $= r(\alpha \circ \iota_i)$ $= r\Phi(\alpha)$. Thus $\Phi$ is an isomorphism of left $R$-modules.

Next we show that $\mathsf{Hom}_R(A, \prod_I B_i) \cong \prod_I \mathsf{Hom}_R(A,B_i)$ as abelian groups. For each $i \in I$, define $\varphi_i : \mathsf{Hom}_R(A, \prod_I B_i) \rightarrow \mathsf{Hom}_R(A,B_i)$ by $\varphi_i(\alpha) = \pi_i \circ \alpha$, where $\pi_i$ denotes the $i$th canonical projection from a direct product. By the universal property of direct products, we have a unique group homomorphism $\Phi : \mathsf{Hom}_R(A,\prod_I B_i) \rightarrow \prod_I \mathsf{Hom}_R(A, B_i)$ such that $\pi_i \circ \Phi = \varphi_i$ for all $i$. We claim that this $\Phi$ is an isomorphism.

Suppose $\alpha \in \mathsf{ker}\ \Phi$. So $\Phi(\alpha) = 0$, and thus $(\pi_i \circ \Phi)(\alpha) = 0$ for all $i$. Thus $\varphi_i(\alpha) = 0$ for all $i$, so $\pi_i \circ \alpha = 0$ for all $i$. That is, every coordinate of any element in the image of $\alpha$ is 0. So $\alpha = 0$. Thus $\mathsf{ker}\ \Phi = 0$, and so $\Phi$ is injective.

Now let $\psi = (\psi_i) \in \prod_I \mathsf{Hom}_R(A,B_i)$. Define $\alpha_\psi : A \rightarrow \prod_I B_i$ by $\alpha_\psi(a)_i = \psi_i(a)$. Certainly $\alpha_\psi$ is a module homomorphism. Note that, for all $i$, $\Phi(\alpha_\psi)_i(a) = \varphi_i(\alpha_\psi)(a)$ $= (\pi_i \circ \alpha_\psi)(a)$ $= \psi_i(a)$. So $\Phi(\alpha_\psi)_i = \psi_i$ for all $i$, and we have $\Phi(\alpha_\psi) = \psi$. So $\Phi$ is surjective, and thus an isomorphism of groups.

Finally, suppose $R$ is commutative. If $r \in R$ and $\alpha : A \rightarrow \prod_I B_i$ is a module homomorphism, then $\Phi(r\alpha) = (\pi_i \circ r\alpha)$ $= (r(\pi_i \circ \alpha))$ $= r(\pi_i \circ \alpha)$ $= r\Phi(\alpha)$. Thus in this case $\Phi$ is an isomorphism of $R$-modules.