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The Hom complex of two two-term complexes

Example

Let $C_{∙} = D_{∙}$ be the two-term identity complex $0 \to Z \overset{1_{Z}}{\to} Z \to 0.$ Then $\underset{}{Hom} (C, D)_{1} ≅ Z, \underset{}{Hom} (C, D)_{0} ≅ Z^{2}, \underset{}{Hom} (C, D)_{- 1} ≅ Z,$ with differentials $\partial_{1} (m) = (m, m), \partial_{0} (a, b) = b - a .$

Facts & Assumptions

Given: The two-term identity complex $C_{∙} = D_{∙}$ in $A b$ .

[L1]

The Hom complex in degree $r$ consists of families $u_{n} : C_{n} \to D_{n + r}$ with differential $(\partial u)_{n} = d_{n + r} u_{n} - (- 1)^{r} u_{n - 1} d_{n}$ (The Hom complex of chain complexes).

[L2]

Degree- $0$ cycles in the Hom complex are exactly chain maps (Zero cocycles in the Hom complex are chain maps).

[L3]

$A b$ is an abelian category (Abelian groups form an abelian category).

Verification

technique · direct

1.1L1L3givenalgebra

A degree- $1$ map has only one possible nonzero component $u_{0} : Z \to Z$ , so it is determined by an integer $m$ . A degree- $0$ map has components $u_{1}, u_{0}$ , hence is determined by $(a, b) \in Z^{2}$ , and a degree $- 1$ map is determined by $u_{1} : Z \to Z$ .

2.1L1L2step 1.1algebra∎

Substituting these components into [L1] gives $\partial_{1} (m) = (m, m), \partial_{0} (a, b) = b - a .$ Therefore $\ker (\partial_{0}) = {(a, a)}$ , exactly the degree- $0$ chain maps from [L2].

Depends on

Used by

Homotopy classes as H-zero of a Hom complex Example

Dependency tree · two levels

12 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)