Free group construction via General Adjoint Functor Theorem in Riehl

• Question 1.

For each $G$, you have a map $f_G\colon S\to U(G)$, the underlying set of $G$, given by sending $S$ ot a generating set of $G$ (you "really" have one such map for each $G$ and each generating set and each set function from $S$ to the generating set).

Because you have a map from $S$ to each $UG$, the universal property of the product (in $\mathsf{Set}$) gives you a unique map from $S$ to $\prod_{G\in\Phi}UG$. But the product of the underlying sets is the same as the underlying set of the product, so this is the same as a map from $S$ to $U(\prod_{G\in\Phi}G)$. This maps sends $s$ in $S$ to the element whose $G$-component is $f_G(s)$. This is the map $\hat{\eta}$.

• Question 2.

The universal property of the free group on $S$, $\Gamma$, is that for every group $G$, if you have a map $S\to U(G)$, this gives a unique map from $\Gamma$ to $G$ that suitably "extends" the map $S\to UG$. If you write down what the universal property of the initial object of $S\downarrow U$, you will see that it is exactly the universal property of the free group on $S$. The fact that $\eta$ is initial in the suitable category yields precisely that $\Gamma$ has the universal property of a free group, hence is free.

• Question 3.

You need to compose with the inclusion $G\hookrightarrow H$. This is usually omitted, since it is just the fact that you are mapping to a subgroup of $H$; we don't usually think of, for example, the map $C_2\to S_3$ mapping onto $\{e,(12)\}$ as consisting of a map into the group $\{e,(12)\}$ followed by the inclusion of that group into $S_3$: we just think of the map being a map into $S_3$. The same is being done here: you think of it as mapping to $G$ to get the factorization, and then formally compose it with the inclusion $G\to H$ to "view it" as a map into $H$.