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 not every topological groupoid is equivalent to a disjoint union of topological groups.
Thinking about a different statement that is true for ordinary groupoids (each ordinary groupoid is equivalent to a skeletal groupoid) and Butz-Moerdijk's paper has led me to the following questions:
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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.
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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?)
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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?
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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?
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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.
Best Answer
First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the "right" notion of Morita equivalence of topological groupoids.
Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.
Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta_x = (s\pi(x),x)$.
Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continuous section of $\pi$.