I'm looking here, and I don't see an example like this: $$\begin{eqnarray} \zeta_1 & \xrightarrow{f} \omega_1 & \xrightarrow{g} \epsilon_1 \\ \zeta_2 & \rightarrow \omega_2 & \rightarrow \epsilon_2 \\ \zeta_3 \end{eqnarray}$$
In particular, I'm asked to give an example where $g$ is surjective but $g\circ f$ is not, and it would seem that if $g$ is surjective, then $g\circ f$ is necessarily surjective. How would I prove this?
So, Mr. Thomas Andrews seems to be saying construct something like this: $$\begin{eqnarray} \cdot \rightarrow \cdot \rightarrow \cdot \\ \cdot \nearrow {\color{red}\cdot} \rightarrow \cdot \\ \cdot \rightarrow \cdot \rightarrow \cdot \end{eqnarray}$$
Let's take a set $S=\{0,1\}$ and study functions $f,g:S\to S$, $f(x) =0$, $g(x)=x$. $g$ is surjective, $f$ is not. $g(f(x))=0$, so $g\circ f$ is not surjective.