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?
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.