Bourbaki

Lemma 4

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

Algebra, chapter X, § 3. Résolutions

Let f:M′→Mf : M' \to M be a homomorphism of AA-modules of type C\mathscr{C}, and p:P→Mp : P \to M a bounded left resolution of PP of type C0\mathscr{C}_0. There exist a bounded left resolution p′:P′→M′p' : P' \to M' of type C0\mathscr{C}_0 and a morphism of complexes u:P′→Pu : P' \to P such that p∘u=f∘p′p \circ u = f \circ p'.

Let us argue by induction on the length nn of PP, the assertion being trivial when the latter is <0< 0. Consider the mapping g:M′×P0→Mg : M' \times P_0 \to M such that

g(x,r′)=f(x)−p0(r′) g(x, r') = f(x) - p_0(r')
for x∈M′,r′∈P0x \in M', r' \in P_0,

and its kernel KK; the AA-module KK is of type C\mathscr{C} since gg is surjective and M′×P0M' \times P_0 and MM are of type C\mathscr{C}. Let h:P0′→Kh : P'_0 \to K be a surjective homomorphism, where P0′P'_0 is of type C0\mathscr{C}_0; let us denote by p0′:P0′→M′p'_0 : P'_0 \to M' (resp. u0:P0′→P0u_0 : P'_0 \to P_0) the homomorphism composed of hh and the projection K→MK \to M (resp. K→P0K \to P_0); the homomorphism p0′p'_0 is surjective and we have a commutative diagram

P0′→u0P0p0′↓↓p0M′→fM. \begin{array}{ccc} P'_0 & \xrightarrow{u_0} & P_0 \\ p'_0 \downarrow & & \downarrow p_0 \\ M' & \xrightarrow{f} & M . \end{array}

It is then enough to apply the induction hypothesis to the homomorphism

Ker⁡p0′→Ker⁡p0 \operatorname{Ker} p'_0 \to \operatorname{Ker} p_0

deduced from u0u_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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Cites

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