[proofplan] We prove the arbitrary-index isomorphism first, since the finite case is a special case. A homomorphism out of a [direct sum](/page/Direct%20Sum) is determined by its restrictions to the summands, and conversely a family of maps from the summands defines a unique map on the direct sum because every element has finite support. For the second isomorphism, the finiteness of $n$ allows a map into the direct sum to be assembled from its finitely many component maps. Finally, we verify that the constructions are additive, mutually inverse, and natural. [/proofplan] [step:Identify a map out of a direct sum with its component restrictions] Let $I$ be an index set, let $(M_i)_{i \in I}$ be a family of left $R$-modules, and let \begin{align*} \iota_i: M_i \to \bigoplus_{j \in I} M_j \end{align*} denote the canonical inclusion of the $i$-th summand. Define \begin{align*} \Phi: \operatorname{Hom}_R\left(\bigoplus_{i \in I} M_i,N\right) \to \prod_{i \in I}\operatorname{Hom}_R(M_i,N) \end{align*} by \begin{align*} \Phi(f) = (f \circ \iota_i)_{i \in I}. \end{align*} For each $f \in \operatorname{Hom}_R(\bigoplus_{i \in I}M_i,N)$ and each $i \in I$, the composite $f \circ \iota_i$ is $R$-linear, so $\Phi$ is well-defined. If $f,g$ are $R$-linear maps from $\bigoplus_{i \in I} M_i$ to $N$, then for every $i \in I$, \begin{align*} \Phi(f+g)_i = (f+g)\circ \iota_i = f\circ \iota_i + g\circ \iota_i = \Phi(f)_i+\Phi(g)_i. \end{align*} Thus $\Phi$ is a homomorphism of abelian groups. [/step] [step:Reconstruct the map from a family of component maps] For an element $x \in \bigoplus_{i \in I}M_i$, write $x_i \in M_i$ for its $i$-th component and define its support by \begin{align*} \operatorname{supp}(x) := \{i \in I : x_i \neq 0\}. \end{align*} By the definition of the direct sum, $\operatorname{supp}(x)$ is finite. Define \begin{align*} \Psi: \prod_{i \in I}\operatorname{Hom}_R(M_i,N) \to \operatorname{Hom}_R\left(\bigoplus_{i \in I}M_i,N\right) \end{align*} as follows. Given a family $(f_i)_{i \in I}$ with $f_i \in \operatorname{Hom}_R(M_i,N)$, set \begin{align*} \Psi((f_i)_{i \in I})(x) := \sum_{i \in \operatorname{supp}(x)} f_i(x_i). \end{align*} The sum is finite, so it is defined in the abelian group underlying $N$. We verify that $\Psi((f_i)_{i \in I})$ is $R$-linear. Let $x,y \in \bigoplus_{i \in I}M_i$ and $r \in R$. Since $\operatorname{supp}(x) \cup \operatorname{supp}(y)$ is finite and contains the support of $x+y$, additivity of each $f_i$ gives \begin{align*} \Psi((f_i)_{i \in I})(x+y)=\sum_{i \in \operatorname{supp}(x)\cup \operatorname{supp}(y)} f_i(x_i+y_i). \end{align*} Hence \begin{align*} \Psi((f_i)_{i \in I})(x+y)=\sum_{i \in \operatorname{supp}(x)\cup \operatorname{supp}(y)} f_i(x_i)+\sum_{i \in \operatorname{supp}(x)\cup \operatorname{supp}(y)} f_i(y_i). \end{align*} Terms with zero component contribute $0$, so this equals \begin{align*} \Psi((f_i)_{i \in I})(x)+\Psi((f_i)_{i \in I})(y). \end{align*} Similarly, because $\operatorname{supp}(rx) \subseteq \operatorname{supp}(x)$ and each $f_i$ is $R$-linear, \begin{align*} \Psi((f_i)_{i \in I})(rx)=\sum_{i \in \operatorname{supp}(x)} f_i(rx_i)=\sum_{i \in \operatorname{supp}(x)} r f_i(x_i)=r\Psi((f_i)_{i \in I})(x). \end{align*} Therefore $\Psi((f_i)_{i \in I})$ is an $R$-[module homomorphism](/page/Module%20Homomorphism). The assignment $\Psi$ is additive in the family variable: for families $(f_i)_{i \in I}$ and $(g_i)_{i \in I}$, and for $x \in \bigoplus_{i \in I}M_i$, \begin{align*} \Psi((f_i+g_i)_{i \in I})(x)=\sum_{i \in \operatorname{supp}(x)} (f_i+g_i)(x_i). \end{align*} Thus \begin{align*} \Psi((f_i+g_i)_{i \in I})(x)=\Psi((f_i)_{i \in I})(x)+\Psi((g_i)_{i \in I})(x), \end{align*} so $\Psi$ is a homomorphism of abelian groups. [guided] The only point that requires care is that $I$ may be infinite. A direct sum is not the full product: an element \begin{align*} x \in \bigoplus_{i \in I}M_i \end{align*} has components $x_i \in M_i$, but only finitely many of them are nonzero. We record this finite set as \begin{align*} \operatorname{supp}(x) := \{i \in I : x_i \neq 0\}. \end{align*} This finite-support condition is exactly what makes the formula \begin{align*} \Psi((f_i)_{i \in I})(x) := \sum_{i \in \operatorname{supp}(x)} f_i(x_i) \end{align*} meaningful: the sum is finite, so it is a legitimate sum in the abelian group underlying $N$. We now verify $R$-linearity. Let $x,y \in \bigoplus_{i \in I}M_i$ and let $r \in R$. The support of $x+y$ is contained in the finite set $\operatorname{supp}(x)\cup \operatorname{supp}(y)$, so we may sum over that finite set without changing the value, because terms with zero component contribute $0$. Since each \begin{align*} f_i: M_i \to N \end{align*} is an $R$-[linear map](/page/Linear%20Map), it is additive. Therefore \begin{align*} \Psi((f_i)_{i \in I})(x+y)=\sum_{i \in \operatorname{supp}(x)\cup \operatorname{supp}(y)} f_i(x_i+y_i). \end{align*} Using additivity of $f_i$ for each index in this finite set gives \begin{align*} \Psi((f_i)_{i \in I})(x+y)=\sum_{i \in \operatorname{supp}(x)\cup \operatorname{supp}(y)} f_i(x_i)+\sum_{i \in \operatorname{supp}(x)\cup \operatorname{supp}(y)} f_i(y_i). \end{align*} Removing zero terms from the first and second sums yields \begin{align*} \Psi((f_i)_{i \in I})(x+y)=\Psi((f_i)_{i \in I})(x)+\Psi((f_i)_{i \in I})(y). \end{align*} For scalar multiplication, $\operatorname{supp}(rx)$ is contained in $\operatorname{supp}(x)$. Since every $f_i$ is $R$-linear, \begin{align*} \Psi((f_i)_{i \in I})(rx)=\sum_{i \in \operatorname{supp}(x)} f_i(rx_i). \end{align*} Using $R$-linearity of each $f_i$ gives \begin{align*} \Psi((f_i)_{i \in I})(rx)=\sum_{i \in \operatorname{supp}(x)} r f_i(x_i). \end{align*} Because scalar multiplication in the left $R$-module $N$ distributes over finite sums, \begin{align*} \Psi((f_i)_{i \in I})(rx)=r\sum_{i \in \operatorname{supp}(x)} f_i(x_i)=r\Psi((f_i)_{i \in I})(x). \end{align*} Hence $\Psi((f_i)_{i \in I})$ is an $R$-module homomorphism. Finally, $\Psi$ is itself additive as a map between Hom groups. If $(f_i)_{i \in I}$ and $(g_i)_{i \in I}$ are two families of $R$-linear maps, then for every $x \in \bigoplus_{i \in I}M_i$, \begin{align*} \Psi((f_i+g_i)_{i \in I})(x)=\sum_{i \in \operatorname{supp}(x)} (f_i+g_i)(x_i). \end{align*} Expanding the pointwise sum in the Hom group gives \begin{align*} \Psi((f_i+g_i)_{i \in I})(x)=\sum_{i \in \operatorname{supp}(x)} f_i(x_i)+\sum_{i \in \operatorname{supp}(x)} g_i(x_i). \end{align*} Thus \begin{align*} \Psi((f_i+g_i)_{i \in I})(x)=\Psi((f_i)_{i \in I})(x)+\Psi((g_i)_{i \in I})(x), \end{align*} so $\Psi$ is a homomorphism of abelian groups. [/guided] [/step] [step:Show the two constructions out of the direct sum are inverse] Let $f \in \operatorname{Hom}_R(\bigoplus_{i \in I}M_i,N)$. For $x \in \bigoplus_{i \in I}M_i$, the finite-support decomposition of $x$ is \begin{align*} x=\sum_{i \in \operatorname{supp}(x)} \iota_i(x_i). \end{align*} Using additivity of $f$ over this finite sum, \begin{align*} (\Psi(\Phi(f)))(x)=\sum_{i \in \operatorname{supp}(x)} (f\circ \iota_i)(x_i)=f\left(\sum_{i \in \operatorname{supp}(x)} \iota_i(x_i)\right)=f(x). \end{align*} Hence $\Psi\circ \Phi=\operatorname{id}$. Conversely, let $(f_i)_{i \in I} \in \prod_{i \in I}\operatorname{Hom}_R(M_i,N)$. For $j \in I$ and $m \in M_j$, the element $\iota_j(m)$ has support contained in $\{j\}$ and has $j$-th component $m$. Therefore \begin{align*} (\Phi(\Psi((f_i)_{i \in I})))_j(m)=\Psi((f_i)_{i \in I})(\iota_j(m))=f_j(m). \end{align*} Since this holds for every $j \in I$ and every $m \in M_j$, we have $\Phi\circ \Psi=\operatorname{id}$. Thus \begin{align*} \operatorname{Hom}_R\left(\bigoplus_{i \in I}M_i,N\right) \cong \prod_{i \in I}\operatorname{Hom}_R(M_i,N) \end{align*} as abelian groups. Taking $I=\{1,\ldots,n\}$ gives the first finite isomorphism. [/step] [step:Identify a map into a finite direct sum with its component projections] Now assume $I=\{1,\ldots,n\}$. For each $i \in \{1,\ldots,n\}$, let \begin{align*} \pi_i: \bigoplus_{j=1}^n M_j \to M_i \end{align*} be the canonical projection onto the $i$-th summand. Define \begin{align*} \Theta: \operatorname{Hom}_R\left(N,\bigoplus_{i=1}^n M_i\right) \to \prod_{i=1}^n \operatorname{Hom}_R(N,M_i) \end{align*} by \begin{align*} \Theta(g)=(\pi_i\circ g)_{i=1}^n. \end{align*} Each composite $\pi_i\circ g$ is $R$-linear, so $\Theta$ is well-defined. Since composition with each projection preserves pointwise addition of homomorphisms, $\Theta$ is a homomorphism of abelian groups. Define \begin{align*} \Lambda: \prod_{i=1}^n \operatorname{Hom}_R(N,M_i) \to \operatorname{Hom}_R\left(N,\bigoplus_{i=1}^n M_i\right) \end{align*} as follows. Given $(g_i)_{i=1}^n$ with $g_i \in \operatorname{Hom}_R(N,M_i)$, let $\Lambda((g_i)_{i=1}^n)$ be the map \begin{align*} \Lambda((g_i)_{i=1}^n): N \to \bigoplus_{i=1}^n M_i \end{align*} defined by \begin{align*} \Lambda((g_i)_{i=1}^n)(y)=(g_1(y),\ldots,g_n(y)). \end{align*} Because the tuple has only finitely many components, it is an element of the finite direct sum. Additivity and $R$-linearity of the coordinate maps $g_i$ imply additivity and $R$-linearity of $\Lambda((g_i)_{i=1}^n)$. Pointwise addition in each coordinate also shows that $\Lambda$ is a homomorphism of abelian groups. [/step] [step:Show the two constructions into the finite direct sum are