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. 



Groupoid as a 2-coequaliser
Your claim is incorrect because you truncated the simplicial diagram too much. Indeed, if what you said were true, then the isomorphism class of a group would be determined by its cardinality, but this is obviously not true. 


What you are missing is the data that tells us about composition. We need to look at the diagram


$$G\_2 \mathrel{\hbox{$\begin{matrix} \smash{\rightarrow} \newline \smash{\rightarrow} \newline \smash{\rightarrow} \end{matrix}$}} G\_1 \mathrel{\hbox{$\begin{matrix} \smash{\rightarrow} \newline \smash{\leftarrow} \newline \smash{\rightarrow} \end{matrix}$}}G\_0$$
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 $\mathcal{C}$.


