Is this functor representable? Fix a group $G_0$ and $R$ a subset of $G_0$. Consider the functor $F$ from $\textbf{Grps}$ to $\textbf{Sets}$, sending every object $G$ in $\textbf{Grps}$ to $F(G)$, the subset of $\varphi \in \text{Hom}(G_0, G)$ such that $\varphi(r) = 1$ for every $r \in R$, and sending each homomorphism $f: G \to G'$ to the map $F(f) : \varphi \mapsto f \circ \varphi$.
Question. Is this functor representable?
 A: Yes, it is representable, consider $G_R$ the normal subgroup of $G_0$ generated by $R$, $F(G)=Hom(G_0/G_R,G)$
A: Let $G_1$ be the normal closure of $R$, the smallest normal subgroup of $G_0$ containing $R$. Since the kernel of a homomorphism is a normal subgroup, any element of $F(G)$ contains $G_1$ in its kernel. Conversely, any homomorphism from $G_0$ to $G$ whose kernel contains $G_1$ is obviously an element of $F(G)$. Hence,$$F(G) = \{\varphi \in \text{Hom}(G_0,G) : \text{Ker}\,\varphi \supset G_1\}.$$But there is a natural identification of the latter set with $\text{Hom}(G_0/G_1, G)$, which we denote by $\varphi \mapsto \overline{\varphi}$.  Thus, we obtain a natural identification of $F(G)$ with $\text{Hom}(G_0/G_1, G)$. Then for any two groups $G$ and $G'$ and any $f \in \text{Hom}(G, G')$, we can easily verify commutativity of the following diagram.
$$\require{AMScd}
\begin{CD}
F(G) @>\varphi \mapsto f \circ \varphi >>  F(G')\\
@V\varphi \mapsto \overline{\varphi} VV     @VV \psi \mapsto \overline{\psi} V   \\
\text{Hom}(G_0/G_1, G) @> \alpha \mapsto f \circ \alpha >> \text{Hom}(G_0/G_1, G')
\end{CD}
$$
Hence, $F$ is naturally isomorphic to $\text{Hom}(G_0/G_1, -)$, or equivalently, $F$ is represented by $G_0/G_1$.
