Free group over a quotient of sets

Question

Given a set $S$, let $F_S$ denote the free group on the set $S$ (not the free abelian group). Let $T \subseteq S$ be a subset. Consider the quotient set $S/T$. Is it true that $F_{S/T} \simeq F_S/F_T$ ?

Answer 1

According to the comment, something similar is true:

Let $T\subseteq S$ and consider the set difference $S\setminus T$, moreover let $(F_T)$ denote the normal subgroup generated by $F_T$ within the free group $F_S$. Then, $$F_S / (F_T) \cong F_{S\setminus T} \ . $$

Proof. Consider the mapping $\varphi_0:S\to F_S$ which maps all $t\in T$ to the unit element $e$ (empty word), but fixes all $s\in S\setminus T$. Since $F_S$ is free group over $S$, any function $S\to G$ to a group extends uniquely to a group homomorphism on $F_S$. In particular $\varphi_0$ induces a homomorphism $\varphi:F_S\to F_S$.

Now, the image of $\varphi$ is easily seen to be $F_{S\setminus T}$ (as embedded in $F_S$), and the kernel of $\varphi$ is going to be $(F_T)$, the collection of words in which, after 'deleting' all letters $t\in T$, what remains is the unit.

Answer 2

The functor $F : \mathsf{Set} \to \mathsf{Grp}$ is left adjoint, therefore preserves all colimits. If $T \subseteq S$ is a subset, and $T$ is non-empty, say $t_0 \in T$, then usually $S/T$ means the coequalizer of the two maps $T \rightrightarrows S$, where the first one is the inclusion and the second one is constant with value some $t_0 \in T$. It follows that $F(S/T)$ is the coequalizer of the two homomorphisms $F(T) \rightrightarrows F(S)$. Therefore, $F(S/T)=F(S)/N$, where $N$ is the normal subgroup generated by $\{t t_0^{-1} : t \in T\}$, or equivalently by $\{t t'^{-1} : t,t' \in T\}$.