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Let $A$ and $B$ be $C^*$-algebras. Suppose that there exists a projection $p$ in $\mathcal{M}(B)$, the multiplier algebra of $B$, such that $A=pBp$. That is, $A$ is a corner of $B$.

Question: Is it true that the corner $p(\mathcal{M}(B))p$ contains the multiplier algebra of $A$ (which we view as a subalgebra of $B$)? In other words, does every multiplier of the corner come from the corner of the multiplier algebra?

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Yes. We have $$ M(A) = M(pBp) = pBp \subseteq pM(B)p. $$ The second equality holds since $pBp$ is a unital C*-algebra.

Corrected version: There is an obvious map $\iota \colon pBp \hookrightarrow p M(B)p$. This inclusion extends to the multiplieralgebra $\bar \iota \colon M(pBp) \to pM(B)p$. The extension exists since $\iota$ is non-degenerate. Indeed, if $(e_n)$ is an approximate unit for $B$, then $pe_np \to p$ strictly, which is the unit of $pM(B)p$. Furthermore, it is clear that $\bar \iota$ is still injective.

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  • $\begingroup$ Why must $pBp$ be unital if we start off with $B$ non-unital? $\endgroup$
    – geometricK
    Jun 10, 2019 at 14:22
  • $\begingroup$ Ah okay, I made a mistake. It is not the case that that $pBp$ is strictly closed in $M(B)$, e.g. $ \mathbb K = 1_{B(H)} \mathbb K 1_{B(H)}$. $\endgroup$
    – Epsilon
    Jun 10, 2019 at 14:49
  • $\begingroup$ I have corrected the proof. Thanks for pointing out. $\endgroup$
    – Epsilon
    Jun 10, 2019 at 14:56

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