The free group on $X$ represents $G\mapsto \mathbf{S}\mathbf{e}\mathbf{t}(X,U(G))$

By [F1], restriction along $i$ is a bijection $\mathbf{G}\mathbf{r}\mathbf{p}(F(X),G)\to \mathbf{S}\mathbf{e}\mathbf{t}(X,U(G))$, with inverse $f\mapsto \widehat{f}$.

Now let $X=\{x\}$. By [L1], every reduced word is either empty, a string of $n\ge 1$ copies of $x$, or a string of $n\ge 1$ copies of $x^{-1}$: a reduced word containing both signs would have an adjacent sign change and hence a cancellable pair. Thus every element is an integer power of $x$, so [F3] makes $F(X)$ cyclic.

If $r:G\to H$ is a group homomorphism, then $r\circ \widehat{f}$ and $\widehat{U(r)\circ f}$ are homomorphisms $F(X)\to H$ with the same restriction to $X$; uniqueness in [F1] makes them equal. Thus the bijections of step 1.1 are natural in $G$.

Steps 1.1 and 2.1, together with [F4], show that $F(X)$ represents the stated functor. By [L2], changing the chosen free-group model changes this representation by the unique generator-compatible isomorphism.

The positive words $x,xx,xxx,\dots$ have different finite lengths and are distinct reduced normal forms by [L1], so $F(X)$ has infinitely many elements. Hence the singleton free group is infinite cyclic.