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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.