The free-group functor $F:\mathbf{Set}\to\mathbf{Grp}$ and free-module functor $R^{(-)}:\mathbf{Set}\to R\text{-}\mathbf{Mod}$

For a function $f:X\to Y$, the composite $X\xrightarrow fY\to F(Y)$ extends uniquely by [L2] to a homomorphism $F(f):F(X)\to F(Y)$.

Construct $R^{(X)}$ explicitly, since [L3] says only what it means for a module to be free and does not build one: let $R^{(X)}$ be the set of functions $a:X\to R$ whose support $\operatorname{supp}(a)=\{x:a_x\ne0\}$ is finite, with pointwise addition and scalar multiplication. Both operations preserve finite support because $\operatorname{supp}(a+b)\subseteq\operatorname{supp}(a)\cup\operatorname{supp}(b)$ and $\operatorname{supp}(ra)\subseteq\operatorname{supp}(a)$, so $R^{(X)}$ is a left $R$-module, and the family $e_x$ with $e_x(x)=1$ and $e_x=0$ elsewhere is a basis: every $a$ is the finite sum $\sum_{x\in\operatorname{supp}(a)}a_xe_x$, and a vanishing finite combination has every coefficient zero by evaluating at each index. So $R^{(X)}$ is free in the sense of [L3]. Now $R^{(f)}$ sends $a$ to the family $y\mapsto\sum_{x\in f^{-1}(y)\cap\operatorname{supp}(a)}a_x$; the index set is finite because it lies in $\operatorname{supp}(a)$, which is what [L4] requires, whereas $f^{-1}(y)$ itself may be infinite. The result again has finite support, contained in $f[\operatorname{supp}(a)]$, and $R^{(f)}$ is additive and $R$-linear because each coefficient is a finite sum of the corresponding coefficients of $a$. On basis elements it sends $e_x$ to $e_{f(x)}$.

Both maps assigned to $1_X$ fix every generator. The uniqueness of the free extensions therefore gives $F(1_X)=1_{F(X)}$ and $R^{(1_X)}=1_{R^{(X)}}$.

For $X\xrightarrow fY\xrightarrow gZ$, the maps $F(gf)$ and $F(g)F(f)$ agree on every generator. The module maps $R^{(gf)}$ and $R^{(g)}R^{(f)}$ likewise send $e_x$ to $e_{g(f(x))}$; finite-sum reindexing gives the same equality in coefficient form.

Hence $X\mapsto F(X)$ and $X\mapsto R^{(X)}$, with the maps above, define functors $\mathbf{Set}\to\mathbf{Grp}$ and $\mathbf{Set}\to R\text{-}\mathbf{Mod}$.