ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

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Naturality of a connecting map under a map of coefficient sequences

Example

Compare the cone sequences of $\times 2$ and $\times 4$ on $Z [0]$ . There is a morphism of short exact sequences $\begin{array}{ccccccccc} 0 & \to & Z [0] & \to & Cone (\times 2) & \to & Z [1] & \to & 0 \\ ↓ \times 2 & ↓ θ & ↓ 1 \\ 0 & \to & Z [0] & \to & Cone (\times 4) & \to & Z [1] & \to & 0, \end{array}$ where $θ$ is the identity in degree $1$ and multiplication by $2$ in degree $0$ . The connecting square commutes because the top connecting map is $\times 2$ and the bottom one is $\times 4$ .

Facts & Assumptions

Given: The morphism of cone sequences displayed in the example.

[L1]

The homology connecting morphism is natural under morphisms of short exact sequences (Naturality of the homology connecting morphism).

[L2]

In module categories, the connecting morphism is computed by the lift-boundary formula (Elementwise formula for the connecting map in module categories).

Verification

technique · direct

1.1L1givenconstruct

The map $θ$ is a chain map because the top differential is multiplication by $2$ and the bottom one is multiplication by $4$ , so $4 \cdot 1 = 2 \cdot 2$ on degree $1$ . Thus the displayed diagram is a morphism of short exact sequences.

2.1L1L2step 1.1algebra∎

By [L2], the top connecting morphism is multiplication by $2$ and the bottom one is multiplication by $4$ . Therefore $(\times 2) \circ \partial_{t o p} = \partial_{b o t t o m} \circ 1$ on $H_{1} (Z [1])$ . This is exactly the commuting square asserted by [L1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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)