# Bicartesian squares of abelian groups

A commuting square is called bicartesian if it is both a pullback and a pushout. I would like to show that given any diagram of abelian groups $A \stackrel{f}{\twoheadrightarrow} B\stackrel{\beta}{\hookrightarrow} C$, I can always embed this as part of a bicartesian square:


That is, I want to show there exists an abelian group $D$, an injection $\alpha: A\hookrightarrow D$, and a surjection $g: D\twoheadrightarrow C$ that makes a commutative square as above. I also know that a square is bicartesian if and only if the induced map on the cokernels (or kernels) is an isomorphism.

Let's call $K = \ker f$ and $G = \mathrm{coker}\,\beta$. I am looking for an abelian group $D$ such that the indicated square is commutative and the following diagram has exact rows and columns:


We can pick off the following short exact sequences from the diagram:

$0 \to A \to D \to C \to 0$ EDIT: As Zhen pointed out, this sequence need not be exact

$0 \to K \to D \to C \to 0$

$0 \to A \to D \to G \to 0$

I think I need to find elements of $\mathrm{Ext}^{1}(C,K)$ and $\mathrm{Ext}^{1}(G,A)$, but I also need them to be mutually "compatible" (in whatever way that makes sense). I'm not sure how to algebraically express that the object $D$ I pick is able to simultaneously fit into all of these short exact sequences. I imagine that I have to use Mayer-Vietoris or the long exact sequence for Ext, but I don't have a good feel for how this should work. Any hints would be much appreciated.

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$0 \to A \to D \to C \to 0$ need not be exact. For example, take $A = 2 \mathbb{Z}$, $B = \mathbb{Z} / 2 \mathbb{Z}$, $C = \mathbb{Z} / 4 \mathbb{Z}$, $D = \mathbb{Z}$. –  Zhen Lin Nov 6 '12 at 8:16
Thanks Zhen! I suspected I may have overconstrained my problem. –  jmracek Nov 6 '12 at 14:55

Myself and a group of friends figured out how to proceed. Write out the long exact sequence for derived functors of $\mathrm{Hom}(G,\cdot)$ applied to the first column to get:
$\cdots \to \mathrm{Ext}^{1}(G,K) \to \mathrm{Ext}^{1}(G,A) \to \mathrm{Ext}^{1}(G,B) \to 0$
The sequence terminates because $\mathbb{Z}$ is a PID so $\mathrm{Ext}^{n}(G,H) = 0$ for any abelian groups $G$ and $H$ when $n > 1$. Now $C \in \mathrm{Ext}^{1}(G,B)$, so since the last map is a surjection it came from an extension $D \in \mathrm{Ext}^{1}(G,A)$. The extension $D$ also comes with the injection $\alpha$ we need, as well as a map between extensions $g: D\to C$. To show that $G$ is a surjection we can just use the snake lemma, so $\mathrm{coker}\,g = 0$ since $f$ is a surjection and the map from $G$ to $G$ at the end is an isomorphism.