Remark 3

Tag 0DD4, permanent. Cite it as [Bourbaki, Tag 0DD4].

Let

(\mathscr{E})

0 \to M' \to M \to M'' \to 0

be an exact sequence of $A$ -modules. By Remark 2 of No. 2 and that of § 8, No. 7, the following diagrams are commutative:

\begin{array}{ccc} H_A^{i-1}(M'') & \xrightarrow{\gamma^{i-1}(M'')} & D(\mathrm{Ext}_A^{d-i+1}(M'', \Omega)) \\ \downarrow & & \downarrow (-1)^{d-i+1}D(\delta^d i(\mathscr{E}, \Omega)) \\ H_A^i(M') & \xrightarrow{\gamma^i(M')} & D(\mathrm{Ext}_A^{d-i}(M', \Omega)) \end{array}

\begin{array}{ccc} \mathrm{Ext}_A^{d-i}(M', \widehat{\Omega}) & \xrightarrow{\delta^i(M')} & D(H_A^i(M')) \\ \downarrow & & \downarrow (-1)^{d-i+1}D(\partial^{i-1}(\mathscr{E})) \\ \mathrm{Ext}_A^{d-i+1}(M'', \widehat{\Omega}) & \xrightarrow{\delta^{i-1}(M'')} & D(H_A^{i-1}(M'')) \end{array}

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

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