Let $G$ be a group acting on a topological space $X$, then the quotient map $X \to X/G$ is open.
I want to ask, whether this fact generalizes to orbit spaces of groupoids. More precisely:
Let $G$ be a topological groupoid. I will denote objects by $G_0$, morphisms by $G_1$, the source map by $s$ and the target map by $t$.
There is an equivalence relation $\sim$ on $G_0$:
$$x\sim y \text{ iff. there exists a morphism } g \in G_1 \text{ such that } s(g) = x \text{ and } t(g) = y.$$
It is easy to check that this is an equivalence relation. The class corresponding to $x$ will be called its orbit. Note that, the orbit of $x$ contains all those points in $G_0$ which are targets of morphisms originating at $x$, or in other words, the orbit of $x$ is the set $t(s^{-1}(x))$.
Then, $G_0/\sim$ equipped with the quotient topology
is called the orbit space of the groupoid $G$.
My question is: is the quotient map $G_0 \to G_0/\sim$ open?
Remark. We can recover the original statement about quotient by a group, by considering the translation gropoid $G \ltimes X$ (since its orbit space is precisely $X/G$).
I haven't made any progress in the general case.
EDIT: I was making things unnecessarily complicated. At least, for the étale case, the proof is laughably trivial; its just a point-set argument similar to what we do in the group case.
First of note that if $p : X \to X/\sim$ is a quotient map and $A\subset X$, then $p^{-1}(p(A))$ is the set of all the elements of $X$ which are related by $\sim $ to some element in $A$.
Now, let $G$ be an étale topological groupoid with orbit space $G_0/\sim$ and quotient map $q : G_0 \to G_0/\sim$. Suppose $U$ is an open subset of $G_0$. We need to show that $q(U)$ is open. It suffices to show that $q^{-1}(q(U))$ is open because $q$ is a quotient map. By the preceding observation, $q^{-1}(q(U))$ is the set of those elements $G_0$ which are in the orbit of some element $U$, or in other words, it is the set of all points in $G_0$ which are targets of morphisms originating in $U$. Thus, $q^{-1}(q(U)) = t(s^{-1}(U))$, which is open because both $s$ and $t$ are local homeomorphism.
So, that does it for the étale case.
Best Answer
Actually, in the proof above, étale requirement is an overkill, since all we ever used was that $t$ is open. So, we actually proved the result for the groupoids which have target map open. nLab calls such groupoids, open topological groupoids. (Note that if the target map is open, then so is the source one).
In particular, the statement is always true for Lie groupoids.
Now, coming to the general case: the assertion is false.
The point is, any equivalence relation on a topological space $X$ can be realized as a topological groupoid (as described in 'Topological groupoids quantales' by Palmigiano and Re). That is to say, if $R \subset X\times X$ is an equivalence relation, then, $R$ can be though of as a groupoid over $X$, with the projection maps (restricted to $R$) as the source and the target maps. The orbit space of this groupoid, $X/\sim$ is precisely the quotient space $X/R$.
So, if $R$ was a non-open relation (ie., the quotient map $X \to X/R$ is not open). Then if we think of $R$ as a groupoid over $X$, then the quotient map $X \to X/\sim$ is not open. For example consider the quotient space $[0, 1]/(0\sim 1)$. We know that the associated quotient map is not open. Thus, the quotient map for the orbit space of the associated groupoid is not open.