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Relative cofibers and quotient complexes

PDF p. 7: Motivic purity at a smooth closed point

Statement

For a cochain map $f : A \to B$ , use cohomological shifts $A [s]^{n} = A^{n + s}$ , $d_{A [s]} = (- 1)^{s} d_{A}$ , and put $Q_{f}^{n} = B^{n} \oplus A^{n + 1}, d (b, a) = (d_{B} b + f (a), - d_{A} a) .$ This is the cone of $f$ , giving the relative cofiber triangle $A \overset{f}{\to} B \to Q_{f} \to A [1]$ in the derived category. If $f$ is termwise injective, projection $(b, a) \mapsto [b]$ gives a quasi-isomorphism $Q_{f} \to B / f (A)$ . Consequently the quotient-complex model for a relative motive is justified whenever an actual termwise monomorphic complex model of its open-inclusion map has separately been supplied; no such monomorphism is asserted for an arbitrary map of motives.

Facts & Assumptions

Given: Complexes of abelian groups and a cochain map $f$ , with termwise injectivity only for the quotient conclusion. The same argument applies to complexes in an abelian category.

Goal: Establish the cone differential and the exact condition allowing its replacement by a quotient.

[F1]

The cochain cone and shift conventions, and the cone triangle in the derived category, are those of Derived category of an abelian category. A quasi-isomorphism induces isomorphisms on every cohomology group.

Proof

1.1F1givenalgebra

Applying the displayed differential twice gives first coordinate $d_{B}^{2} b + d_{B} f (a) - f d_{A} a = 0$ and second coordinate $d_{A}^{2} a = 0$ . Thus it is a cochain complex. Its form is exactly the cone in [F1], so the displayed relative triangle is its cone triangle.

2.1step 1.1givenconstruct

Assume $f$ termwise injective. The projection to $B / f (A)$ is a surjective cochain map because $d [b] = [d_{B} b]$ . Its kernel identifies with $A^{n} \oplus A^{n + 1}$ via $(a_{0}, a_{1}) \mapsto (f a_{0}, a_{1})$ , and its differential in these coordinates is $(a_{0}, a_{1}) \mapsto (d_{A} a_{0} + a_{1}, - d_{A} a_{1})$ .

3.1step 2.1algebra∎

The map $h (a_{0}, a_{1}) = (0, a_{0})$ has degree $- 1$ on this kernel. One computes $d h (a_{0}, a_{1}) = (a_{0}, - d_{A} a_{0})$ and $h d (a_{0}, a_{1}) = (0, d_{A} a_{0} + a_{1})$ , so $d h + h d = id$ . Thus the kernel is contractible and has zero cohomology. The short exact sequence of this kernel, $Q_{f}$ and the quotient gives that projection induces isomorphisms on cohomology. It is the asserted quasi-isomorphism, under precisely the stated injectivity hypothesis.

Depends on

Derived category of an abelian category

Used by · 2

Dependency tree · 3 levels

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