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If I have a functor $X\to Y$ between topological groupoids with appropriate Haar measures, such that $X_0 \to Y_0$ is injective and a homeomorphism onto its image, then I should have (or rather, I hope to have) a non-degenerate $\ast$-homomorphism $C^\ast(X) \to M(C^\ast(Y))$, where the latter is the multiplier algebra of $C^\ast(Y)$. I've been hunting for a description of this map but have been unsuccessful, can anyone point me to a treatment?

Edit In fact I think I only need the case that $X_0 \to Y_0$ is in fact a homeomorphism, since I can factor the general case that I'm interested in into that case, followed by something I think I understand.


Addendum: in the special case that the groupoids have only one object, and so can be identified with groups, I think what I'm after is the (extension of the) pushforward map that takes an integrable function on the domain and gives back a measure on the codomain. But I'm aware that generalising to groupoids sometimes isn't quite so straightforward.

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  • $\begingroup$ While I'm not confident in the addendum any more, the comments before Lemma 7 in Localization of Cofibration Categories and Groupoid $C^\ast$-algebras (arxiv.org/pdf/1609.03805.pdf) claim something similar for discrete groupoids (though without mentioning the multiplier algebra, so I'm a bit wary). $\endgroup$
    – David Roberts
    Apr 10, 2018 at 5:51
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    $\begingroup$ In the discrete case the algebras are unital, so the multiplier algebra does not enter the picture. $\endgroup$ Apr 10, 2018 at 12:23
  • $\begingroup$ @MateuszWasilewski hang on, what if the set of units of the groupoid is uncountable? $\endgroup$
    – David Roberts
    Apr 10, 2018 at 22:14
  • $\begingroup$ ...or even just not finite? $\endgroup$
    – David Roberts
    Apr 11, 2018 at 1:35
  • $\begingroup$ I am sorry, you are right; I too hastily assumed that the situation in this respect is the same as for groups. $\endgroup$ Apr 11, 2018 at 14:46

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This is not a complete answer, but that was way too long for a comment:

First I started almost sure that this sort of things would be in the literature, but it seems harder than I thought to find something talking about this.

A pretty good lead is the paper "A universal property for groupoid C* algebras,I" by Alcides Buss, Rohit Holkar and Ralf Meyer which might answer your question with a little bit of work.

Their main theorem (3.23) says that given a $C^*$-algebra $D$ and a Hilbert $D$-module $F$, representations of $C^*(G)$ on $F$ corresponds to a notion of "representations of $G$ on $F$". This is an extention of the theory of integration and desintegration of representations with value in Hilbert space to representation with value in a Hilbert module.

As morphisms to the multiplier algebra $A \rightarrow M(D)$ are exactly the same as representations of $A$ on the Hilbert $D$-module $D$. This tells you precisely what you need to have in order to get such a morphisms.

I believe this is a good approach to that problem as this universal property is the simplest way to construct the map you are looking for in the case of group:

The universal property of the maximal group C* algebra $C^*(G)$ is that a morphisms from $C^*(G)$ to $M(A)$ is the same as unitary action of $G$ by multiplier on $A$. In particular, any morphism from $G$ to $H$ induced a unitary action by multiplier of $G$ on $C^*(H)$ by simply restricting the action of $H$ and this corresponds to a morphisms $C^*(G) \rightarrow M(C^*(H))$.

Now I haven't read their paper in much details yet so I'm not totally sure how their notion of 'representation' would behave with respect to composition of morphisms of groupoids, but it should anyway give you a good understanding of when this kind of morphisms exists or not.

I'm especially a little worried in general about a possible condition of "compatiblity" of the Haar measure on the source and the target which might makes the results false in general, but true in lots of cases (like étale groupoids, Lie groupoids, or groups, where the choice of a Haar measure is not really a problem)

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  • $\begingroup$ Thanks, Simon! This helps. I think I managed to see the idea of how this is supposed to work in the group case after a long discussion with a local, and him trying to track down useful references. The compatibility issue for measures is of course where the groupoid case differs, since there isn't a uniqueness result as for locally compact groups. $\endgroup$
    – David Roberts
    Apr 12, 2018 at 22:57

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