Given an adjunction $$ , $G$ is faithful $G$ is faithful if and only if $\varphi^{-1}$ carries epis to epis I have the next doubt about the solution of this problem
Given an adjunction $<F,G,\varphi>:X\rightarrow A$ Prove that $G$ is faithful if and only if $\varphi^{-1}$ carries epimorphism to epimorphism.
$(\Rightarrow)$ Let it be $f:x\rightarrow Ga$ an epimorphism we want to show that $\varphi^{-1}(f)$ is and epimorphism, so let it be $g_{1}, g_{1}: a\rightarrow b$ morphisms such that:
$g_{1}\circ \varphi^{-1}(f)=g_{2}\circ \varphi^{-1}(f)$ 
by definition of $\varphi^{-1}(f)$ we have:
$g_{1}\circ \epsilon_{a}\circ F(f)=g_{2}\circ \epsilon_{a}\circ F(f)$
we have that $g_{1}\circ \epsilon_{a}\circ F(f):F(x)\rightarrow b$ if we apply $\varphi$ to $g_{1}\circ \epsilon_{a}\circ F(f)$ by definition of $\varphi$ we have
$\begin{eqnarray}
\varphi(g_{1}\circ \epsilon_{a}\circ F(f))&=&G(g_{1}\circ \epsilon_{a}\circ F(f))\circ \eta_{x}\\
&=&G(g_{1})\circ G(\epsilon_{a})\circ GF(f)\circ \eta_{x}\\
&=&G(g_{1})\circ G(\epsilon_{a})\circ \eta_{Ga}\circ f\\
&=&G(g_{1})\circ id\circ f\\
&=&G(g_{1})\circ f
\end{eqnarray}$
The third line  by naturality of $\eta$ it follows  and the fourth one by the triangular identity. In the same way for $g_{2}\circ \epsilon_{a}\circ F(f)$ we will have the same. Therefore
$G(g_{1})\circ f=G(g_{2})\circ f$
Because $f$ is an epimorphism and $G$ is faithful, it follows that $g_{1}=g_{2}$, therefore $\varphi^{-1}$ carries epimorphism to epimorphism.
I hope the idea is correct for this part, for  the $(\Leftarrow)$ part i do not know how to use the hipothesis because if $G(g_{1})=G(g_{1})$ for morphisms $g_{1}, g_{2}:a\rightarrow b$ the only thing i have is if we apply $\varphi^{-1}$ an we use the naturality of $\epsilon$, i will have:
$g_{1}\circ \epsilon_{a}=g_{2}\circ \epsilon_{a}$
Thank you for your time.
 A: Consider the composition $[a,b]\overset{G}{\rightarrow}[Ga,Gb]\overset{\varphi ^{-1}}{\rightarrow}[FGa,b]$. 
By Yoneda, it is determined by $(\varphi ^{-1}\circ G)(id _a)=\varepsilon _a$ and since, $\varphi ^{-1}$ carries epis to epis, and $G(id_a)$ is epic, $\varepsilon _a$ is an epi. But then, using the fact that
$f:b\rightarrow a$ is monic (epic) $\Leftrightarrow f^{*}: [a,-]\rightarrow [b,-]$ is epic (monic), we have 
$\varphi ^{-1}\circ G$ is a monic, and this only happens when $G$ is faithful.
Remark: we can use this idea to get a fast proof of the forward direction: note that since $F$ is a left adjoint, is preserves epimorphisms (because it preserves colimits and hence pushouts). Now if $G$ is faithful, then $\varphi ^{-1}\circ G$ is injective and so by the above argument, $\varepsilon _a$ is epi. Then the result is immediate since if $f$ is epi, then $$g_{1}\circ \varphi^{-1}(f)=g_{2}\circ \varphi^{-1}(f)\Rightarrow g_{1}\circ \epsilon_{a}\circ F(f)=g_{2}\circ \epsilon_{a}\circ F(f)\Rightarrow g_1\circ \varepsilon _a=g_2\circ \varepsilon _a\Rightarrow g_1=g_2$$.
