Bourbaki

Exercise 8, § 2

Algebra, X, p. 173

Let 0→M′→uM→vM′′→00 \to M' \xrightarrow{u} M \xrightarrow{v} M'' \to 0 be an exact sequence of A-modules, which can be considered as an exact sequence of complexes (zero in degrees ≠0\neq 0). Now prove that the morphism of complexes φ:Con (u)→M′′\varphi : \mathrm{Con}\,(u) \to M'' (X, p. 39) is a homotopy equivalence if and only if the exact sequence 0→M′→M→M′′→00 \to M' \to M \to M'' \to 0 is split.

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.