For every ring R, the category R-Mod is complete and cocomplete

A set-indexed Cartesian product of left $R$-modules, with componentwise addition and scalar multiplication, is a module. Coordinatewise pairing gives its product universal property. The empty product is the zero module.

For parallel linear maps $f,g:M\rightrightarrows N$, their agreement set is $\ker (f-g)$ and is a submodule by [F1] and [F2]. Corestriction through its inclusion gives the equalizer universal property. Hence [L1] gives completeness.

For $f,g:M\rightrightarrows N$, quotient $N$ by the submodule generated by $\{f(m)-g(m):m\in M\}$. The quotient map equalizes $f,g$, and [F4] says every equalizing linear map factors through it uniquely. Thus this is a coequalizer.

For a set $S$, let $R^{(S)}$ be the set of functions $\lambda :S\to R$ vanishing outside some finite subset, with pointwise addition and scalar multiplication; these are again finitely supported, so [F1] makes $R^{(S)}$ a left $R$-module. For $s\in S$ let $e_s$ take the value $1$ at $s$ and $0$ elsewhere. Each $\lambda$ is the finite sum ${\sum }_s\lambda (s)e_s$ over its support, and evaluating any such expression at a point of $S$ returns its coefficient there, so the expression is unique; hence $\{e_s:s\in S\}$ is a basis and $R^{(S)}$ is free in the sense of [F3]. Given a left $R$-module $P$ and any function $g:S\to P$, the assignment $\lambda \mapsto {\sum }_s\lambda (s)g(s)$, summed over the finite support, is $R$-linear by [F1] and sends $e_s$ to $g(s)$; and any linear map agreeing with $g$ on every $e_s$ agrees with it on every finite $R$-combination of the $e_s$, hence everywhere. So a linear map out of $R^{(S)}$ may be prescribed arbitrarily on the basis and is determined by that prescription.

For modules $(M_i)$, let $S={\coprod }_i\{i\}\times \mathrm{\mid }{M}_i\mathrm{\mid }$. In the free module $R^{(S)}$ of step 1.4, quotient by the submodule generated by $[i,0]$, $[i,x+y]-[i,x]-[i,y]$, and $[i,rx]-r[i,x]$. The maps $M_i\to R^{(S)}\mathrm{/}N$ induced by $x\mapsto [i,x]+N$ are linear.

A family of linear maps $M_i\to P$ extends uniquely from $S$ to the free module by step 1.4, kills the displayed generators, and therefore factors uniquely through the quotient by [F4]. This is the coproduct universal property. For an empty family, it returns the zero module.

Steps 2.1, 3.1, and 1.3 give all small coproducts and coequalizers, so [L1] gives cocompleteness.