Lemma 13.27.7. Let $\mathcal{A}$ be an abelian category. Let $0 \to A \to Z \to B \to 0$ and $0 \to B \to Z' \to C \to 0$ be short exact sequences in $\mathcal{A}$. Denote $[Z] \in \mathop{\mathrm{Ext}}\nolimits ^1(B, A)$ and $[Z'] \in \mathop{\mathrm{Ext}}\nolimits ^1(C, B)$ their classes. Then $[Z] \circ [Z'] \in \mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {A}(C, A)$ is $0$ if and only if there exists a commutative diagram

$\xymatrix{ & & 0 \ar[d] & 0 \ar[d] \\ 0 \ar[r] & A \ar[r] \ar[d]^1 & Z \ar[r] \ar[d] & B \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A \ar[r] & W \ar[r] \ar[d] & Z' \ar[r] \ar[d] & 0 \\ & & C \ar[r]^1 \ar[d]& C \ar[d]\\ & & 0 & 0 }$

with exact rows and columns in $\mathcal{A}$.

Proof. Omitted. Hints: You can argue this using the result of Lemma 13.27.5 and working out what it means for a $2$-extension class to be zero. Or you can use that if $[Z] \circ [Z'] \in \mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {A}(C, A)$ is zero, then by the long exact cohomology sequence of $\mathop{\mathrm{Ext}}\nolimits$ the element $[Z] \in \mathop{\mathrm{Ext}}\nolimits ^1(B, A)$ is the image of some element in $\mathop{\mathrm{Ext}}\nolimits ^1(W', A)$. $\square$

Comment #7367 by Shizhang on

Maybe the vertical arrows should all go downward?

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).