Bourbaki

Theorem 2

Tag 0DX7, permanent. Cite it as [Bourbaki, Tag 0DX7].

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

The right-unbounded sequence of homomorphisms of kk-modules

0⟶HomA(M′′,N)→Hom(s,1)HomA(M,N)→Hom(r,1)HomA(M′,N)0 \longrightarrow \mathrm{Hom}_A(M'', N) \xrightarrow{\mathrm{Hom}(s, 1)} \mathrm{Hom}_A(M, N) \xrightarrow{\mathrm{Hom}(r, 1)} \mathrm{Hom}_A(M', N)
→δ0(F,N)ExtA1(M′′,N)→⋯→δn−1(F,N)ExtAn(M′′,N)→Extn(s,1)ExtAn(M,N)\xrightarrow{\delta^0(\mathscr{F}, N)} \mathrm{Ext}_A^1(M'', N) \to \cdots \xrightarrow{\delta^{n-1}(\mathscr{F}, N)} \mathrm{Ext}_A^n(M'', N) \xrightarrow{\mathrm{Ext}^n(s, 1)} \mathrm{Ext}_A^n(M, N)
→Extn(r,1)ExtAn(M′,N)→δn(F,N)ExtAn+1(M′′,N)→⋯\xrightarrow{\mathrm{Ext}^n(r, 1)} \mathrm{Ext}_A^n(M', N) \xrightarrow{\delta^n(\mathscr{F}, N)} \mathrm{Ext}_A^{n+1}(M'', N) \to \cdots
is exact.

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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Cites

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Translated with machine help, not yet reviewed by a person.