Proposition 12

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

Algebra, chapter X, § 5. Modules d’extensions

The homomorphisms (14), (15), (16) are morphisms of complexes.

Let us prove it for example for the homomorphism (14). Denote

\kappa : \operatorname{Homgr}_A(C', C'') \otimes_k \operatorname{Homgr}_A(C, C') \to \operatorname{Homgr}_A(C, C')

this homomorphism. Let $f \in \operatorname{Homgr}_A(C', C'')_p$ and $g \in \operatorname{Homgr}_A(C, C')_q$ ; then by definition one has $\kappa(f \otimes g) = f \circ g$ . Moreover :

\mathrm{D}(f \otimes g) = \mathrm{D}f \otimes g + (-1)^p f \otimes \mathrm{D}g = (d'' \circ f) \otimes g - (-1)^p (f \circ d') \otimes g + + (-1)^p f \otimes (d' \circ g) - (-1)^{p+q} f \otimes (g \circ d),

whence

\kappa(\mathrm{D}(f \otimes g)) = d'' \circ f \circ g - (-1)^p f \circ d' \circ g + + (-1)^p f \circ d' \circ g - (-1)^{p+q} f \circ g \circ d \\ = d'' \circ f \circ g - (-1)^{p+q} f \circ g \circ d = \mathrm{D}(f \circ g) = \mathrm{D}(\kappa(f \otimes g)).

One proves analogously that the homomorphisms (15) and (16) are morphisms of complexes.

One deduces from the morphism (14) homomorphisms of $k$ -modules (X, p. 62)

(17) $\mathrm{H}^p(\operatorname{Homgr}_A(C', C'')) \otimes_k \mathrm{H}^q(\operatorname{Homgr}_A(C, C')) \to \mathrm{H}^{p+q}(\operatorname{Homgr}_A(C, C''))$ .

Taking $C = A$ , one sees that the homomorphism :

(18) $\operatorname{Homgr}_A(C', C'') \otimes_k C' \to C''$ which maps $f \otimes x$ to $f(x)$ is a morphism of complexes of left $A$ -modules ; there is associated to it a canonical homomorphism ( $X$ , p. 80) of graded $A$ -modules $\gamma : H(\mathrm{Homgr}_A(C', C'')) \otimes_k H(C') \to H(C'')$ , which corresponds to the canonical homomorphism of $k$ -modules

\lambda : H(\mathrm{Homgr}_A(C', C'')) \to \mathrm{Homgr}_A(H(C'), H(C''))

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

Cited by

Nothing.

Cites

Nothing.