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Let $A$ and $B$ be unital $C^*$-algebras, so we can view these as operator systems, and it makes sense to consider their injective envelopes $I(A)$ and $I(B)$. These injective envelopes become $C^*$-algebras for the Choi-Effros product.

Given a unital $*$-morphism $f: A \to B$, is it true that there exists a unique unital $*$-morphism $\overline{f}: I(A) \to I(B)$ that extends $f$?

Once the above question is answered positively (if the answer is positive), the following will probably be easy:

Is this construction functorial? I.e. is $I(-)$ a functor from the category of unital $C^*$-algebras to the category of unital $C^*$-algebras (with morphisms unital $*$-homomorphisms?

A reference is more than enough for me to be satisfied with an answer.

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One can view $A$ and $B$ as sitting completely isometrically inside their injective envelopes $I(A)$ and $I(B)$. Then by injectivity a unital *-homomorphism (or more generally a unital completely positive map) $f:A\rightarrow B\subseteq I(B)$ extends to a unital completely positive map $\overline f:I(A) \rightarrow I(B)$.

[Edit: this should work]

Paulsen in this paper, Proposition 3.5, points out that any C$^*$-algebra containing $K(H)$ has injective envelope $B(H)$. Then $A = K(H) + \mathbb C I$ has $I(A) = B(H)$.

Consider the $*$-homomorphism $f:A\rightarrow \mathbb C$ given by $f(k+\alpha I) = \alpha$. Note that $I(\mathbb C) = \mathbb C$ and that any state of the Calkin algebra $B(H)/K(H)$ precomposed with the quotient map $q:B(H)\rightarrow B(H)/K(H)$ extends the map $f$. Therefore, $\overline f$ need not be a $*$-homomorphism or unique.

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  • $\begingroup$ Thanks. I also got this far. I can show that if $f$ is a $*$-isomorphism, then $\overline{f}$ is also a $*$-isomorphism, but my proof uses a big gun. $\endgroup$
    – Andromeda
    Dec 1 '21 at 21:56
  • $\begingroup$ The fact that nothing is mentioned in Paulsen's book or any of the foundational papers makes me suspicious. The injective envelope is a very slippery beast. $\endgroup$ Dec 1 '21 at 21:59
  • $\begingroup$ Indeed, but Paulsen's treatment on the topic lacks some fundamentals (for example, he does not prove/mention that the $C^*$-algebra structure on an injective operator system obtained from the Choi-Effros product is unique, but uses this implicitly later in the chapter). $\endgroup$
    – Andromeda
    Dec 1 '21 at 22:04
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    $\begingroup$ I know that the above answers the problem in the negative, but (because I'm curious) what if one restricts to unital $\ast$-monomorphisms $f\colon A \to B$? By Hamana's construction of injective envelopes, we obtain an embedding $I(A) \to I(B)$ which extends $f: A \to B \subseteq I(B)$ so a $\ast$-homomorphism exists in this case (which was the obstruction above). I doubt it is unique though. $\endgroup$
    – Jamie Gabe
    Dec 1 '21 at 22:34
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    $\begingroup$ Oh no, $I(A)$ sits inside $I(B)$ as an operator system, not as a C*-subalgebra in this case! Thanks for clearing that up for me! $\endgroup$
    – Jamie Gabe
    Dec 1 '21 at 22:48
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As mentioned by Chris, injective envelopes are brutal when seen as C$^*$-algebras.

Let $A=\text{UHF}(2^\infty)$ and $B$ the hyperfinite II$_1$ factor. Take $f$ to be the inclusion map. We have $I(B)=B$, while $I(A)$ is a wild AW$^*$ factor of type III.

If $g:I(A)\to B$ is a $*$-homomorphism and $\tau$ is the trace on $B$, then $\gamma=\tau\circ g$ is a trace on $I(A)$. In a type III AW$^*$-factor any projection $p$ can be halved, so there exist $p_1,p_2$ with $p=p_1+p_2$ and $p\sim p_1\sim p_2$, which gives us the usual
$$ \gamma(p)=\gamma(p_1)+\gamma(p_2)=2\gamma(p), $$ and so $\gamma(p)=0$ for any projection $p$. Thus $\gamma=0$. As $\tau$ is faithful, $g=0$.

In summary, $f$ is a $*$-monomorphism that admits no extension to a $*$-homomorphism, and in fact the only $*$-homomorphism $I(A)\to I(B)$ is the zero homomorphism.

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