Question on an induced map into the pullback of epimorphisms Let
\begin{eqnarray}
W&\to& Z\\
\downarrow &&\downarrow\\
X&\to&Y
\end{eqnarray}
be a pullback diagram in a category $C$ where all the involved morphisms are epimorphisms. Let $S$ be an object of $C$ and $S\to Z$, $S\to X$ two epimorphisms making the outer diagram commute.
Does the induced map $s:S\to W$ have to be an epimorphism?

Consider the category $\mathbf{Set}$ of sets and functions. Take $A = \{x, y\}$, $X = Z = A + A$ (disjoint union), $Y = A$ and 
$$
f(0, a) = a\\
f(1, a) = x
$$
as a function $f : A + A \to A$ and take the kernel pair of $f$, i.e. pullback of $f$ along itself.
Then $W = \{((0, a), (0, a))\ |\ a \in A\} \cup \{((1, a), (1, a'))\ |\ a, a' \in A\}$ and the projections $W \to X$ and $W \to Z$ are surjections.
Take $S = A + A$ and the morphisms $S \to X$ and $S \to Z$ to be identities. Then the unique morphism into $W$ is given by $s \mapsto (s, s)$ and the pair $((1, x), (1, y))$ is not in its image, hence it is not an epi.
The property therefore does not hold in general.
