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Nov 3, 2020 at 15:25 vote accept CommunityBot
Nov 3, 2020 at 11:51 comment added Ruy This is Proposition 2.5 in Lance's book, but it takes a bit of interpretation. You need to take the Hilbert module $E$ mentioned there to be $E=B$, and then you need to know that $\mathcal L(E)$ is the multiplier algebra of $B$ (actually this is Lance's definition of the multiplier algebra).
Nov 3, 2020 at 10:54 comment added Ulrich Pennig This is proven if I remember correctly in the book by Lance, "Hilbert $C^*$-modules: A toolkit for operator algebraists".
S Nov 3, 2020 at 8:48 history suggested gmvh
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Nov 3, 2020 at 8:48 comment added user167952 @Ruy Do you have a reference?
Nov 3, 2020 at 8:48 review Suggested edits
S Nov 3, 2020 at 8:48
Nov 3, 2020 at 2:41 answer added Nik Weaver timeline score: 3
Nov 3, 2020 at 2:25 comment added Ruy There is a general result that says that you get your extension provided your morphism $\phi:A\to B$ satisfies $\overline{\phi(A)B}=B$. Since this holds in your case, your extension exists.
Nov 2, 2020 at 23:36 history asked user167952 CC BY-SA 4.0