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The Hom-complex differential squares to zero

Statement

For every pair of chain complexes $C_{∙}, D_{∙}$ , the differential $\partial$ of The Hom complex of chain complexes satisfies $\partial_{r - 1} \partial_{r} = 0$ for every $r \in Z$ .

Facts & Assumptions

Given: A degree- $r$ graded morphism $u = (u_{n})_{n} \in \underset{}{Hom} (C, D)_{r}$ .

[L1]

The differential on the Hom complex is $(\partial u)_{n} = d_{n + r}^{D} u_{n} - (- 1)^{r} u_{n - 1} d_{n}^{C}$ (The Hom complex of chain complexes).

Proof

technique · direct

1.1L1givenalgebra

Using [L1] twice, the $n$ th component of $\partial (\partial u)$ is $d_{n + r - 1}^{D} d_{n + r}^{D} u_{n} - (- 1)^{r} d_{n + r - 1}^{D} u_{n - 1} d_{n}^{C} - (- 1)^{r - 1} d_{n + r - 1}^{D} u_{n - 1} d_{n}^{C} - u_{n - 2} d_{n - 1}^{C} d_{n}^{C} .$

2.1step 1.1algebra∎

The middle two terms cancel because $- (- 1)^{r} - (- 1)^{r - 1} = 0$ , and the outer two terms vanish because successive differentials in $C$ and $D$ compose to zero. Hence every component of $\partial (\partial u)$ is zero, so $\partial^{2} = 0$ .

The Hom-complex differential squares to zero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources