TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Every split coequalizer is a coequalizer and an absolute colimit

Statement

Every split coequalizer diagram (Split coequalizer diagrams) is a coequalizer diagram (Equalizers and coequalizers as limits and colimits of a parallel pair), and its coequalizer is an absolute colimit (Absolute colimits).

Facts & Assumptions

Given: A split coequalizer diagram $x ⇉^{f, g} y \overset{h}{\to} z$ with splitting maps $t : y \to x$ and $s : z \to y$ .

[L1]

A split coequalizer diagram has maps $f, g : x ⇉ y$ , $h : y \to z$ , $t : y \to x$ , and $s : z \to y$ satisfying $h f = h g$ , $h s = 1_{z}$ , $g t = 1_{y}$ , and $f t = s h$ (Split coequalizer diagrams).

[L2]

A colimit is absolute when every functor preserves it (Absolute colimits).

Proof

technique · direct

1.1

Let $k : y \to w$ satisfy $k f = k g$ . Define $u : = k s : z \to w$ . Then $u h = k s h = k f t = k g t = k$ , using $f t = s h$ , the equality $k f = k g$ , and $g t = 1_{y}$ .

L1construct

1.2

Let $H$ be any functor with the given category as domain. Functoriality sends the four equations in [L1] to $H h H f = H h H g$ , $H h H s = 1_{H z}$ , $H g H t = 1_{H y}$ , and $H f H t = H s H h$ , so the image diagram is again split.

L1given

2.1

If $v : z \to w$ also satisfies $v h = k$ , then $v = v h s = k s = u$ because $h s = 1_{z}$ . Thus $h$ has the coequalizer universal property, including when $f = g$ or $h$ is an identity.

step 1.1L1algebra

3.1

Repeating steps 1.1 and 2.1 in the target of $H$ shows that $H h$ is a coequalizer of $H f$ and $H g$ . Since $H$ was arbitrary, every functor preserves this coequalizer, so it is absolute by [L2].

step 1.1step 2.1step 1.2L2∎

Depends on

Used by

Dependency tree · two levels

5 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

E. Riehl, Category Theory in Context, 2nd ed., Lemma 5.4.6 (standard reference, not scraped)
S. Mac Lane, Categories for the Working Mathematician, 2nd ed., Section VI.6 (standard reference, not scraped)