Remark 2

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

Algebra, chapter X, § 4. Produits de torsion

The complex morphism $p_M \otimes p_N : L(M) \otimes_A L(N) \to M \otimes_A N$ induces on the degree 0 homology the isomorphism

\gamma_{M,N}^{-1} : \operatorname{Tor}_0^A(M, N) \to M \otimes_A N

inverse to $\gamma_{M,N}$ .

We have $L(1_M) = 1_{L(M)}$ , $L(1_N) = 1_{L(N)}$ , hence by passage to the homology:

(8)

\operatorname{Tor}^A(1_M, 1_N) = 1_{\operatorname{Tor}^A(M, N)} .

If $f' : M' \to M''$ (resp. $g' : N' \to N''$ ) is a homomorphism of right $A$ -modules (resp. left $A$ -modules), we have $L(g' \circ g) = L(g') \circ L(g)$ and $L(f' \circ f) = L(f') \circ L(f)$ , hence

(9)

\operatorname{Tor}^A(f' \circ f, g' \circ g) = \operatorname{Tor}^A(f', g') \circ \operatorname{Tor}^A(f, g) .

Consider the morphisms of $k$ -complexes

L(M) \otimes_A N \xleftarrow{1 \otimes p_N} L(M) \otimes_A L(N) \xrightarrow{p_M \otimes 1} M \otimes_A L(N)

and the $k$ -homomorphisms they induce in homology:

H(L(M) \otimes_A N) \xleftarrow{\psi_M(N)} \mathrm{Tor}^A(M, N) \xrightarrow{\overline{\psi}_N(M)} H(M \otimes_A L(N)) ;

according to Proposition 4 of X, p. 67, $1 \otimes p_N$ and $p_M \otimes 1$ are homology morphisms. Therefore:

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

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