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 ith 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 ith 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 ith 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.

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