Bourbaki

Corollary 1

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

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

If CC is split and H(C)H(C) projective (resp. if C′C' is split and Hn(C′)H_n(C') injective for each nn), then the canonical homomorphism

λ(C,C′): H(HomgrA(C,C′))→HomgrA(H(C),H(C′)) \lambda(C,C'):\ H(\mathrm{Homgr}_A(C,C'))\to\mathrm{Homgr}_A(H(C),H(C'))

is bijective.

Suppose for example C′C' split and H(C′)H(C') injective for each nn, the case where CC is split and H(C)H(C) projective being proved in an analogous manner. By X, p. 35, def. 6, there exists a homotopy equivalence u′:C′→H(C′)u':C'\to H(C') ; by Proposition 3, HomgrA(1,u′)\mathrm{Homgr}_A(1,u') is a homotopy equivalence from HomgrA(C,C′)\mathrm{Homgr}_A(C,C') onto HomgrA(C,H(C′))\mathrm{Homgr}_A(C,H(C')) ; since

HomgrA(1,H(u′))∘λ(C,C′)=λ(C,H(C′))∘H(HomgrA(1,u′)) \mathrm{Homgr}_A(1,H(u'))\circ\lambda(C,C')=\lambda(C,H(C'))\circ H(\mathrm{Homgr}_A(1,u'))

and that HomgrA(1,H(u′))\mathrm{Homgr}_A(1,H(u')) and H(HomgrA(1,u′))H(\mathrm{Homgr}_A(1,u')) are bijective, it is enough for us to prove that λ(C,H(C′))\lambda(C, H(C')) is bijective, which brings us back to the case where C′C' is injective and has zero differential.

The canonical exact sequences (X, p. 28)

(III)

0→B(C)→iC→δC/B(C)→0 0 \to B(C) \xrightarrow{i} C \xrightarrow{\delta} C/B(C) \to 0
(IV)
0→H(C)→jC/B(C)→δB(C)→0 0 \to H(C) \xrightarrow{j} C/B(C) \xrightarrow{\delta} B(C) \to 0

give exact sequences (X, p. 83, Proposition 2, a))

0→HomgrA(C/B(C),C′)→l′HomgrA(C,C′)→lHomgrA(B(C),C′)→0 0 \to \mathrm{Homgr}_A(C/B(C), C') \xrightarrow{l'} \mathrm{Homgr}_A(C, C') \xrightarrow{l} \mathrm{Homgr}_A(B(C), C') \to 0
0→HomgrA(B(C),C′)→ΔHomgrA(C/B(C),C′)→jHomgrA(H(C),C′)→0. 0 \to \mathrm{Homgr}_A(B(C), C') \xrightarrow{\Delta} \mathrm{Homgr}_A(C/B(C), C') \xrightarrow{j} \mathrm{Homgr}_A(H(C), C') \to 0 .

Since dC=i∘δ∘pd_C = i \circ \delta \circ p, the differential DD of HomgrA(C,C′)\mathrm{Homgr}_A(C, C') is given by Dn=(−1)n+1Pn∘Δn∘lnD_n = (-1)^{n+1} P_n \circ \Delta_n \circ l_n; we then have

Z(HomgrA(C,C′))=Ker (P∘Δ∘I)=Ker I=Im P, \mathrm{Z}(\mathrm{Homgr}_A(C, C')) = \mathrm{Ker}\,(P \circ \Delta \circ I) = \mathrm{Ker}\,I = \mathrm{Im}\,P,
B(HomgrA(C,C′))=Im (P∘Δ∘I)=P(Im Δ)=P(Ker J); \mathrm{B}(\mathrm{Homgr}_A(C, C')) = \mathrm{Im}\,(P \circ \Delta \circ I) = P(\mathrm{Im}\,\Delta) = P(\mathrm{Ker}\,J) ;

whence an isomorphism φ:H(HomgrA(C,C′))→HomgrA(H(C),C′)\varphi : H(\mathrm{Homgr}_A(C, C')) \to \mathrm{Homgr}_A(H(C), C') such that, if a∈HomgrA(C/B(C),C′)a \in \mathrm{Homgr}_A(C/B(C), C'), the image of the class of P(a)P(a) under φ\varphi is J(a)J(a); we immediately verify that φ\varphi coincides with the canonical homomorphism λ\lambda.

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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