False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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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.
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FALSE: the boundaries of a complex are a quotient of its cycles

Statement

For every chain complex $C_{∙}$ and degree $n$ , the boundary object $B_{n} (C)$ is a quotient of the cycle object $Z_{n} (C)$ .

Facts & Assumptions

Given: In the category $A b$ , the two-term complex $0 \to Z \overset{ι}{\to} Q \to 0,$ with $Z$ in degree $1$ , $Q$ in degree $0$ , and $ι$ the inclusion.

[L1]

Boundaries are the images of the incoming differentials, and cycles are the kernels of the outgoing differentials (Cycle and boundary subobjects of a complex).

Refutation

technique · direct

1.1L1given

In degree $0$ , the outgoing differential is $0 : Q \to 0$ , so [L1] gives $Z_{0} (C) = \ker (0) = Q .$ The incoming differential is $ι : Z \to Q$ , so $B_{0} (C) = im (ι) = Z .$

1.2algebra

Let $ϕ : Q \to Z$ be any group homomorphism. Then for every positive integer $n$ , $ϕ (1) = ϕ (n \cdot \frac{1}{n}) = n ϕ (\frac{1}{n}),$ so $ϕ (1)$ is divisible by every positive integer. The only integer with that property is $0$ , hence $ϕ (1) = 0$ . Therefore $ϕ (\frac{m}{n}) = m ϕ (\frac{1}{n}) = 0$ for every rational number $\frac{m}{n}$ , so $ϕ = 0$ . Thus no epimorphism $Q \to Z$ exists.

2.1step 1.1step 1.2∎

Step 1.1 shows that in this complex $B_{0} (C) = Z$ and $Z_{0} (C) = Q$ , while step 1.2 shows that $Z$ is not a quotient of $Q$ . Therefore the statement is false.

Depends on

Cycle and boundary subobjects of a complex

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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)