[proofplan] We construct, for an arbitrary left $R$-module $M$, a free left $R$-module generated by the underlying set of $M$. The canonical map from this free module to $M$ sends the basis element indexed by $m \in M$ to $m$, and it is surjective because every element of $M$ is hit by its own basis vector. Finally, we prove directly that this free module is projective by lifting maps along surjective homomorphisms basis element by basis element. [/proofplan] [step:Construct the free module generated by the underlying set of $M$] Let $M$ be a left $R$-module. Let $S$ denote the underlying set of $M$. Define the left $R$-module \begin{align*} F := \bigoplus_{s \in S} R e_s, \end{align*} where $e_s$ denotes the standard generator in the copy of $R$ indexed by $s$. Equivalently, every element of $F$ has a unique finite expression \begin{align*} \sum_{i=1}^{n} r_i e_{s_i}, \end{align*} with $n \in \mathbb{N}$, $r_i \in R$, and $s_i \in S$. Define a map \begin{align*} \pi: F &\to M \\ \sum_{i=1}^{n} r_i e_{s_i} &\mapsto \sum_{i=1}^{n} r_i s_i. \end{align*} The uniqueness of the coordinate expression in the direct sum makes $\pi$ well-defined. For $r \in R$ and $x,y \in F$, the definition gives \begin{align*} \pi(x+y) &= \pi(x)+\pi(y), \\ \pi(rx) &= r\pi(x), \end{align*} so $\pi$ is an $R$-[linear map](/page/Linear%20Map). [/step] [step:Show that the canonical map onto $M$ is surjective] Let $m \in M$. Since $m \in S$, the basis element $e_m \in F$ is defined. By construction, \begin{align*} \pi(e_m) = 1_R m = m. \end{align*} Thus every element $m \in M$ lies in the image of $\pi$, so $\pi: F \to M$ is surjective. [/step] [step:Prove that the constructed free module is projective] We verify the lifting property for $F$. Let $q: N \to L$ be a surjective $R$-linear map of left $R$-modules, and let $f: F \to L$ be an $R$-linear map. For each $s \in S$, surjectivity of $q$ gives an element $n_s \in N$ such that \begin{align*} q(n_s) = f(e_s). \end{align*} Define \begin{align*} \widetilde{f}: F &\to N \\ \sum_{i=1}^{n} r_i e_{s_i} &\mapsto \sum_{i=1}^{n} r_i n_{s_i}. \end{align*} Again, the direct-sum coordinate expression makes this map well-defined, and the displayed formula shows that $\widetilde{f}$ is $R$-linear. For every basis element $e_s$, we have \begin{align*} (q \circ \widetilde{f})(e_s) = q(n_s) = f(e_s). \end{align*} Since both $q \circ \widetilde{f}$ and $f$ are $R$-linear maps from $F$ and agree on the basis $\{e_s : s \in S\}$, they agree on all of $F$. Hence \begin{align*} q \circ \widetilde{f} = f. \end{align*} Therefore every $R$-linear map from $F$ lifts along every surjective $R$-linear map, so $F$ is projective. [/step] [step:Conclude that every module admits an epimorphism from a projective module] For the arbitrary left $R$-module $M$, we have constructed a projective left $R$-module $F$ and a surjective $R$-linear map \begin{align*} \pi: F \to M. \end{align*} Thus every object of $R\operatorname{-Mod}$ receives an epimorphism from a projective object. Therefore $R\operatorname{-Mod}$ has enough projectives. [/step]