Lemma 2

Tag 0DS6, permanent. Cite it as [Bourbaki, Tag 0DS6].

(Cyl $(u)$ , $\overline{D}$ ) and (Con $(u)$ , $D$ ) are complexes of A-modules.

Indeed, one has

\overline{D} \circ \overline{D}(x', y', x) = \overline{D}(d'x' + y', -d'y', dx - u(y')) =

= (d'(d'x' + y') - d'y', -d'(-d'y'), d(dx - u(y')) - u(-d'y')) = 0

since

d' \circ d' = 0

d \circ d = 0

and

d \circ u = u \circ d'

. Similarly

D \circ D = 0

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

Cited by

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