Remarque

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

Algebra, chapter X, § 2. Complexes de A-modules

Considérons le diagramme

\begin{array}{ccccccccc} \cdots & \longrightarrow & H_n(C') & \xrightarrow{H_n(\tilde{u})} & H_n(\mathrm{Cyl}\,(u)) & \xrightarrow{H_n(\pi)} & H_n(\mathrm{Con}\,(u)) & \xrightarrow{\partial_n(\tilde{u},\pi)} & H_{n-1}(C') \longrightarrow \cdots \\ & & \downarrow 1 & & \downarrow H_n(\beta) & & \downarrow 1 & & \downarrow 1 \\ \cdots & \longrightarrow & H_n(C') & \xrightarrow{H_n(u)} & H_n(C) & \xrightarrow{H_n(\pi)} & H_n(\mathrm{Con}\,(u)) & \xrightarrow{H_n(\delta)} & H_{n-1}(C') \longrightarrow \cdots \end{array}

où la première ligne (resp. la seconde) est la suite exacte d'homologie associée à la suite exacte (6) (resp. (7)). Les applications $H_n(\beta)$ sont bijectives (prop. 7, c)) et le diagramme est commutatif, puisque a) $u = \beta \circ \tilde{u}$ (prop. 7, a)) donc $H_n(u) = H_n(\beta) \circ H_n(\tilde{u})$ , b) $H_n(\beta) = H_n(\alpha)^{-1}$ et $\pi = \tilde{\pi} \circ \alpha$ (prop. 7, a) et c)), donc $H_n(\tilde{\pi}) = H_n(\pi) \circ H_n(\beta)$ , c) $H_n(\delta) = \partial_n(\tilde{u},\tilde{\pi})$ (lemme 3, b)).

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

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