Remark 3

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

If $g : N \to N'$ is a homomorphism of left A-modules, then

(1_{L(M)} \otimes g) \circ (1_{L(M)} \otimes 1_N) = (1_{L(M)} \otimes 1_N) \circ (1_{L(M)} \otimes L(g)) ,

therefore the diagram

\begin{array}{ccc} \mathrm{Tor}^A(M, N) & \xrightarrow{\psi_{M(N)}} & H(L(M) \otimes_A N) \\ \mathrm{Tor}^A(1, g) \downarrow & & \downarrow H(1 \otimes g) \\ \mathrm{Tor}^A(M, N') & \xrightarrow{\psi_{M(N')}} & H(L(M) \otimes_A N') \end{array}

is commutative.

Analogously, if $f : M \to M'$ is a homomorphism of right A-modules, one has a commutative diagram :

\begin{array}{ccc} \mathrm{Tor}^A(M, N) & \xrightarrow{\overline{\psi}_{N(M)}} & H(M \otimes_A L(N)) \\ \mathrm{Tor}^A(f, 1) \downarrow & & \downarrow H(f \otimes 1) \\ \mathrm{Tor}^A(M', N) & \xrightarrow{\overline{\psi}_{N(M')}} & H(M' \otimes_A L(N)) . \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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