# Groupoid as a 2-coequaliser

Let $G=(G_1, G_0, s, t, u, i,\circ)$ be a groupoid, where $s, t$ are source and target maps, $i$ is the inverse, $u$ is the unit, and $\circ$ is the composition.

Denote $\underline{G_1}, \underline{G_0}$ the trivial groupoid on $G_1, G_0$ respectively. There are morphisms of Lie groupoids $s:\underline{G_1}\to \underline{G_0}$ and $t: \underline{G_1}\to \underline{G_0}$, and $u:\underline{G_0}\to G$. There is a natural transformation $us\Rightarrow ut:\underline{G_1}\to G$ given by $id: G_1\to G_1$.

Is it true that the 2-commutative diagram

$$\underline{G_1}⇉^s_t \underline{G_0}\xrightarrow{u} G$$

is a 2-coequaliser diagram in the 2-category of groupoids?

If this is not the case, how to correct it? I learn from Thomason’s homotopy colimit theorem that the 2-colimit of the nerve $\Delta^{op}\to NG$ and $G$ has the same homotopy type. A relevant statement is a quotient stack is a 2-coequalizer.

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If it were true, then $G$ would not depends on the composition law... this would be weird, doesn't it ? If you add the data in degree 2 then it work. –  Simon Henry Jan 17 '14 at 13:06
@SimonHenry The composition law show up in the natural transformation. –  Ma Ming Jan 17 '14 at 14:29
@SimonHenry I see the point. It did not show up, and my guess should be wrong (Let $G_0=pt$, consider a space with 2 different group structures.) –  Ma Ming Jan 17 '14 at 14:45

$$G_2 \mathrel{\hbox{\begin{matrix} \smash{\rightarrow} \newline \smash{\rightarrow} \newline \smash{\rightarrow} \end{matrix}}} G_1 \mathrel{\hbox{\begin{matrix} \smash{\rightarrow} \newline \smash{\leftarrow} \newline \smash{\rightarrow} \end{matrix}}}G_0$$ where $G_2$ is the set of composable pairs of morphisms in $\mathcal{G}$, and the arrows are the evident face and degeneracy operators of the nerve of $\mathcal{G}$. Let $\mathcal{C}$ be the pseudocolimit of this diagram. By definition, that means the exponential $[\mathcal{C}, \mathcal{D}]$ is isomorphic to the pseudolimit of the following diagram: $$[G_2, \mathcal{D}] \mathrel{\hbox{\begin{matrix} \smash{\leftarrow} \newline \smash{\leftarrow} \newline \smash{\leftarrow} \end{matrix}}} [G_1, \mathcal{D}] \mathrel{\hbox{\begin{matrix} \smash{\leftarrow} \newline \smash{\rightarrow} \newline \smash{\leftarrow} \end{matrix}}} [G_0, \mathcal{D}]$$ There is a natural comparison functor $[\mathcal{G}, \mathcal{D}] \to [\mathcal{C}, \mathcal{D}]$, and if you squint for a while you will see that it is (half of) an equivalence of categories (though usually not an isomorphism). Thus $\mathcal{G}$ is equivalent to $\mathcal{C}$.