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Apr 12, 2018 at 12:24 answer added Simon Henry timeline score: 2
Apr 12, 2018 at 11:35 comment added David Roberts @Simon yes, I only care in the case that X_0 is a disjoint summand of Y_0 (so a special case of taking a topological cofibration), but I think I can reduce to the case that the functor is the identity on objects.
Apr 12, 2018 at 8:21 comment added Simon Henry Don't you also need $X_0$ to be open in $Y_0$ ? Other wise I don't see how you hope to get that for "discrete" groupoids (I mean discrete in the sense of $G_1=G_0$).
Apr 12, 2018 at 8:16 comment added Simon Henry The reason Lemma 7 of the reference you quote does not use multiplier is because they only construct that morphism in the case of a cofibrations of groupoid, which are injective on objects.
Apr 12, 2018 at 6:05 history edited David Roberts CC BY-SA 3.0
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Apr 11, 2018 at 14:46 comment added Mateusz Wasilewski I am sorry, you are right; I too hastily assumed that the situation in this respect is the same as for groups.
Apr 11, 2018 at 1:35 comment added David Roberts ...or even just not finite?
Apr 10, 2018 at 22:14 comment added David Roberts @MateuszWasilewski hang on, what if the set of units of the groupoid is uncountable?
Apr 10, 2018 at 12:23 comment added Mateusz Wasilewski In the discrete case the algebras are unital, so the multiplier algebra does not enter the picture.
Apr 10, 2018 at 5:51 comment added David Roberts 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).
Apr 7, 2018 at 3:19 history asked David Roberts CC BY-SA 3.0