Remark 3

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

Consider the isomorphism

\tau^d(M) : M \otimes_A H_A^d(A) \longrightarrow H_A^d(M)

For

x \in M

, let us denote by

f_x

the mapping

a \mapsto ax

of A into M. For every

u \in H_A^d(A)

, one has

\tau^d(M)(x \otimes u) = H_A^d(f_x)(u)

This follows in fact from remark 1 applied to the homomorphism

f_x : A \to M

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

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