Hom commutes with finite direct products in both slots

Let R be a ring with 1 and let A, B, and M be unital left R-modules. Prove that \mathsf{Hom}_R(A \times B, M) \cong_R \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M) and \mathsf{Hom}_R(M,A \times B) \cong_R \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B). (Where \cong_R means “isomorphic as R-modules”.)


Given R-module homomorphisms \alpha : A \rightarrow M and \beta : B \rightarrow M, define \Phi(\alpha,\beta) : A \times B \rightarrow M by \Phi(\alpha,\beta)(a,b) = \alpha(a) + \beta(b). Note that, for all a_1,a_2 \in A, b_1,b_2 \in B, and r \in R, we have

\Phi(\alpha,\beta)((a_1,b_1) + r \cdot (a_2,b_2))  =  \Phi(\alpha,\beta)((a_1 + r \cdot a_2, b_1 + r \cdot b_2))
 =  \alpha(a_1 + r \cdot a_2) + \beta(b_1 + r \cdot b_2)
 =  \alpha(a_1) + r \cdot \alpha(a_2) + \beta(b_1) + r \cdot \beta(b_2)
 =  \Phi(\alpha,\beta)(a_1,b_1) + r \cdot \Phi(\alpha,\beta)(a_2,b_2).

So \Phi(\alpha,\beta) : A \times B \rightarrow M is an R-module homomorphism. Consider the map \Phi : \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M) \rightarrow \mathsf{Hom}_R(A \times B, M). We claim that \Phi is an R-module homomorphism. To see this, let \alpha_1,\alpha_2 : A \rightarrow M and \beta_1,\beta_2 : B \rightarrow M be R-module homomorphisms and let r \in R. Now

\Phi((a_1,b_1) + r \cdot (a_2,b_2))(a,b)  =  \Phi(\alpha_1 + r \cdot \alpha_2, \beta_1 + r \cdot \beta_2)(a,b)
 =  (\alpha_1 + r \cdot \alpha_2)(a) + (\beta_1 + r \cdot \beta_2)(b)
 =  \alpha_1(a) + r \cdot \alpha_2(a) + \beta_1(b) + r \cdot \beta_2(b)
 =  \Phi(\alpha_1,\beta_1)(a,b) + r \cdot \Phi(\alpha_2,\beta_2)(a,b)
 =  (\Phi(\alpha_1,\beta_1) + r \cdot \Phi(\alpha_2,\beta_2))(a,b).

So \Phi is an R-module homomorphism. Now suppose (\alpha,\beta) \in \mathsf{ker}\ \Phi. In particular, 0 = \Phi(\alpha,\beta)(a,0) = \alpha(a) for all a \in A, so that \alpha = 0. Similarly, \beta = 0. Thus (\alpha,\beta) = 0, and in fact \mathsf{ker}\ \Phi = 0. Hence \Phi is injective. Finally, let \theta : A \times B \rightarrow M be an R-module homomorphism, and define \alpha_\theta : A \rightarrow M and \beta_\theta : B \rightarrow M by \alpha_\theta(a) = \theta(a,0) and \beta_\theta(b) = \theta(0,b). Note that \alpha_\theta(a_1 + r \cdot a_2) = \theta(a_1 + r \cdot a_2,0 = \theta(a_1,0) + r \cdot \theta(a_2,0) = \alpha_\theta(a_1) + r \cdot \alpha_\theta(a_2), so that \alpha_\theta is an R-module homomorphism. Similarly, \beta_\theta is an R-module homomorphism. Since \Phi(\alpha_\theta, \beta_\theta)(a,b) = \alpha_\theta(a) + \beta_\theta(b) = \theta(a,0) + \theta(0,b) = \theta(a,b), we have \Phi(\alpha_\theta,\beta_\theta) = \theta. Hence \Phi is surjective. Thus, we have \mathsf{Hom}_R(A \times B, M) \cong_R \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M).

Now let \pi_A : A \times B \rightarrow A and \pi_B : A \times B \rightarrow B be the left and right coordinate projections; certainly these are R-module homomorphisms. Define \Psi : \mathsf{Hom}_R(M,A \times B) \rightarrow \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B) by \Psi(\varphi) = (\pi_A \circ \varphi, \pi_B \circ \varphi). If \varphi, \psi \in \mathsf{Hom}_R(M,A \times B) and r \in R, then

\Psi(\varphi + r \cdot \psi)  =  (\pi_A \circ (\varphi + r \cdot \psi), \pi_B \circ (\varphi + r \cdot \psi))
 =  (\pi_A \circ \varphi + r \cdot (\pi_A \circ \psi), \pi_B \circ \varphi + r \cdot (\pi_B \circ \psi))
 =  (\pi_A \circ \varphi, \pi_A \circ \psi) + r \cdot (\pi_B \circ \varphi, \pi_B \circ \psi)
 =  \Psi(\varphi) + r \cdot \Psi(\psi).

Thus \Psi is an R-module homomorphism. Now suppose \varphi \in \mathsf{ker}\ \Psi. In particular, if m \in M, then 0 = \Psi(\varphi)(m,m) = ((\pi_A \circ \varphi)(m), (\pi_b \circ \psi)(m)) = (\pi_A(\varphi(m)), \pi_B(\varphi(m)) = \varphi(m). So \varphi = 0, hence \mathsf{ker}\ \Psi = 0, and thus \Psi is injective. Finally, suppose (\alpha,\beta) \in \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B). Define \varphi_{\alpha,\beta} : M \rightarrow A \times B by \varphi_{\alpha,\beta}(m) = (\alpha(m), \beta(m). Certainly \varphi_{\alpha,\beta} is an R-module homomorphism. Moreover, we have \Psi(\varphi_{\alpha,\beta})(m) = ((\pi_A \circ \varphi_{\alpha,\beta})(m), (\pi_B \circ \varphi_{\alpha,\beta})(m)) = (\alpha(m), \beta(m)). Thus \Psi(\varphi_{\alpha,\beta}) = (\alpha,\beta), and so \Psi is surjective. Hence \mathsf{Hom}_R(M, A \times B) \cong_R \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B).

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