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The subject line says it all, but perhaps it would be more reasonable to split the question into two parts: 1) can a pullback diagram also be a pushout diagram?; if so, 2) can necessary and sufficient conditions be given for this to happen?

Thanks!

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1) Sure, a square that is both a pull-back (cartesian) and a push-out (cocartesian) is called a bicartesian square (Freyd called them push-me pull-you first, then later Doolittle diagrams...). 2) I don't think that there is a general characterization without further assumptions on the morphisms and the surrounding category. For example, in additive categories you can show that the push-out under a kernel is also a pull-back if and only if the push-out of the kernel is a monic. –  t.b. Nov 12 '11 at 13:08
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For the first question see also en.wikipedia.org/wiki/Pulation_square (and references give there.) –  Martin Sleziak Nov 12 '11 at 13:14
    
I know this question is old and you are asking about the direction "pullback$\implies$pushout", but I thought to myself I just comment here what I found out recently: $\require{AMScd}$ If the square $$\small\begin{CD} A @>i>>X \\ @VVV@VVV \\ Y@>j>>Z \end{CD}$$ is a pushout and $i$ is injective, then $j$ is injective and the square is a pullback. This also holds in the category of weak Hausdorff $k$-spaces: If $i$ is a closed embedding, then so is $j$, the pushout is constructed as in the category of sets, and the square is also a pullback. –  Stefan Hamcke Sep 13 at 16:54

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