ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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The snake lemma applied to multiplication by an integer

Example

Fix $n \geq 1$ and apply multiplication by $n$ to the short exact sequence $0 \to Z \overset{\times n}{\to} Z \to Z / n \to 0.$ The snake lemma produces $0 \to 0 \to 0 \to Z / n \overset{δ}{\to} Z / n \overset{0}{\to} Z / n \overset{1}{\to} Z / n \to 0,$ so the connecting morphism is an isomorphism, and under the standard identifications it is the identity.

Facts & Assumptions

Given: The multiplication-by- $n$ endomorphism of the short exact sequence above.

[L1]

Abelian groups form an abelian category (Abelian groups form an abelian category).

[L2]

The snake lemma gives the exact sequence attached to that ladder (Snake lemma in an abelian category).

Verification

1.1L1algebra

Multiplication by $n$ on $Z$ has zero kernel and cokernel $Z / n$ , while the induced map on the quotient term $Z / n$ is zero. The induced map $coker (\times n) \to coker (\times n)$ is multiplication by $n$ on $Z / n$ , hence is $0$ , and the induced map $coker (\times n) \to coker (0)$ is the identity of $Z / n$ .

2.1L2step 1.1algebra

Substituting those terms into [L2] gives the displayed exact sequence. Exactness at the first copy of $Z / n$ forces $δ$ to be an isomorphism. In the standard snake construction, the class of $1$ in $\ker (0) = Z / n$ lifts to $1 \in Z$ and then maps to the class of $1$ in $coker (\times n) = Z / n$ , so under these standard identifications $δ$ is the identity.

3.1step 2.1∎

This is the concrete snake sequence for multiplication by an integer.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

Saunders Mac Lane, Categories for the Working Mathematician, Lemma VIII.4.5 (standard reference, not scraped)