Category theory - Prove that $\operatorname{Hom}$ preserves representations for quasi-inverse functors 
Let $F: \mathcal C \to \mathcal D$ and $G: \mathcal D \to \mathcal C$ be quasi-inverse functors, and let $H : \mathcal C \to Set$ be a representable (contravariant) functor with representative $X \in \mathcal C$. Prove that $H \circ G$ is representable by $F(X)$.

$\DeclareMathOperator\Hom{Hom}$As ismorphisms are transitive, it suffices to consider the case when $H = \Hom( -, X)$. To this end, we wish to find $\phi : \Hom(-,X) \circ G \to \Hom(-,F(X))$ an ismorphism, from which we quickly deduce that for any $f: B \to A$ and $g: GA \to X$ it must be that $\phi_A(g) \circ f = \phi_B(g \circ Gf)$. I am not sure how to dind such a $\phi$ though. It seems like I have to somehow use the fact that $F$ and $G$ are quasi-inverses...


$\newcommand\cat\mathscr\DeclareMathOperator\id{id}$Let $F:\cat C\rightleftarrows\cat D:G$ be quasi-inverse functors.
  Then $F,G$ are fully faithful and there exists natural isomorphisms $\varepsilon:F\circ G\to\id_{\cat C}$ and $\eta:\id_{\cat D}\to G\circ F$ such that
  \begin{align}
&\eta_GG(\varepsilon)=1_G&
&F(\eta)\varepsilon_F=1_F
\end{align}

Proof.
Let $\bar\eta:\id_{\cat D}\to G\circ F$ be a natural isomorphism.
The functor $F$ is faithful, for if $u,v:A\rightrightarrows B$ and $F(u)=F(v)$ then
\begin{align}
u\bar\eta_B
&=\bar\eta_A(G\circ F)(u)\\
&=\bar\eta_A(G\circ F)(v)\\
&=v\bar\eta_B
\end{align}
which implies $u=v$. Similarly, $G$ is faithful.
The functor $F$ is full, for if $y:F(A)\to F(B)$ and $x=\bar\eta_AG(y)\bar\eta_B^{-1}$, then
\begin{align}
\bar\eta_A(G\circ F)(x)
&=x\bar\eta_B\\
&=\bar\eta_AG(y)
\end{align}
which implies $y=F(x)$ (since $G$ is faithful).
Let $\varepsilon:F\circ G\to\id_{\cat C}$ be a natural isomorphism.
Since $F$ is full and faithful, for each object $A$ in $\cat C$ there exists one and only one isomorphism $\eta_A:A\to (G\circ F)(A)$ such that $F(\eta_A)=\varepsilon_{F(A)}^{-1}$.
Then $\eta:\id_{\cat C}\to G\circ F$ is a natural isomorphism (again using faithfulness of $F $) and $F(\eta)\varepsilon_F=1_F$.
By naturalness of $\varepsilon$, we have $\varepsilon_{F\circ G}\varepsilon=(F\circ G)(\varepsilon)\varepsilon$ from which we get $\varepsilon_{F\circ G}=(F\circ G)(\varepsilon)$.
Consequently,
\begin{align}
F(\eta_GG(\varepsilon))
&=F(\eta_G)(F\circ G)(\varepsilon)\\
&=F(\eta_G)\varepsilon_{F\circ G}\\
&=1_{F\circ G}\\
&=F(1_G)
\end{align}
from which $\eta_GG(\varepsilon)=1_G$. $\square$
$\DeclareMathOperator\Hom{Hom} $For all objects $A$ of $\cat C$ we define
\begin{align}
&\varphi_A:\Hom_{\cat C}(G(A),X)\to\Hom_{\cat D}(A,F(X))&
&f\mapsto\varepsilon_A^{-1}F(f)
\end{align}
and
\begin{align}
&\psi_A:\Hom_{\cat D}(A,F(X))\to\Hom_{\cat C}(G(A),X)&
&g\mapsto G(g)\eta_X^{-1}
\end{align}
We have to show that $\varphi_A$ is natural in $A$ and $\varphi_A\circ\psi_A$ and $\psi_A\circ\varphi_A$ are identity functions.
For all $f:G(A)\to X$ we have
\begin{align}
(\psi_A\circ\varphi_A)(f)
&=G(\varepsilon_A^{-1}F(f))\eta_X^{-1}\\
&=G(\varepsilon_A)^{-1}(G\circ F)(f)\eta_X^{-1}\\
&=\eta_{G(A)}(G\circ F)(f)\eta_X^{-1}\\
&=f\eta_X\eta_X^{-1}\\
&=f
\end{align}
For all $g:A\to F(X)$ we have
\begin{align}
(\varphi_A\circ\psi_A)(g)
&=\varepsilon_A^{-1}F(G(g)\eta_X^{-1})\\
&=\varepsilon_A^{-1}(F\circ G)(g)F(\eta_X)^{-1}\\
&=\varepsilon_A^{-1}(F\circ G)(g)\varepsilon_{F(X)}\\
&=\varepsilon_A^{-1}\varepsilon_Ag\\
&=g
\end{align}
Let $u:B\to A$ be a morphism in $\cat C$.
Then naturalness of $\varphi_A$ means 
$$\require{AMScd}
\begin{CD}
\Hom_{\cat C}(G(A),X) @>\varphi_A>> \Hom_{\cat D}(A,F(X))\\
@VVV @VVV\\
\Hom_{\cat C}(G(B),X) @>>\varphi_B> \Hom_{\cat D}(B,F(X))
\end{CD}$$
For all $f:G(A)\to X$ we have
\begin{align}
\varphi_B(G(u)f)
&=\varepsilon_B^{-1}F(G(u)f)\\
&=\varepsilon_B^{-1}(F\circ G)(u)F(f)\\
&=u\varepsilon_A^{-1}F(f)\\
&=u\varphi_A(f)
\end{align}
