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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?

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up vote 1 down vote accepted

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.

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An even simpler example would be to take $Y=1$ and $X = Z$ any set with at least two elements. Then the pullback is the product $X \times Z$. Taking $S$ to be $X$ with morphisms to $X$ and $Z$ identities the unique map is the diagonal map, which is not an epi. –  Aleš Bizjak Dec 1 '13 at 16:50
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