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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:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} A & \ra{\alpha} & D \\ \da{f} & &\da{g} \\ B & \ra{\beta} & C \end{array} $$

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:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} & &0 & &0 \\ & &\da{} & &\da{} \\ & & K & \ra{\simeq} & K \\ & & \da{k} & & \da{}\\ 0 & \ra{}& A & \ra{\alpha} & D &\ra{} &G& \ra{} & 0\\ & & \da{f} & &\da{g} & &\da{\simeq}\\ 0& \ra{}& B & \ra{\beta} & C & \ra{}& G & \ra{} & 0 \\ & &\da{} & & \da{} \\ & & 0& & 0 \end{array} $$

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

1 Answer 1

up vote 1 down vote accepted

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.

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