I am stuck on a part of an exercise in Vakil's Algebraic Geometry notes.
We define groupification of a semgroup $S$ to be a map of abelian semigroups $p:S\rightarrow G$ such that any map of abelian semigroups from $S$ to an abelian group $G'$ factors uniquely through $G$.
Now, there is a groupification functor $H$ from the category of abelian semigroups to the category of abelian groups. The construction is given by $H(S)=(S\times S)/\sim$,where $(a,b)\sim(c,d)$ if and only if there exists an $e\in S$ such that $a+d+c=b+c+e$. Addition is given componentwise. The inverse of $[(a,b)]$ is $[(b,a)]$, the identity is $[(0,0)]$ and the map $p:S\rightarrow (S\times S)/\sim$ is given by $p(s)=[(s,0])$, which yields the groupification of $S$.
Let $F$ be the forgetful functor from the category of abelian groups to semigroups. My question is why is $H$ left adjoint to $F$? Specifically, why is $\text{Hom}(H(S),G)\cong \text{Hom}(S,F(G))$, where $G$ is a group and $S$ is a semigroup? I am not sure how to even construct a map between the two. Since $F(G)$ is only a semigroup, I can't use the universal property of groupification on the map from $S$ to $F(G)$.
Any map of abelian semigroups from $S$ to an abelian group $G$ is formally a morphism of semigroups $S \rightarrow F(G)$, where $F:\mathsf{Ab}\rightarrow\mathsf{AbSgrp}$ is the forgetful functor. To consider $G$ as an abelian semigroup, you have to forget that every element has an inverse...