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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:

  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.

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    $\begingroup$ People often use a different definition of equivalence for topological groupoids because (1) is false. See for example mathoverflow.net/questions/111801/… $\endgroup$ Commented Mar 15, 2022 at 18:53
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    $\begingroup$ Regarding Q4: Well, the topos itself is a a site. 2-functoriality of the equivariant sheaves construction ensures that (strongly) equivalent topological groupoids yield the same topos. The converse is false due to the Morita equivalence phenomenon. $\endgroup$
    – Zhen Lin
    Commented Mar 15, 2022 at 22:20
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    $\begingroup$ Regarding Q5: It occurs to me that there is not even "the" Butz–Moerdijk groupoid associated to one topos. There are a number of arbitrary choices that have to be made: first, you have to choose a pre-bound, then you have to choose a jointly surjective set of points, and then you have the choose a sufficiently large cardinal. It is not obvious to me that the groupoid you get is independent of these choices in any meaningful sense other than Morita equivalence. $\endgroup$
    – Zhen Lin
    Commented Mar 16, 2022 at 7:59
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    $\begingroup$ @user478652: Section 3.2 explains the difference between the morphisms you described and the morphisms Benjamin Steinberg pointed out above. $\endgroup$ Commented Mar 16, 2022 at 9:22
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    $\begingroup$ Yes, I am referring to the notion of equivalence you defined in Q1. Other notions considered in this context are weaker, so this one is "strong". $\endgroup$
    – Zhen Lin
    Commented Mar 16, 2022 at 16:59

2 Answers 2

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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 $\pi$.

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  • $\begingroup$ Thanks for the example! But I don't agree with "note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use" without further evidence. I just wrote down the canonical way of talking about internal equivalences for internal categories. "Moerdijk has papers with the right notion of Morita equivalence of topological groupoids." - As you say yourself, this notion has a different name, "Morita equivalence", but I wasn't talking about Morita equivalence! $\endgroup$
    – user478652
    Commented Mar 16, 2022 at 9:08
  • $\begingroup$ @Benjamin Steinberg: in the third paragraph, do you mean "skeletal subcategory equivalent to $\mathcal{G}$"? $\endgroup$ Commented Mar 16, 2022 at 9:37
  • $\begingroup$ Note that $\mathcal{G}$ in this example is always Morita equivalent to $X/R$. $\endgroup$ Commented Mar 16, 2022 at 9:39
  • $\begingroup$ @KonradWaldorf, thanks for the correction $\endgroup$ Commented Mar 16, 2022 at 11:03
  • $\begingroup$ @BenjaminSteinberg Can you say anything about questions (2) and (3)? $\endgroup$
    – user478652
    Commented Mar 20, 2022 at 17:55
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Maybe not quite what you want, but a "locally trivial" topological groupoid always turns out to be Morita equivalent to a skeletal topological groupoid.

A topological groupoid $X := (X_1\rightrightarrows X_0)$ is said to be locally trivial if for every point $p\in X_0$ there is a neighbourhood $U$ of $p$ and a lift of the inclusion $\{p\}\times U \hookrightarrow X_0 \times X_0$ through $(s,t)\colon X_1\to X_0\times X_0$. This definition is due to Ehresmann, and dates from 1959.

For a topological groupoid $X$, denote by $X[X_0^\delta]$ the topological groupoid with the same objects with the discrete topology, and the same hom spaces, so that the morphism space is now a disjoint union of the hom-spaces, indexed by pairs of objects. There is a canonical continuous functor $X[X_0^\delta]\to X$, and this is a Morita equivalence if and only if $X$ is locally trivial (this is Lemma 4.35 in my thesis, I've not seen it elsewhere, but it might have been known).

Now we can choose a single representative from each isomorphism class in $X$, and take the full subgroupoid of $X[X_0^\delta]$ on those objects. This skeletal topologial groupoid is equivalent (in the usual, strong sense) to $X[X_0^\delta]$, and a fortiori is Morita equivalent to the original groupoid.

Now it's possible that a topological groupoid is skeletal without having a discrete space of objects (take, for instance, any bundle of groups on any space eg the trivial one). But if a topological groupoid is Morita equivalent to a skeletal topological groupoid with a discrete space of objects, then we can reverse the above argument, and it should be locally trivial.

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