Let $A\rightrightarrows X$ be a groupoid, where $X$ is the set of objects and $A$ is the set of arrows. 
My favorite example of a groupoid is an action groupoid. If a group $G$ acts on the left on a set $X$,
we set 
$$
A=\{(x,g,y)\mid x,y\in X, g\in G,\ y=g\*x\},
$$
then $A\rightrightarrows X$ with the evident maps is called the action groupoid corresponding to the action of $G$ on $X$. 
It is often denoted by $G\ltimes X$.


Let $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ be a morphism of groupoids (a functor). 
We say that $F$ is an *equivalence of groupoids* if it is an equivalence of categories.


Let $x\in X$. We denote by $A(x)$ the set of arrows $a\colon x\to x$. 
We denote by $\pi\_0(X)$ the set of connected components of $X$ 
(i.e., the set of equivalence classes in $X$ with respect to the equivalence relation induced by $A$).
We say that a morphism $F$ as above is a *weak equivalence of groupoids* (or a *quasi-isomorphism*) if 
$\pi\_0(F)\colon \pi\_0(X)\to \pi\_0(Y)$ is a bijection and, for any $x\in X$, the induced homomorphism
$F\_x\colon A(x)\to B(y)$ is an isomorphism, where $y=F(x)$.



> 
> **Question 1.** Is it true that any weak equivalence of groupoids is an equivalence?
> 
> 
> 


Now assume that a group $\Gamma$ acts on our groupoid $A\rightrightarrows X$. We say that $A\rightrightarrows X$ is a $\Gamma$-groupoid.
My favorite example of a $\Gamma$-groupoid comes from an action of an algebraic group $\mathcal{G}$, defined over a field $k$, on a $k$-variety 
$\mathcal{X}$. Let $k\_s$ denote a separable closure of $k$, then we set $\Gamma:={\rm Gal}(k\_s/k)$, and we consider the action groupoid 
$\mathcal{G}(k\_s)\ltimes\mathcal{X}(k\_s)$, on which $\Gamma$ acts.


By a *weak equivalence of $\Gamma$-groupoids* we mean a $\Gamma$-functor $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ 
that is a weak equivalence of groupoids. 
By an *equivalence of $\Gamma$-groupoids* we mean a $\Gamma$-functor $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$
such that there exists a a $\Gamma$-functor $F'$ in the opposite direction and each of the composite functors $F\circ F'$ and $F'\circ F$ is 
$\Gamma$-naturally-isomorphic to the corresponding identity functor.



> 
> **Question 2.** Is it true that any weak equivalence of $\Gamma$-groupoids is an equivalence of $\Gamma$-groupoids?
> 
> 
> 


I expect the answer "No" to Question 2, but I cannot construct a counter-example.



Equivalence and weak equivalence of groupoids
The answer to question 1 is “yes” since every weak equivalence is essentially surjective (let $y\in Y$, $[y]$ the connected component of $y$, which is in the image of $\pi\_0(F)$, thus there exists $y^\prime\in [y]$ which is an element of the image of $F$ and isomorphic to $y$) and full and faithful (since $F\_x$ is an isomorphism, the mappings $F\_{xy}\colon A(x,y)\to B(F(x),F(y))$ are also bijective, choose $f\in A(y,x)$ then $\alpha\colon A(x,y)\to A(x), g\mapsto fg$ and $\beta\colon B(F(x),F(y))\to B(F(x)), g\mapsto F(f)g$ are bijections and $F\_{xy}=\beta^{-1}\circ F\_x\circ \alpha$ is bijective).


As a counter example for question 2 consider: $\Gamma=C\_2$, $X$ the “complete” (in the sense of graph theory) groupoid consisting of two objects and four isomorphisms. The action of $\Gamma$ on $X$ is defined to swap the objects. Let $Y$ be the terminal groupoid consisting of a single isomorphism. $\Gamma$ acts trivially on $Y$. The unique morphism $X\to Y$ is a weak equivalence of $\Gamma$-groupoids. But there does not exist a $\Gamma$-morphism $Y\to X$.


