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 $\mathop{\mathrm{Ext}}\nolimits ^1(Z', A)$.
$\square$

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