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Cofiber bounds imply low-degree cohomology isomorphisms

PDF p. 4: equivalence in degrees at most three

Statement

Let f:A→B be a morphism in D(Z) and let C be its cofiber, represented by a mapping cone. If Hj(C)=0 for every j≤n, then Hi(f) is an isomorphism for every i≤n and is injective for i=n+1. If, in addition, Hn+1(C)=0, it is an isomorphism also for i=n+1.

Facts & Assumptions

Given: A derived-category morphism and a cone with the stated vanishing.

Goal: Establish the stated low-degree consequences.

[F1]

Cohomology is a functor on D(Z) (Cohomology descends to the derived category).

Proof

1.1F1construct

Represent the derived morphism by a chain map after replacing its source by a quasi-isomorphic complex. Its mapping cone fits in a degreewise split short exact sequence of complexes 0→B→Cone⁡(f)→A[1]→0. Taking cycles and boundaries gives the long exact sequence Hi−1(C)→Hi(A)→Hi(B)→Hi(C). The result is independent of representatives by [F1].

2.1step 1.1algebra∎

For i≤n, both adjacent cone groups in step 1.1 vanish, giving an isomorphism. For i=n+1, the left group Hn(C) vanishes, giving injectivity; if also Hn+1(C)=0, the right group vanishes and the map is surjective as well. This proves every claimed range without a condition on Hn+2(C).

Depends on

Used by · 1

Dependency tree · 3 levels

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