TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30‡ rests on unproved material (inherited)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Euler-Poincare formula for finite free complexes

Statement

Let $C_{∙}$ be a bounded chain complex of finite-rank free abelian groups. Assume that every homology group $H_{n} (C)$ is free of finite rank. Then $χ (C) = \sum_{n \in Z} (- 1)^{n} rank (H_{n} (C)) .$

Facts & Assumptions

Given: A bounded chain complex $C_{∙}$ of finite-rank free abelian groups whose homology groups are free of finite rank.

[L1]

The Euler characteristic is the finite alternating sum $\sum_{n} (- 1)^{n} rank (C_{n})$ (Euler characteristic of a finite complex of finite-rank free abelian groups).

[L2]

Homology is the quotient $Z_{n} (C) / B_{n} (C)$ (Homology object of a chain complex).

[L3]

Abelian groups and $Z$ -modules are the same objects and morphisms (Abelian groups and $Z$ -modules have the same objects and morphisms).

[L4]

The ring $Z$ is a principal ideal domain: it is a commutative ring (The integers form a commutative ring), it has no zero divisors (The integers have no zero divisors; multiplicative cancellation), and every ideal is generated by one integer because every subgroup of $(Z, +)$ is cyclic (Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ , Principal ideal domain).

[L5]

A submodule of a finite-rank free module over a PID is again finite free (A submodule of a free module of finite rank over a PID is free of no larger rank).

[L6]

A finite free module over a commutative ring is projective (Under the stated choice boundary, free modules are projective and hence flat).

[L7]

A short exact sequence ending in a projective module splits (Equivalent characterizations of projective modules).

[L8]

A free module of rank $r$ has a basis of $r$ elements (The free module on a set and its standard basis, Invariant basis number and the rank of a free module).

Proof

technique · direct

1.1

By [L3] and [L4], each $C_{n}$ is a finite-rank free $Z$ -module over a PID. Therefore [L5] makes every cycle group $Z_{n} (C) \leq C_{n}$ and every boundary group $B_{n} (C) \leq C_{n}$ finite free. Since the homology groups are finite free by the hypothesis, all ranks appearing below are defined.

L3L4L5given

2.1

By [L2], the short exact sequence $0 \to Z_{n} (C) \to C_{n} \to B_{n - 1} (C) \to 0$ has projective quotient $B_{n - 1} (C)$ by steps 1.1 and [L6], so [L7] splits it. Likewise $0 \to B_{n} (C) \to Z_{n} (C) \to H_{n} (C) \to 0$ splits because $H_{n} (C)$ is finite free by hypothesis and hence projective by [L6]. Using [L8], split exactness gives $rank (C_{n}) = rank (Z_{n} (C)) + rank (B_{n - 1} (C))$ and $rank (Z_{n} (C)) = rank (B_{n} (C)) + rank (H_{n} (C)) .$

L2L6L7L8step 1.1algebra

3.1

Substitute the second identity of step 2.1 into the first: $rank (C_{n}) = rank (B_{n} (C)) + rank (H_{n} (C)) + rank (B_{n - 1} (C)) .$ Multiply by $(- 1)^{n}$ and sum over all $n$ . The boundedness in [L1] makes this a finite sum, and the two boundary sums cancel after the index shift $m = n - 1$ : $\sum_{n} (- 1)^{n} rank (B_{n - 1} (C)) = - \sum_{n} (- 1)^{n} rank (B_{n} (C)) .$ Hence only the homology contribution remains.

L1step 2.1algebra

4.1

The resulting identity is $\sum_{n} (- 1)^{n} rank (C_{n}) = \sum_{n} (- 1)^{n} rank (H_{n} (C)),$ which is exactly the formula in the statement by [L1].

L1step 3.1∎

Depends on

Used by

Euler-Poincare for a finite complex Example

Dependency tree · two levels

70 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

P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)
K. Conrad, Modules over a PID, Theorem 2.2 and Corollary 2.6 (standard reference, not scraped)