[proofplan] We define the candidate map by sending a coordinate vector $(r_1,\ldots,r_n)$ to the finite $R$-linear combination $\sum_{i=1}^{n} r_i y_i$ in $N$. The module axioms in $N$ show that this formula is additive and compatible with the left $R$-action, hence gives an $R$-[module homomorphism](/page/Module%20Homomorphism). The formula sends each standard basis vector to the prescribed element because only one coordinate is nonzero. Uniqueness follows from the unique coordinate expansion of every element of $R^n$ in the standard basis. [/proofplan] [step:Define the homomorphism by the prescribed linear combination] Define a function $f:R^n\to N$ as follows. For an element $x=(r_1,\ldots,r_n)\in R^n$, set \begin{align*} f(x)=\sum_{i=1}^{n} r_i y_i. \end{align*} This is a well-defined element of $N$, since each $r_i\in R$, each $y_i\in N$, the scalar product $r_i y_i$ belongs to $N$, and the sum is finite in the underlying abelian group of $N$. [/step] [step:Verify that the constructed map is $R$-linear] Let $x=(r_1,\ldots,r_n)$ and $x'=(s_1,\ldots,s_n)$ be elements of $R^n$, and let $a\in R$. By coordinatewise addition in $R^n$ and distributivity of the left $R$-action on $N$, \begin{align*} f(x+x')=\sum_{i=1}^{n}(r_i+s_i)y_i=\sum_{i=1}^{n}(r_i y_i+s_i y_i)=\sum_{i=1}^{n}r_i y_i+\sum_{i=1}^{n}s_i y_i=f(x)+f(x'). \end{align*} By coordinatewise scalar multiplication in $R^n$ and associativity of the left $R$-action on $N$, \begin{align*} f(ax)=\sum_{i=1}^{n}(a r_i)y_i=\sum_{i=1}^{n}a(r_i y_i)=a\sum_{i=1}^{n}r_i y_i=af(x). \end{align*} Thus $f$ is additive and preserves scalar multiplication, so $f:R^n\to N$ is an $R$-module homomorphism. [guided] We must check that the formula defining $f$ is compatible with the two operations in a left $R$-module: addition and scalar multiplication. Take two arbitrary elements $x=(r_1,\ldots,r_n)$ and $x'=(s_1,\ldots,s_n)$ of $R^n$. Their sum is computed coordinatewise, so \begin{align*} x+x'=(r_1+s_1,\ldots,r_n+s_n). \end{align*} Applying the definition of $f$ gives \begin{align*} f(x+x')=\sum_{i=1}^{n}(r_i+s_i)y_i. \end{align*} For each index $i$, the module distributive law in $N$ gives $(r_i+s_i)y_i=r_i y_i+s_i y_i$. Therefore \begin{align*} f(x+x')=\sum_{i=1}^{n}(r_i y_i+s_i y_i)=\sum_{i=1}^{n}r_i y_i+\sum_{i=1}^{n}s_i y_i=f(x)+f(x'). \end{align*} This proves additivity. Now take an arbitrary scalar $a\in R$. Scalar multiplication in the free module $R^n$ is coordinatewise, so \begin{align*} ax=(a r_1,\ldots,a r_n). \end{align*} Using the definition of $f$ again, \begin{align*} f(ax)=\sum_{i=1}^{n}(a r_i)y_i. \end{align*} Because $N$ is a left $R$-module, the scalar action is associative: $(a r_i)y_i=a(r_i y_i)$ for every $i$. Hence \begin{align*} f(ax)=\sum_{i=1}^{n}a(r_i y_i)=a\sum_{i=1}^{n}r_i y_i=af(x). \end{align*} Thus $f$ is additive and compatible with scalar multiplication. These are exactly the defining properties of an $R$-module homomorphism, so $f:R^n\to N$ is an $R$-module homomorphism. [/guided] [/step] [step:Check that the standard basis vectors have the prescribed images] Fix $j\in\{1,\ldots,n\}$. The standard basis vector $e_j\in R^n$ has $j$th coordinate $1_R$, the multiplicative identity of $R$, and all other coordinates $0_R$, the additive identity of $R$. Therefore \begin{align*} f(e_j)=\sum_{i=1}^{n}(e_j)_i y_i=1_R y_j+\sum_{\substack{1\le i\le n, i\ne j}}0_R y_i=y_j. \end{align*} The last equality uses the unital module axiom $1_R y_j=y_j$ and the zero-scalar identity $0_R y_i=0_N$ in the left $R$-module $N$. Since $j$ was arbitrary, $f(e_i)=y_i$ for every $i\in\{1,\ldots,n\}$. [/step] [step:Prove uniqueness from the standard basis expansion] Let $g:R^n\to N$ be an $R$-module homomorphism satisfying $g(e_i)=y_i$ for every $i\in\{1,\ldots,n\}$. For an arbitrary element $x=(r_1,\ldots,r_n)\in R^n$, the standard basis expansion in the free module $R^n$ is \begin{align*} x=\sum_{i=1}^{n} r_i e_i. \end{align*} Using additivity and scalar compatibility of $g$, \begin{align*} g(x)=g\left(\sum_{i=1}^{n}r_i e_i\right)=\sum_{i=1}^{n}g(r_i e_i)=\sum_{i=1}^{n}r_i g(e_i)=\sum_{i=1}^{n}r_i y_i=f(x). \end{align*} Thus $g(x)=f(x)$ for every $x\in R^n$, so $g=f$. Hence the homomorphism constructed above is unique, completing the proof. [/step]