TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-04

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The Euler-Poincare formula for a finite simplicial complex with free homology

Statement

Let $K$ be a finite simplicial complex. Assume that each simplicial homology group $H_{n}^{s i m p} (K)$ is free of finite rank. Then $χ (K) = \sum_{n \geq 0} (- 1)^{n} rank H_{n}^{s i m p} (K) .$

Proof

Given: A finite simplicial complex $K$ whose simplicial homology groups are free of finite rank.

1.1given

For each $n$ , the chain group $C_{n} (K)$ is free abelian on the oriented $n$ -simplices of $K$ , so $rank C_{n} (K) = f_{n} (K)$ . Since $K$ is finite, only finitely many of these groups are nonzero.

2.1step 1.1

Therefore the simplicial chain complex of $K$ is a bounded chain complex of finite-rank free abelian groups, and its homology groups are free of finite rank by hypothesis. The finite-free Euler-Poincare theorem applies and gives $\sum_{n \geq 0} (- 1)^{n} rank C_{n} (K) = \sum_{n \geq 0} (- 1)^{n} rank H_{n}^{s i m p} (K)$ .

3.1step 1.1step 2.1∎

Replace $rank C_{n} (K)$ by $f_{n} (K)$ using step 1.1, and replace the left-hand side by $χ (K)$ by definition. This yields the stated formula.

Depends on

Used by

Dependency tree · two levels

20 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

Allen Hatcher, Algebraic Topology (standard reference, not scraped)
Vidit Nanda, Computational Algebraic Topology, Lecture 03: Homology (standard reference, not scraped)