Bourbaki

Exercise 16, § 3

Algebra, X, p. 182

a) Let 0→M′→M→M′′→00 \to M' \to M \to M'' \to 0 be an exact sequence of A-modules, (P', p') (resp. (P'', p'')) a projective resolution of M' (resp. M''). Now prove that there exists a projective resolution (P, p) of M and a commutative diagram:

0→P′→P→P′′→0↓p′↓p↓p′′0→M′→M→M′′→0 \begin{array}{ccccccccc} 0 & \to & P' & \to & P & \to & P'' & \to & 0 \\ & & \downarrow p' & & \downarrow p & & \downarrow p'' & & \\ 0 & \to & M' & \to & M & \to & M'' & \to & 0 \end{array}
where the horizontal lines are exact. b) Let 0→N′→N→N′′→00 \to N' \to N \to N'' \to 0 be a second exact sequence of A-modules, (Q, q) (resp. (Q', q'), resp. (Q'', q'')) a projective resolution of N (resp. N', resp. N''), in such a way that there exists a commutative diagram with exact lines:
0→Q′→Q→Q′′→0↓q′↓q↓q′′0→N′→N→N′′→0 \begin{array}{ccccccccc} 0 & \to & Q' & \to & Q & \to & Q'' & \to & 0 \\ & & \downarrow q' & & \downarrow q & & \downarrow q'' & & \\ 0 & \to & N' & \to & N & \to & N'' & \to & 0 \end{array}
Let on the other hand:
0→M′→M→M′′→0↓u′↓u↓u′′0→N′→N→N′′→0. \begin{array}{ccccccccc} 0 & \to & M' & \to & M & \to & M'' & \to & 0 \\ & & \downarrow u' & & \downarrow u & & \downarrow u'' & & \\ 0 & \to & N' & \to & N & \to & N'' & \to & 0 . \end{array}
be a commutative diagram of exact sequences, and let u~′:P′→Q′\tilde{u}' : P' \to Q' and u~′′:P′′→Q′′\tilde{u}'' : P'' \to Q'' be morphisms of complexes such that u′∘p′=q′∘u~′u' \circ p' = q' \circ \tilde{u}' and u′′∘p′′=q′′∘u~′′u'' \circ p'' = q'' \circ \tilde{u}''. Now prove that there exists a morphism u~:P→Q\tilde{u} : P \to Q such that u∘p=q∘u~u \circ p = q \circ \tilde{u} and that the diagram:
0→P′→P→P′′→0↓u~′↓u~↓u~′′0→Q′→Q→Q′′→0 \begin{array}{ccccccccc} 0 & \to & P' & \to & P & \to & P'' & \to & 0 \\ & & \downarrow \tilde{u}' & & \downarrow \tilde{u} & & \downarrow \tilde{u}'' & & \\ 0 & \to & Q' & \to & Q & \to & Q'' & \to & 0 \end{array}
is commutative.

c) Let u~1′:P′→Q′,u~1:P→Q\tilde{u}_1' : P' \to Q', \tilde{u}_1 : P \to Q and u~1′′:P′′→Q′′\tilde{u}_1'' : P'' \to Q'' be three morphisms of complexes such that

u′∘p′=q′∘u~1′,u∘p=q∘u~1,u′′∘p′′=q′′∘u~1′′ u' \circ p' = q' \circ \tilde{u}_1', \quad u \circ p = q \circ \tilde{u}_1, \quad u'' \circ p'' = q'' \circ \tilde{u}_1''
and making the diagram (*) commutative; let s′s' (resp. s′′s'') be a homotopy connecting u~′\tilde{u}' to u~1′\tilde{u}_1' (resp. u~′′\tilde{u}'' to u~1′′\tilde{u}_1''). Now prove that there exists a homotopy ss connecting u~\tilde{u} to u~1\tilde{u}_1, such that the diagram:

0→P′→P→P′′→0↓s′↓s↓s′′0→Q′→Q→Q′′→0 \begin{array}{ccccccccc} 0 & \to & P' & \to & P & \to & P'' & \to & 0 \\ & & \downarrow^{s'} & & \downarrow^s & & \downarrow^{s''} & & \\ 0 & \to & Q' & \to & Q & \to & Q'' & \to & 0 \end{array}

is commutative.

d) State and prove the results corresponding to a),b),c)a), b), c) for injective resolutions.

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.