inverse] Let $g \in \operatorname{Hom}_R(N,\bigoplus_{i=1}^n M_i)$ and let $y \in N$. The element $g(y)$ is the finite tuple of its components: \begin{align*} g(y)=((\pi_1\circ g)(y),\ldots,(\pi_n\circ g)(y)). \end{align*} Therefore \begin{align*} (\Lambda(\Theta(g)))(y)=g(y), \end{align*} so $\Lambda\circ\Theta=\operatorname{id}$. Conversely, let $(g_i)_{i=1}^n \in \prod_{i=1}^n\operatorname{Hom}_R(N,M_i)$. For each $j \in \{1,\ldots,n\}$ and each $y \in N$, \begin{align*} (\Theta(\Lambda((g_i)_{i=1}^n)))_j(y)=\pi_j(g_1(y),\ldots,g_n(y))=g_j(y). \end{align*} Thus $\Theta\circ\Lambda=\operatorname{id}$. Hence \begin{align*} \operatorname{Hom}_R\left(N,\bigoplus_{i=1}^n M_i\right) \cong \prod_{i=1}^n\operatorname{Hom}_R(N,M_i) \end{align*} as abelian groups. [/step] [step:Verify naturality of the isomorphisms] The first isomorphism is natural in the target module $N$ because it is defined by precomposition with the canonical inclusions $\iota_i$ and its inverse is defined by the finite-support summation formula. Explicitly, if \begin{align*} h: N \to N' \end{align*} is an $R$-module homomorphism, then for every $f: \bigoplus_{i \in I}M_i \to N$ and every $i \in I$, \begin{align*} \Phi(h\circ f)_i=h\circ f\circ \iota_i=h\circ \Phi(f)_i. \end{align*} Thus applying $h$ after the reconstructed map agrees componentwise with applying $h$ to each component map. We also verify naturality in the summand variables. Let $(M_i')_{i \in I}$ be another family of left $R$-modules, and for each $i \in I$ let \begin{align*} a_i: M_i' \to M_i \end{align*} be an $R$-module homomorphism. Let \begin{align*} a: \bigoplus_{i \in I} M_i' \to \bigoplus_{i \in I} M_i \end{align*} be the induced direct-sum homomorphism, so that $a\circ\iota_i'=\iota_i\circ a_i$ for every $i \in I$, where $\iota_i':M_i'\to\bigoplus_{j\in I}M_j'$ is the canonical inclusion. For every $f: \bigoplus_{i \in I}M_i \to N$ and every $i \in I$, \begin{align*} \Phi(f\circ a)_i=f\circ a\circ\iota_i'=f\circ\iota_i\circ a_i=\Phi(f)_i\circ a_i. \end{align*} This is exactly the componentwise commutativity of the naturality square in the variables $(M_i)_{i \in I}$. The second isomorphism is natural in the source module $N$ because it is defined by postcomposition with the canonical projections $\pi_i$ and its inverse is the coordinatewise tuple map. Explicitly, if \begin{align*} k: N' \to N \end{align*} is an $R$-module homomorphism, then for every $g: N \to \bigoplus_{i=1}^n M_i$ and every $i \in \{1,\ldots,n\}$, \begin{align*} \Theta(g\circ k)_i=\pi_i\circ g\circ k=\Theta(g)_i\circ k. \end{align*} For naturality in the finite family of target summands, let $b_i:M_i\to M_i'$ be $R$-module homomorphisms and let $b:\bigoplus_{i=1}^n M_i\to\bigoplus_{i=1}^n M_i'$ be the induced direct-sum homomorphism. For every $g:N\to\bigoplus_{i=1}^n M_i$ and every $j\in\{1,\ldots,n\}$, \begin{align*} \Theta(b\circ g)_j=\pi_j'\circ b\circ g=b_j\circ\pi_j\circ g=b_j\circ\Theta(g)_j, \end{align*} where $\pi_j':\bigoplus_{i=1}^n M_i'\to M_j'$ is the canonical projection. Therefore both displayed isomorphisms are natural, and the proof is complete. [/step]