5
$\begingroup$

This is same as saying a connected groupoid is equivalent to a group. But I have no idea how to construct such a group, can't even get started.

Any help is appreciated.

$\endgroup$
2
  • $\begingroup$ What do you mean by equivalent? $\endgroup$ – Dori Bejleri Oct 15 '14 at 2:16
  • 1
    $\begingroup$ equivalence of categories $\endgroup$ – user112564 Oct 15 '14 at 2:23
12
$\begingroup$

Let $\mathcal{Gr}$ be your groupoid category. Take any object $X \in \mathcal{Gr}$. Consider the subcategory $G$ consisting of just the single object $X$ and with $\operatorname{Hom}_G(X,X) = \operatorname{Hom}_\mathcal{Gr}(X,X)$. The category $G$ is a group because every element of $\operatorname{Hom}_G(X,X)$ can be composed by any other element, they're all invertible because $\mathcal{Gr}$ is a groupoid, and there is an identity morphism $X \to X$.

Now consider the inclusion functor $G \hookrightarrow \mathcal{Gr}$. This functor is fully faithful by construction since $\operatorname{Hom}_G(X,X) = \operatorname{Hom}_\mathcal{Gr}(X,X)$, and it is essentially surjective since $\mathcal{Gr}$ is is connected so for any $Y \in \mathcal{Gr}$, there exists a morphism $X \to Y$ which is necessarily an isomorphism since $\mathcal{Gr}$ is a groupoid. Therefore the inclusion is an equivalence of categories.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks. Did you use some theorems about equivalence of categories? Like essentially surjective + fully faithful implies equivalence? I don't know any of these theorems so I am a little puzzling.. $\endgroup$ – user112564 Oct 15 '14 at 5:02
  • 4
    $\begingroup$ @user112564 Yes, (assuming the axiom a choice) a functor is an equivalence iff it is fully faithful and essentially surjective. See this for example: en.wikipedia.org/wiki/… $\endgroup$ – Najib Idrissi Oct 15 '14 at 7:30
4
$\begingroup$

Using a different terminology, but basically similar reasoning as Dori:

Every category is equivalent to its skeleton

Every connected groupoid is a category

In a skeleton you fuse together all isomorphic objects, therefore the skeleton of a connected groupoid is a category with a single element and invertible morphisms (a group)

So every connected groupoid is equivalent to a group

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.