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Nov 23, 2016 at 20:50 comment added Adrián González Pérez My apologies, @Ruy. I missread your argument. I just read the beginning and thought that you were using that if $A$ is exact and $\alpha$ amenable, $A \rtimes_\alpha G$ is exact and that $C^\ast G$ is not exact for nonamenable $G$. You cannot embbed a non exact algebra in an exact one. Your argument works fine since we are not considering all possible embeddings, just $\iota_G$. My bad.
Nov 23, 2016 at 18:58 comment added Ruy @Adrian, if $A$ is unital, the argument works regardless of exactness. In the nonunital case I guess one would have to extend the conditional expectation to the multiplier algebra and, if this works, exactness of $A$ will not be needed. Can you explain your argument?
Nov 23, 2016 at 14:58 comment added Adrián González Pérez If I am not mistaken your argument shows that $\iota_G$ is never injective when $A$ is exact and $\alpha$ is an amenable action of a nonamenable group. It will also be interesting to know whether $\iota_G$ can be injective for nonamenable actions.
Sep 15, 2016 at 14:09 vote accept Eusebio Gardella
Sep 14, 2016 at 21:16 comment added Ruy Perhaps I should add that, by a result of Ozawa (arXiv:math/0002185), every exact group, such as $F_2$, admits an amenable action on a compact space.
Sep 14, 2016 at 21:06 history answered Ruy CC BY-SA 3.0