Exercise 17, § 3
Algebra, X, p. 183
It is set in small type as supplementary.
In this exercise, if is a bicomplex (X, p. 174, exercise 11), we shall denote by the complex . Every complex will be considered as a bicomplex, zero in degree for .
Let be a complex of -modules. A projective resolution of is a pair , where is a bicomplex and a morphism of bicomplexes, such that for and that the complex (resp. ) defines a projective resolution of (resp. ) for every . For to be a projective resolution of , it is sufficient that the complexes and define for every projective resolutions of and .
(a) Now prove that every -complex admits a projective resolution (choose for every projective resolutions and of and ; using exercise 16, construct projective resolutions of and of as well as exact sequences:
(b) Let be another complex, a projective resolution of , a morphism of complexes. Now prove that there exists a morphism of bicomplexes such that (for every , choose morphisms and ; construct as previously, using exercise 16, b)).
(c) Let be a morphism of complexes homotopic to , and let be a morphism of bicomplexes such that . Now prove that and are homotopic (in the sense of X, p. 174, exercise 11, c); use exercise 16, c)).
d) Let be a subset of , an -complex such that for . Prove that there exists a projective resolution of such that for .
e) Suppose that there exists an integer such that every -module admits a projective resolution of length . Prove that every -complex admits a projective resolution such that for .
f) Let be an -complex. An injective resolution of is a pair , where is a bicomplex and a morphism of bicomplexes, such that for and such that the complex (resp. ) defines an injective resolution of (resp. ) for every . State and prove the analogues of results a) to e) for injective resolutions.
Solution
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