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Two-out-of-three for quasi-isomorphisms in a short exact sequence diagram

Statement

Consider a morphism of short exact sequences of complexes $\begin{array}{ccccccccc} 0 & \to & A_{∙} & \to & B_{∙} & \to & C_{∙} & \to & 0 \\ ↓ a & ↓ b & ↓ c \\ 0 & \to & A_{∙}^{'} & \to & B_{∙}^{'} & \to & C_{∙}^{'} & \to & 0. \end{array}$ If any two of $a$ , $b$ , and $c$ are quasi-isomorphisms, then so is the third.

Facts & Assumptions

Given: A morphism of short exact sequences of complexes.

[L1]

Such a ladder induces a morphism between the associated long exact homology sequences (The long exact homology sequence is natural).

[L2]

In a morphism of long exact sequences, if the four surrounding comparison maps in a five-term window are isomorphisms, then the middle one is an isomorphism (Five lemma for a morphism of long exact sequences).

[L3]

A quasi-isomorphism is a chain map inducing isomorphisms on all homology objects (Quasi-isomorphism).

Proof

technique · direct

1.1L1L2L3givenalgebra

Assume $b$ and $c$ are quasi-isomorphisms. In the long exact ladder from [L1], center the five-term window at $H_{n} (A)$ . The four surrounding comparison maps come from $H_{n + 1} (b)$ , $H_{n + 1} (c)$ , $H_{n} (b)$ , and $H_{n} (c)$ , so they are isomorphisms by [L3]. Hence $H_{n} (a)$ is an isomorphism by [L2].

1.2L1L2L3givenalgebra

Assume $a$ and $c$ are quasi-isomorphisms. Center the five-term window at $H_{n} (B)$ . The surrounding comparison maps come from $H_{n + 1} (c)$ , $H_{n} (a)$ , $H_{n} (c)$ , and $H_{n - 1} (a)$ , so [L2] gives that $H_{n} (b)$ is an isomorphism.

2.1L1L2L3givenalgebra∎

Assume $a$ and $b$ are quasi-isomorphisms. Center the five-term window at $H_{n} (C)$ . The surrounding comparison maps come from $H_{n} (a)$ , $H_{n} (b)$ , $H_{n - 1} (a)$ , and $H_{n - 1} (b)$ , so [L2] yields that $H_{n} (c)$ is an isomorphism. By [L3], the missing map is therefore a quasi-isomorphism in every case.

Depends on

Used by

Short five lemma for quasi-isomorphisms Corollary
Two-out-of-three for a diagram of finite complexes Example

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)
The Stacks Project, Section 12.13: Complexes (standard reference, not scraped)