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Let $\mathcal G$ be an étale groupoid with a locally compact, Hausdorff unit space $\mathcal G^{(0)}$. If $U⊆\mathcal G$ is an open subset, which is Hausdorff in the induced topology, and if $f$ is any member of $C_c(U)$ (continuous, compactly supported, complex valued functions on $U$), one may view $f$ as a function defined on $\mathcal G$ by extending it to be zero on $\mathcal G\setminus U$. In that way $C_c(U)$ may be viewed as a subset of the set of all complex valued functions on $\mathcal G$.

In order to define the C*-algebra of $\mathcal G$, one first looks at the set $C_c(\mathcal G)$ linearly spanned by the union of all $C_c(U)$, where $U$ is an open, Hausdorff subset of $\mathcal G$, as above.

One of the trickiest issues of that theory is that the members of $C_c(\mathcal G)$ are not necessarily continuous functions on $\mathcal G$! En passant, many people have rightly objected that the notation $C_c(\mathcal G)$ is somewhat inappropriate!

Question: Suppose that some $f$ in $C_c(\mathcal G)$ vanishes outside the unit space $\mathcal G^{(0)}$. Is $f$ necessarily continuous on $\mathcal G^{(0)}$?

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    $\begingroup$ It seems if you take my favorite non-Hausdorf etale groupoid, where you take a compact space $X$ and then turn some non-isolated point into a group, then everything that vanishes outside $G^0$ is continuous on $G^0$. But I haven't quite been able to generalize the technique for this example to the general case so maybe one just needs to look at more complicated non-Hausdorff groupoids. $\endgroup$ Commented Oct 5, 2018 at 19:45
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    $\begingroup$ If it was true the argument should be something like if $f\in C_c(G)$ and we define $\Phi(f)$ to be the function on $G^0$ that sends $x$ to the sum of $f(g)$ where $g$ ranges over isotropy group elements that are limits of sequences converging to $x$, then we would like $\Phi(f)$ to be continuous. If it was, then clearly $\Phi(f)=f$ for all $f$ vanishing outside $\mathcal G^0$ and so $f$ would be continuous. This argument works for the above example but I don't know in general. $\endgroup$ Commented Oct 5, 2018 at 19:48
  • $\begingroup$ @Ben, that is a nice idea. I'll think about it. $\endgroup$
    – Ruy
    Commented Oct 6, 2018 at 4:27
  • $\begingroup$ I think I can prove it for inverse semigroup universal groupoids. $\endgroup$ Commented Oct 6, 2018 at 10:39
  • $\begingroup$ Hi @Ben, I think $\Phi(f)$ is not necessarily continuous. A counter-example is given by the function $f_x$ appearing in page 899 of [Non-Hausdorff etale groupoids, PAMS vol 139 (2011), 897-907]. $\endgroup$
    – Ruy
    Commented Oct 6, 2018 at 21:12

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Here is a positive answer to my own question based on Ben's comment above. For each $f$ in $C_c(\mathcal G)$, and for each $x$ in $\mathcal G^{(0)}$, define $$ \Phi(f)|_x = \sum_{\gamma\in s^{-1}(x)} f(\gamma), $$ where $s$ denotes the source map. I next claim that $\Phi(f)$ is continuous on $\mathcal G^{(0)}$.

By the definition of $C_c(\mathcal G)$, we may assume that $f$ lies in $C_c(V)$, where $V$ is some open Hausdorff subset of $\mathcal G$.

Recalling that the open bissections form a basis for the topology of $\mathcal G$, and using a partition of unity argument, we may further assume that the compact support of $f$ is contained in some open bissection $U\subseteq V$.

For $x$ in $s(U)$ there exists a unique $\gamma$ in $U$ for which $s(\gamma)=x$, namely $\gamma=s^{-1}(x)$, where $s^{-1}$ is the inverse of the restriction of $s$ to $U$, which is a homeomorphism because $U$ is a bissection. One then sees that $$ \Phi(f)|_x = \left\{ \matrix{ f(s^{-1}(\gamma)), & \hbox{if } x\in s(U), \cr 0, & \hbox{otherwise,}} \right. $$ from where it easily follows that $\Phi(f)$ is continuous.

To conclude, pick $f$ in $C_c(\mathcal G)$ vanishing outside $\mathcal G^{(0)}$. Then one easily checks that $f=\Phi(f)$, hence $f$ is continuous.

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