Remark

Tag 0DY1, permanent. Cite it as [Bourbaki, Tag 0DY1].

Algebra, chapter X, § 6. Utilisation de résolutions non canoniques

Similarly, consider commutative diagrams of morphisms of complexes

\begin{array}{ccc} R_1 & \xrightarrow{a_1} & M \\ \tilde{f} \downarrow & & \downarrow f \\ R'_1 & \xrightarrow{a'_1} & M' \end{array}

\begin{array}{ccc} S_1 & \xrightarrow{b_1} & P \\ \tilde{g} \downarrow & & g \downarrow \\ S'_1 & \xrightarrow{b'_1} & P' \end{array}

\begin{array}{ccc} N' & \xrightarrow{c'_1} & E' \\ h \downarrow & & \tilde{h} \downarrow \\ N & \xrightarrow{c_1} & E_1 \end{array}

where the complexes $R_1, R'_1, S_1, S'_1$ are projective and zero on the right and where the complexes $E_1, E'_1$ are injective and zero on the left. Then

\mathrm{Tor}^A(g, f) \circ \psi'(S_1, R_1) = \psi'(S'_1, R'_1) \circ H(\tilde{g} \otimes \tilde{f})

\varphi'(R_1, E_1) \circ \mathrm{Ext}_A(f, h) = H(\mathrm{Homgr}_A(\tilde{f}, \tilde{h})) \circ \varphi'(R'_1, E'_1)

as is proved analogously to prop. 2.

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

Cited by

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Cites

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