Let $S$ be a set and $\langle S \rangle$ denote the free group generated by the set $S$.
If we take a subset $T$ of $S$ and consider the quotient group $\langle S \rangle / \operatorname{nc}(T)$, where $\operatorname{nc}(T)$ denotes the normal closure of $\langle T \rangle$ in $\langle S \rangle$, is it true, in general, that it is isomorphic to $\langle S \setminus T \rangle$?
My intuition says it should be, but I’m having a hard time actually closing in on a proof.
My first approach was to take $\pi: \langle S \rangle \to \langle S \rangle / \operatorname{nc}(T)$ to be the canonical projection, and take a reduced product $p = s_1^{\epsilon_1} ... s_n^{\epsilon_n}$. Then:
$\pi(p) = p\operatorname{nc}(T) = \pi(s_1^{\epsilon_1} ... s_n^{\epsilon_n}) = \pi(s_1^{\epsilon_1}) ... \pi(s_n^{\epsilon_n})$
Now this becomes a product of equivalence classes, where every term $\pi(s_i^{\epsilon_i})$ with $s_i \in T$ will cancel, thus yielding a product in $S \setminus T$. Since $\pi$ is surjective, we know $S \setminus T$ generates this quotient group (modulo $\operatorname{nc}(T)$).
I now have a hard time proving this a free group.
I also tried a second approach. Let’s define $f : S \to \langle S \setminus T \rangle$ by $s \mapsto s$ if $s \notin T$ and $s \mapsto 1$ if $s \in T$. This extends into a homomorphism $F : \langle S \rangle \to \langle S \setminus T \rangle$ which is surjective because of the definition of $f$.
Thus, we get $\langle S \rangle / \operatorname{ker}(F) \simeq \langle S \setminus T \rangle$, but I can’t seem to figure out if this kernel is indeed $\operatorname{nc}(T)$. This seems easier, but I am stuck nonetheless.
Am I correct in my belief? If so, how do I actually prove this result?
Thanks a lot in advance!