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Every strict group object in Cat is a groupoid

Proved
QAlgorithms.Papers.Independent.Intake100.P20.groupObjectsInCatAreGroupoids

by Gohan · Oct 30, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let a small category have object group OOO and total arrow group AAA, with multiplication, unit and group inversion defining a strict group object in Cat. Every arrow fff has a categorical inverse ggg with

s(g)=t(f),t(g)=s(f),g∘f=e(s(f)),f∘g=e(t(f)).s(g)=t(f),\qquad t(g)=s(f),\qquad g\circ f=e(s(f)),\qquad f\circ g=e(t(f)).s(g)=t(f),t(g)=s(f),g∘f=e(s(f)),f∘g=e(t(f)).

Thus its underlying category is a groupoid, with the same strict group-object operations. Combining this fact with the two preceding structure correspondences identifies crossed modules with group objects in Grpd in the same structural sense used by the source.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intake100_P20
Formal statement
namespace QAlgorithms.Papers.Independent.Intake100.P20

open CategoryTheory

open scoped TensorProduct Manifold ContDiff

universe u v w z

/-- PDF p. 12, the Brown-Spencer strengthening as explicitly stated in this
paper: every arrow of a strict group object in Cat has a two-sided categorical
inverse. This is categorical inversion, distinct from the ambient group inverse.
Together with the two structure correspondences above, this gives the stated
crossed-module correspondence with group objects in Grpd. The source supplies
no further hypothesis and does not require its cited proof as input. -/
theorem groupObjectsInCatAreGroupoids {A : Type u} {O : Type v}
    [Group A] [Group O] (D : CategoryMaps A O) (h : GroupObjectInCat D) :
    ∀ f : A, ∃ g : A,
      D.source g = D.target f ∧ D.target g = D.source f ∧
      D.compose g f = D.identity (D.source f) ∧
      D.compose f g = D.identity (D.target f) := by sorry

end QAlgorithms.Papers.Independent.Intake100.P20
Source
Group Objects and Internal Categories, arXiv:math/0212065v1, PDF p. 12, §3, Brown–Spencer strengthening
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let AAA and OOO be arbitrary types equipped with group structures, and let DDD consist of maps s,t:A→Os,t:A\to Os,t:A→O, e:O→Ae:O\to Ae:O→A, and a total function c:A×A→Ac:A\times A\to Ac:A×A→A, where c(g,f)c(g,f)c(g,f) denotes the specified composition of fff followed by ggg; this function is defined even when t(f)≠s(g)t(f)\ne s(g)t(f)=s(g). Suppose these maps satisfy all the following conditions: for every x∈Ox\in Ox∈O, s(e(x))=t(e(x))=xs(e(x))=t(e(x))=xs(e(x))=t(e(x))=x; for every f,g∈Af,g\in Af,g∈A with t(f)=s(g)t(f)=s(g)t(f)=s(g), s(c(g,f))=s(f)s(c(g,f))=s(f)s(c(g,f))=s(f) and t(c(g,f))=t(g)t(c(g,f))=t(g)t(c(g,f))=t(g); for every f∈Af\in Af∈A, c(e(t(f)),f)=fc(e(t(f)),f)=fc(e(t(f)),f)=f and c(f,e(s(f)))=fc(f,e(s(f)))=fc(f,e(s(f)))=f; and for every f,g,k∈Af,g,k\in Af,g,k∈A with t(f)=s(g)t(f)=s(g)t(f)=s(g) and t(g)=s(k)t(g)=s(k)t(g)=s(k), c(k,c(g,f))=c(c(k,g),f)c(k,c(g,f))=c(c(k,g),f)c(k,c(g,f))=c(c(k,g),f). Suppose also that multiplication preserves these maps: for every a,b∈Aa,b\in Aa,b∈A, s(ab)=s(a)s(b)s(ab)=s(a)s(b)s(ab)=s(a)s(b) and t(ab)=t(a)t(b)t(ab)=t(a)t(b)t(ab)=t(a)t(b); for every x,y∈Ox,y\in Ox,y∈O, e(xy)=e(x)e(y)e(xy)=e(x)e(y)e(xy)=e(x)e(y); and for every f,g,a,b∈Af,g,a,b\in Af,g,a,b∈A with t(f)=s(g)t(f)=s(g)t(f)=s(g) and t(a)=s(b)t(a)=s(b)t(a)=s(b), c(gb,fa)=c(g,f)c(b,a)c(gb,fa)=c(g,f)c(b,a)c(gb,fa)=c(g,f)c(b,a). Assume further that s(1A)=t(1A)=1Os(1_A)=t(1_A)=1_Os(1A​)=t(1A​)=1O​, e(1O)=1Ae(1_O)=1_Ae(1O​)=1A​, and c(1A,1A)=1Ac(1_A,1_A)=1_Ac(1A​,1A​)=1A​. Finally, suppose inversion preserves the maps: for every a∈Aa\in Aa∈A, s(a−1)=s(a)−1s(a^{-1})=s(a)^{-1}s(a−1)=s(a)−1 and t(a−1)=t(a)−1t(a^{-1})=t(a)^{-1}t(a−1)=t(a)−1; for every x∈Ox\in Ox∈O, e(x)−1=e(x−1)e(x)^{-1}=e(x^{-1})e(x)−1=e(x−1); and for every f,g∈Af,g\in Af,g∈A with t(f)=s(g)t(f)=s(g)t(f)=s(g), c(g,f)−1=c(g−1,f−1)c(g,f)^{-1}=c(g^{-1},f^{-1})c(g,f)−1=c(g−1,f−1). Then for every f∈Af\in Af∈A there exists g∈Ag\in Ag∈A such that

s(g)=t(f),t(g)=s(f),c(g,f)=e(s(f)),c(f,g)=e(t(f)).s(g)=t(f),\qquad t(g)=s(f),\qquad c(g,f)=e(s(f)),\qquad c(f,g)=e(t(f)).s(g)=t(f),t(g)=s(f),c(g,f)=e(s(f)),c(f,g)=e(t(f)).
Human review
  • Endorsed by Shuze Chen · Oct 31, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Oct 31, 2026

    Confirmed by the mission captain (proposal self-audit).

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