Some statements that are true for ordinary groupoids fail for topological groupoids (by which I mean groupoids internal to the category of topological spaces): for instance, every ordinary groupoid is equivalent to the disjoint union of one-object groupoids, but <a href="https://math.stackexchange.com/questions/2954664/is-every-topological-groupoid-equivalent-to-a-disjoint-union-of-topological-grou">not every topological groupoid is equivalent to a disjoint union of topological groups</a>.

Thinking about a different statement that is true for ordinary groupoids (each ordinary groupoid is equivalent to a skeletal groupoid) and <a href="http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=756C6C16BD72D7117B5F1AAA6D1679D1?doi=10.1.1.12.6700&rep=rep1&type=pdf">Butz-Moerdijk's paper</a> has led me to the following questions:

1. Is every topological groupoid equivalent to a skeletal topological groupoid?
 
     For the record, a *functor* between two topological groupoids $\mathcal C$ and $\mathcal D$ is an ordinary functor $$(\mathcal F_0\colon \rm Ob(\mathcal C)\to \rm Ob(\mathcal D), \mathcal F_1\colon \rm Mor(\mathcal C)\to\rm Mor(\mathcal D))$$ between the underlying ordinary groupoids such that $\mathcal F_1$ and $\mathcal F_2$ are continuous. A *natural transformation* between two functors $\mathcal F, \mathcal G\colon \mathcal C\to\mathcal D$ of topological groupoids is a natural transformation $$\eta\colon \rm Ob(\mathcal C)\to\rm Mor(\mathcal D)$$ between the underlying ordinary functors such that $\eta$ is continuous. We say $\eta$ is a *natural isomorphism* if there is a natural transformation $\epsilon\colon \mathcal G\to\mathcal F$ such that for all $X\in\rm Ob(\mathcal C)$, $\eta_X\circ \epsilon_X=\rm id$ and $\epsilon_X\circ \eta_X=\rm id$. Two topological groupoids $\mathcal C$ and $\mathcal D$ are said to *equivalent* if there are functors $\mathcal F\colon\mathcal C\to\mathcal D$ and $\mathcal G\colon\mathcal D\to\mathcal C$ such that $\mathcal F\circ \mathcal G$ and $\mathcal G\circ\mathcal F$ are naturally isomorphic to the identity functors.
2. It feels to me the following condition could be relevant in (1): say that two points $x$ and $y$ of a topological space $X$ are *topologically equivalent* if for each open set $U\subseteq X$, $x\in U$ if and only if $y\in U$.

     If a topological groupoid is equivalent to a skeletal topological groupoid, are isomorphic objects topologically equivalent?

     (Side questions: Is there a name in the literature for "topologically equivalent" points? Has the condition "all topologically equivalent points are isomorphic" any relevance in the study of topological groupoids?)

3. Suppose the anwer to the first question in (2) is "yes". Let $\mathcal C$ and $\mathcal D$ be topological groupoids in which isomorphic objects are topologically equivalent. Are $\mathcal C$ and $\mathcal D$ equivalent if and only if their skeletons are *isomorphic* as topological groupoids?

4. For a topological groupoid $\mathcal C$ is there a canonical site $(C_\mathcal C, J_\mathcal C)$ such that the category of equivariant sheaves on $\mathcal C$ is equivalent to the category of sheaves on $(C_\mathcal C, J_\mathcal C)$?

    If two topological groupoids $\mathcal C$ and $\mathcal D$ are equivalent, does it follow that the sites $(C_\mathcal C, J_\mathcal C)$ and $(C_\mathcal D, J_\mathcal D)$ are equivalent? Does the converse hold?

5. If $\mathcal E$ and $\mathcal E'$ are equivalent topoi, is the Butz-Moerdijk groupoid associated to $\mathcal E$ *equivalent* or *isomorphic* to the Butz-Moerdijk groupoid associated to $\mathcal E'$, as topological groupoids?


**Remark on how to answer.** I feel a bit guilty for writing such an overwhelming amount of questions. (1)-(3) have mainly the purpose to make sure that there aren't any pitfalls or phenomena that occur in the topological groupoid world that don't occur in the ordinary groupoid world (it seems this is a topic one could easily make mistakes). So for these questions a simple yes/no would totally suffice - I can work out the proofs as an exercise.