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

Comment #7367 by Shizhang on

Maybe the vertical arrows should all go downward?

Comment #8784 by on

Typo: In the proof, last sentence, we didn't define what $W'$ is. It should be $\operatorname{Ext}^1(Z',A)$ instead of $\operatorname{Ext}^1(W',A)$.

A notational issue: if we denote $[Z']:C\to B[1]$ and $[Z]:B\to A[1]$ as explained after Definition 13.27.4, shouldn't we write $[Z][1]\circ [Z']$ instead of $[Z]\circ [Z']$ in the statement and the proof? (Alternatively, $[Z][-1]\circ [Z'][-2]:C[-2]\to A$, by the isomorphism $\operatorname{Hom}_{D(\mathcal{A})}(C,A[2])=\operatorname{Hom}_{D(\mathcal{A})}(C[-2],A)$).

Comment #9309 by on

@#9295 Anytime :)

In the commit you linked there's a typo: 'Yoneday' instead of 'Yoneda.'

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