LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Roof equivalence is an equivalence relation

Statement

For any two objects $X, Y$ of a category with a two-sided multiplicative system $S$ , common refinement is an equivalence relation on left roofs from $X$ to $Y$ .

Facts & Assumptions

Given: For any two objects $X, Y$ of a category with a two-sided multiplicative system $S$ , common refinement is an equivalence relation on left roofs from $X$ to $Y$ .

[F1]

The common-refinement conditions are equality of the composite denominators in $S$ and equality of the numerators (Common refinement equivalence of roofs).

[F2]

The two Ore axioms and the two cancellation directions hold for the given multiplicative system (Multiplicative system in a category).

Proof

1.1F1given

For a roof $(s, f)$ , take both refinement legs to be its vertex identity. This gives reflexivity. Interchanging the two refinement legs gives symmetry. These arguments also cover identity roofs and coincident vertices.

1.2F1F2

For transitivity suppose $(s, f) \sim (t, g)$ via $a, b$ and $(t, g) \sim (u, h)$ via $c, d$ . Thus $r = s a = t b \in S$ and $q = t c = u d \in S$ , while $f a = g b$ and $g c = h d$ . Apply Ore to $r$ and $q$ : obtain $v$ in $S$ and $w$ with $r v = q w \in S$ .

2.1F2step 1.2algebra∎

Now $t b v = t c w$ . Cancellation of the postcomposed denominator $t$ supplies $e \in S$ such that $b v e = c w e$ . Therefore $s a v e = u d w e = r v e \in S$ and $f a v e = g b v e = g c w e = h d w e$ . The legs $a v e, d w e$ witness $(s, f) \sim (u, h)$ . Only $v, e$ , and the displayed composite denominator were asserted to lie in $S$ .

Depends on

Dependency tree · 2 levels

Sources

10.3.1–10.3.14, pp. 379–384