Let $G$ be a group with a presentation $\langle S | R \rangle $ ( $G = \langle S|R\rangle$), with an associated map $f\, : S \rightarrow G$ Let $\bar{N}_{R}$ be the normal closure of $R$ in $F_{S}$.
Show that if $H$ is a group, and $h:S \rightarrow H$ a map which has an associated group morphism $\hat{h} \, : F_{S} \rightarrow H$ sending all elements of $R$ to $e_{H}$, then there exists a unique group morphism $g \,:G\rightarrow H$ such that $h = g \circ f$.
Attempt at Solution:
(1) Since $G$ is a presentation we know $G \cong F_{S}/\ \bar{N_{R}}$. Considering the situation for $H$
$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
\lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& H\\
& \diaguparrow{h} @AA \exists! \,\hat{h} A \\
S @>>\iota> F_{S}
\end{CD}$$
The following diagram commutes. Since $ker(\hat{h}) \lhd F_{S}$ then $\bar{N}_{R} \lhd ker(\hat{h})$. $H \supseteq Im(H) \cong F_{S} /\ ker(\psi) \subset F_{S} /\ \bar{N}_{S} = G $. Thus $G \subseteq H$
So we are then left with the fact that we have a map $ G \hookrightarrow H$, which gives us then that there is a unique $ \hat{h}' : F_{S}\rightarrow H$ such that $\hat{h'} = (i \circ f) \circ \iota $, where $\iota$ is the inclusion $ \iota\, : S \rightarrow F_{S}$ and $i$ the inclusion $ i \, : G \hookrightarrow H$. But we then want to show that $i \circ f$ is $h$ and I'm not sure how I can do this.
I strongly suspect I've overcomplicated things - is there a simpler way to deal with this?
I think you have a map turned around somewhere. In particular, you should get just a map $G \to H$, there is no guarantee that it is an inclusion. For instance, taking $G$ to be a free group on $n$ letters and $H = \mathbb{Z}^n$.
For the existence, we have to get our hands a little dirtier than you have. The existence of the surjection $\hat f\colon F_S \to G$ and homomorphism $\hat h\colon F_S \to H$ allows us to thing of group elements in $G$ as equivalence classes of words in $F_S$, and similarly for elements of $H$ in the image of $\hat h$. I will write $K$ for the kernel of $\hat h$, and $R$ for your $\bar N_R$.
I claim the map $g$ is given by $wR \mapsto wK$, where $wR$ is an element of $G$ and $w$ is a word in $F_S$ with $\hat f(w) = wR$. Firstly, this is well-defined. If $wR = w'R$, then $w^{-1}w' \in R$, so $\hat{h}(w^{-1}w') = e_H$, and thus $$wK = wK e_H = wK(w^{-1}w')K = w'K.$$
Moreover, the map $g$ is a homomorphism. I think this makes a cute exercise; I'll leave it to you.
Now that we've shown $g$ exists, we just need to check that $h = g\circ f$ as maps $S \to H$. I'll also leave this to you. But this means that $\hat h = g\circ \hat f$ as maps $F_S \to H$, so uniqueness of $\hat h$ tells us if there were another map $g'$ satisfying these conditions, we must have $g = g'$.