Corollary

Tag 0DZV, permanent. Cite it as [Bourbaki, Tag 0DZV].

Algebra, chapter X, § 7. Produit de composition

Let $0 \to N \xrightarrow{f} R \xrightarrow{g} M \to 0$ and $0 \to N \xrightarrow{f'} R' \xrightarrow{g'} M \to 0$ be two exact sequences, $\theta$ and $\theta'$ their associated classes in $\mathrm{Ext}^1(M, N)$ . For $\theta = \theta'$ to hold, it is necessary and sufficient that there exists an A-homomorphism $h : R \to R'$ making the diagram

\begin{array}{ccc} & & R \\ N & \xrightarrow{f} & R \\ & \downarrow h & \downarrow g \\ & & M \\ & \xrightarrow{f'} & R' \\ & & \xrightarrow{g'} \end{array}

commutative. Such a homomorphism is necessarily an isomorphism.

The condition is sufficient according to cor. 1 of prop. 4. If $\theta = \theta'$ , there exists a commutative diagram with exact rows:

\begin{array}{ccccccccc} 0 & \longrightarrow & N & \longrightarrow & R & \longrightarrow & M & \longrightarrow & 0 \\ & & ^{1_N} & & ^{h'} & & ^{1_M} & & \\ 0 & \longrightarrow & N & \longrightarrow & R'' & \longrightarrow & M & \longrightarrow & 0 \\ & & ^{1_N} & & ^{h''} & & ^{1_M} & & \\ 0 & \longrightarrow & N & \longrightarrow & R' & \longrightarrow & M & \longrightarrow & 0 . \end{array}

The homomorphisms $h'$ and $h''$ are isomorphisms according to X, p. 7, cor. 3, and $h = h'' \circ {h'}^{-1}$ answers the question. The last assertion follows from loc. cit.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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