Bourbaki

Proposition 3

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

Algebra, chapter X, § 2. Complexes de A-modules

If ff and gg are two homotopic morphisms from CC into C′C', then H(f)=H(g)\mathrm{H}(f) = \mathrm{H}(g). Let ss be a homotopy connecting ff to gg. One has

(g−f)(Z(C))=(d′∘s+s∘d)(Z(C))=(d′∘s)(Z(C))⊂B(C′), (g - f)(\mathbf{Z}(C)) = (d' \circ s + s \circ d)(\mathbf{Z}(C)) = (d' \circ s)(\mathbf{Z}(C)) \subset \mathbf{B}(C') ,
hence H(g−f)=0\mathrm{H}(g - f) = 0 and H(g)=H(f)\mathrm{H}(g) = \mathrm{H}(f).

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

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Cites

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