# When is a pullback also a pushout?

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!

-
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
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 '14 at 16:54
@Stephan, How can I prove this. Can you take a look at this question here – qartal Nov 6 '15 at 20:49

## 1 Answer

$\require{amsCD}$ Yes. The most familiar setting where this happens is in abelian categories, where the commutative square

$$\begin{CD} A @>f_b>> B\\ @VVf_cV @VVg_bV\\ C @>g_c>> D \end{CD}$$

is a pullback square iff the corresponding sequence

$$0 \to A \xrightarrow{f_b \oplus f_c} B \oplus C \xrightarrow{g_c - g_b} D$$

is exact. Similarly, the square is a pushout square iff the sequence

$$A \xrightarrow{f_b \oplus f_c} B \oplus C \xrightarrow{g_c - g_b} D \to 0$$

is exact. Hence the square is both a pullback and a pushout square iff the sequence

$$0 \to A \xrightarrow{f_b \oplus f_c} B \oplus C \xrightarrow{g_c - g_b} D \to 0$$

is exact.

-
What about Top? – Ivan Jan 7 '15 at 23:14
@Ivan: things already don't work out very well in $\text{Set}$. Looking at the cardinalities involved in a pushout vs. pullback diagram of finite sets you'll see that there are hardly any examples. – Qiaochu Yuan Jan 8 '15 at 3:30