[proofplan] We construct a free module with one basis element for each element of the underlying set of $M$. A finitely supported formal linear combination of these basis elements maps to the corresponding finite $R$-linear combination in $M$. The basis element indexed by $m \in M$ maps to $m$, so the resulting homomorphism is surjective. [/proofplan] [step:Build the free module indexed by the elements of $M$] Let $I$ denote the underlying set of the left $R$-module $M$. Define \begin{align*} F := \{a: I \to R \mid \operatorname{supp}(a) \text{ is finite}\}, \end{align*} where $\operatorname{supp}(a) := \{i \in I : a(i) \neq 0_R\}$. Give $F$ the structure of a left $R$-module by pointwise addition and scalar multiplication: \begin{align*} (a+b)(i) &:= a(i)+b(i), \\ (r a)(i) &:= r a(i), \end{align*} for all $a,b \in F$, $r \in R$, and $i \in I$. For each $i \in I$, define $e_i: I \to R$ by \begin{align*} e_i(j) := \begin{cases} 1_R, & j=i,\\ 0_R, & j \neq i. \end{cases} \end{align*} Then $e_i \in F$, and every $a \in F$ has the finite expansion \begin{align*} a = \sum_{i \in \operatorname{supp}(a)} a(i)e_i. \end{align*} This expansion is unique because evaluating both sides at each $i \in I$ recovers the coefficient $a(i)$. Hence $\{e_i : i \in I\}$ is a basis of $F$, so $F$ is a free left $R$-module. [/step] [step:Define the homomorphism by sending each basis element to its index] Define a map \begin{align*} \pi: F &\to M \\ a &\mapsto \sum_{i \in \operatorname{supp}(a)} a(i)\, i. \end{align*} The sum is finite by the definition of $F$, so $\pi(a)$ is a well-defined element of $M$. We verify that $\pi$ is a left $R$-module homomorphism. Let $a,b \in F$ and $r \in R$. Since \begin{align*} \operatorname{supp}(a+b) \subseteq \operatorname{supp}(a) \cup \operatorname{supp}(b), \end{align*} we may sum over the finite set $\operatorname{supp}(a) \cup \operatorname{supp}(b)$ and obtain \begin{align*} \pi(a+b) &= \sum_{i \in \operatorname{supp}(a) \cup \operatorname{supp}(b)} (a(i)+b(i))\, i \\ &= \sum_{i \in \operatorname{supp}(a) \cup \operatorname{supp}(b)} a(i)\, i + \sum_{i \in \operatorname{supp}(a) \cup \operatorname{supp}(b)} b(i)\, i \\ &= \pi(a)+\pi(b). \end{align*} Similarly, \begin{align*} \pi(r a) &= \sum_{i \in \operatorname{supp}(a)} r a(i)\, i \\ &= r \sum_{i \in \operatorname{supp}(a)} a(i)\, i \\ &= r\pi(a). \end{align*} Thus $\pi: F \to M$ is a left $R$-module homomorphism. [/step] [step:Show that every element of $M$ is hit] Let $m \in M$. Since $m \in I$, the basis element $e_m \in F$ is defined. By the definition of $\pi$, \begin{align*} \pi(e_m) = 1_R m = m, \end{align*} using the unital left $R$-module axiom. Therefore every $m \in M$ lies in the image of $\pi$, so $\pi$ is surjective. Hence $M$ is a quotient of the free left $R$-module $F$. [/step]