## § Weird free group construction from adjoint functor theorem

- We wish to construct the free group on a set $S$. Call the free group $\Gamma S$.
- Call the forgetful functor from groups to sets as $U$.
- The defining property of the free group is that if we are given a mapping $\phi: S \to UG$, a map which tells us where the generators go, there is a unique map $\Gamma \phi: \Gamma S \to G$ which maps the generators of the free group via a group homomorphism into $G$. Further, there is a bijection between $\phi$ and $\Gamma \phi$.
- Written differently, there is a bijection $\hom_\texttt{Set}(S, UG) \simeq \hom_\texttt{Group}(\Gamma S, G)$. This is the condition for an adjunction.
- The idea to construct $\Gamma S$ is roughly, to take all possible maps $f_i: S \to UG$ for all groups $G$, take the product of all such maps, and define $\Gamma S \equiv im(\pi_i f_i)$. The details follow.
- First off, we can't take all groups, that's too large. So we need to cut down the size somehow. We do this by considering groups with at most $|S|$ generators, since that's all the image of the maps $f_i$ can be anyway. We're only interested in the image at the end, so we can cut down the groups we consider to be set-sized.
- Next, we need to somehow control for isomorphisms. So we first take
*isomorphism classes * of groups with at most $|S|$ generators. Call this set of groups $\mathcal G$We then construct all possible maps $f_i: S \to UG$ for all possible maps $f$, for all possible $G \in \mathcal G$. - This lets us construct the product map $f : S \to \prod_{G \in \mathcal G} UG$ given by $f(s) \equiv \prod_{G \in \mathcal G} f_i(s)$.
- Now we define the free group $\gamma S \equiv im(f)$. Why does this work?
- Well, we check the universal property. Suppose we have some map $h: S \to UH$. This must induce a map $\Gamma h: \Gamma S \to H$.
- We can cut down the map, by writing the map as $h_{im}: S \to im(h)$. This maps into some subset of $UH$, from which we can generate a group $H_{im} \subseteq H$.
- First off, there must be some index $k$ such that $f_k = h_{im}$, since the set of maps $\{ f_i \}$ covers all possible maps from $S$into groups with those many generators.
- This implies we can project the group $\Gamma S$ at the $k$th index to get a map from $\Gamma S$ into $H_{im}$.
- We can then inject $H_{im}$ into $H$, giving us the desired map!