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The answers to the question "Is the injective envelope functorial" resoundingly remind us that the injective envelope of a C$^*$-algebra really belongs in the category of completely positive maps.

My question then is whether there is ever a non-trivial exception:

Suppose $\mathcal A$ is unital C$^*$-algebra such that every unital $*$-homomorphism $\phi : \mathcal A \rightarrow \mathcal B$ extends to a unital $*$-homomorphism $\Phi : I(\mathcal A) \rightarrow I(\mathcal B)$.

Does this imply $\mathcal A = I(\mathcal A)$ or can one find a non-trivial exception?

Feel free to simplify this by fixing the target $\mathcal B$, maybe make it $B(\mathcal H)$.

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  • $\begingroup$ Here's a sketch for a (possible) counter-example. For $X$ with $C(X)$ injective and $x,y\in X$ not isolated, $A=\{ f \in C(X) : f(x)=f(y)\} = C(Y)$, $Y=X/x=y$, is (probably) no longer isometrically injective, but any $\ast$-homomorphism from $C(Y)$ into $B(H)$ extends on the C*-algebra of bounded Borel functions on $Y$., which (probably) naturally contains $C(X)$. $\endgroup$ Dec 2, 2021 at 23:48
  • $\begingroup$ @NarutakaOZAWA This could indeed work. However, among other questions I have, why is the injective envelope of $C(Y)$ equal to $C(X)$? $\endgroup$ Dec 3, 2021 at 22:57
  • $\begingroup$ That's because $C(X)$ is injective and there is no proper intermediate subalgebra between $C(Y)\subset C(X)$. $\endgroup$ Dec 3, 2021 at 23:12
  • $\begingroup$ @NarutakaOZAWA But the injective envelope need not be realized as a C$^*$-subalgebra of $C(X)$ but rather as an intermediate operator system with some other multiplication that turns it into a C$^*$-algebra. Or perhaps things simplify in commutative C$^*$-algebras? $\endgroup$ Dec 3, 2021 at 23:30
  • $\begingroup$ Actually, here it is fine. One can prove that there are no proper intermediate operator systems between $A$ and $C(X)$. Thus, $I(A) = C(X)$ $\endgroup$ Dec 4, 2021 at 15:27

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