ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-29

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The kernel-cokernel sequence of a composite of module maps

Example

In $Z - M o d$ , take $f : Z \to Z, f (n) = 2 n, g : Z \to Z / 2, g (n) = n m o d 2.$ Then $g f = 0$ , and the sequence of The kernel-cokernel sequence of a composite becomes $0 \to 0 \to Z \overset{\times 2}{\to} 2 Z \overset{0}{\to} Z / 2 \overset{1_{Z / 2}}{\to} Z / 2 \to 0 \to 0,$ which is exact.

Facts & Assumptions

Given: The maps $f$ and $g$ in $Z - M o d$ .

[L1]

Module categories are abelian (Modules over a ring form an abelian category).

[L2]

Every composite has the kernel-cokernel exact sequence (The kernel-cokernel sequence of a composite).

Verification

technique · direct

1.1L1algebra

Here $\ker (f) = 0$ , $\ker (g f) = Z$ , $\ker (g) = 2 Z$ , $coker (f) ≅ Z / 2$ , $coker (g f) ≅ Z / 2$ , and $coker (g) = 0$ .

2.1L2step 1.1algebra

The map $\ker (g f) \to \ker (g)$ is just $f$ , so it is multiplication by $2$ onto the subgroup $2 Z$ . The connecting map $\ker (g) \to coker (f)$ is zero because it is induced by the cokernel map $q_{f} : Z \to Z / 2$ , which kills every even integer. The map $coker (f) \to coker (g f)$ is the identity on $Z / 2$ , and the remaining arrows are the obvious zero maps. This is the displayed sequence.

3.1step 2.1algebra∎

That concrete sequence is exact by direct inspection.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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, Exercise VIII.4.6 (standard reference, not scraped)