left adjoints preserve pushouts I need to show that left adjoints preserve epimorphisms.
I want to follow the following idea: $f: A\longrightarrow B$ is epic if and only if the diagram:
$$
\begin{array}{ccc}
A & \overset{f}\longrightarrow & B \\
{\scriptstyle{f}}{\downarrow} & & {\downarrow}\scriptstyle{1_B} \\
B & \underset{1_B}{\longrightarrow} & B
\end{array}
$$
is a pushout. So I suppose that $A,B\in \mathcal{C}$, $f$ in $\mathcal{C}$ and that the diagram above is a pushout. I consider the functors $F:\mathcal{C}\longrightarrow\mathcal{D}$ and $G:\mathcal{D}\longrightarrow\mathcal{C}$ such that $F\dashv G$ and I want to show that the diagram
$$
\begin{array}{ccc}
F(A) & \overset{F(f)}\longrightarrow & F(B) \\
{\scriptstyle{F(f)}}{\downarrow} & & {\downarrow}\scriptstyle{F(1_B)} \\
F(B) & \underset{F(1_B)}{\longrightarrow} & F(B)
\end{array}
$$
is a pushout. Hence the conclusion will follow.
However, I don't think this way of reasoning is totally fine: it turns out that I don't use the hypothesis of $F$ being left adjoint to $G$ in the proof that the second diagram above is a pushout. I know it is really simple and basic, but I feel like I am not seeing something. Can anyone please help me?
(For the record, this is the sketch of my proof that the second diagram is a pushout. For sure it commutes ($F$ respects composition, being a functor). I consider an object $Y\in \mathcal{D}$ and maps such that the diagram
$$
\begin{array}{ccc}
F(A) & \overset{F(f)}\longrightarrow & F(B) \\
{\scriptstyle{F(f)}}{\downarrow} & & {\downarrow}\scriptstyle{y'} \\
F(B) & \underset{y}{\longrightarrow} & Y
\end{array}
$$
commutes. I need to find a unique $\overline{y}:F(B)\longrightarrow Y$ such that $\overline{y}\circ 1_{F(B)}=y'$. Take $\overline{y}\equiv y'=y$.)

That the image diagram is a pushout follows from the general fact that left adjoints preserve all colimits.
Assuming you haven't proven that yet, you can compute
$$ \begin{align*} \mathcal{D}(F(\operatorname{colim} X_j), Y) 
&\cong \mathcal{C}(\operatorname{colim} X_j, GY) 
\\&\cong \lim \mathcal{C}(X_j, GY) 
\\&\cong \lim \mathcal{D}(FX_j, Y)
\end{align*} $$
which shows that $F(\operatorname{colim} X_j) \cong \operatorname{colim} F(X_j)$.
