Proposition 10

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

Let us denote by $(\mathscr F^\circ)$ the exact sequence of left $A$ -modules

0\longrightarrow M'\xrightarrow{r}M\xrightarrow{s}M''\longrightarrow0.

The diagram

\begin{array}{ccccc} \operatorname{Tor}^A(M'',N)&\xrightarrow{\partial(\mathscr F,N)}&\operatorname{Tor}^A(M',N)\\ \Big\downarrow{\sigma_{M'',N}}&&\Big\downarrow{\sigma_{M',N}}\\ \operatorname{Tor}^{A^\circ}(N^\circ,{M''}^\circ)&\xrightarrow{\partial(N^\circ,\mathscr F^\circ)}&\operatorname{Tor}^{A^\circ}(N^\circ,{M'}^\circ) \end{array}

is commutative.

Indeed, this follows from X, p. 31, prop. 2, applied to the commutative diagram

\begin{array}{ccccccccc} 0&\longrightarrow&M'\otimes_A L(N)&\xrightarrow{r\otimes1}&M\otimes_A L(N)&\xrightarrow{s\otimes1}&M''\otimes_A L(N)&\longrightarrow&0\\ &&\Big\downarrow{\sigma(M',L(N))}&&\Big\downarrow{\sigma(M,L(N))}&&\Big\downarrow{\sigma(M'',L(N))}\\ 0&\longrightarrow&L(N^\circ)\otimes_{A^\circ}{M'}^\circ&\xrightarrow{1\otimes r}&L(N^\circ)\otimes_{A^\circ}M^\circ&\xrightarrow{1\otimes s}&L(N^\circ)\otimes_{A^\circ}{M''}^\circ&\longrightarrow&0 . \end{array}

We shall see later other commutation relations (cf. X, p. 131, cor. 1*).*

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

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