Is the injective envelope functorial? 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.
 A: 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.
A: 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.
