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Rational cofiber implies finite-coefficient equivalence

PDF p. 6: Finite-coefficient comparison

Statement

Let $f : A \to B$ be a morphism in $D (Z)$ whose cofiber is isomorphic in $D (Z)$ to the underlying complex of a complex of $Q$ -vector spaces. For every positive integer $n$ , $f \otimes_{Z}^{L} Z / n$ is an isomorphism in $D (Z)$ , and consequently induces an isomorphism on $H^{i}$ for every integer $i$ .

Facts & Assumptions

Given: $f$ and $n$ as stated.

Goal: Prove the finite-coefficient equivalence and its cohomological consequence.

[F1]

A rational complex has zero derived quotient by $n$ . Finite derived quotients of rational complexes vanish

[F2]

Taking the derived quotient by $n$ commutes with mapping cones. Derived finite coefficients preserve mapping cones

Proof

1.1F1F2given

By [F1] and [F2], the mapping cone of $f \otimes^{L} Z / n$ is zero in $D (Z)$ , equivalently it is acyclic.

2.1step 1.1∎

The long exact cohomology sequence of the mapping cone, obtained from the degreewise split sequence $B \to Cone (f) \to A [1]$ , shows that a chain map with acyclic cone induces an isomorphism on every cohomology group. Thus the finite-coefficient map is a quasi-isomorphism and hence an isomorphism in $D (Z)$ . Applying $H^{i}$ gives the stated group isomorphism.

Depends on

Used by · 1

Finite-coefficient motivic to cdh comparison

Dependency tree · 3 levels

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