A connected groupoid is equivalent to a groupoid with one object

Question

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.

Answer 1

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.

Answer 2

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