$\DeclareMathOperator\Hom{Hom}$For which unital $C^{\ast}$-algebras $A$ does it hold that for all compact Hausdorff $S$ we have the bijection:
\begin{align\*}
\Hom(A, C(S)) \cong \Hom(S, \Hom (A, \mathbb{C}))?
\end{align\*}
This holds for all commutative unital $C^{\ast}$, but it seems to hold for all $C^{\ast}$ algebras with $\Hom(A, \mathbb{C}) = \varnothing$.


Does this hold in general? Are there some easy counter-example? Are there some characterizations of the algebras for which this property holds?



Adjunction via Gelfand duality
$\DeclareMathOperator\Hom{Hom}$


Yes, this is true, and the proof is elementary: let us write $\Omega(A):=\Hom(A,\mathbb{C})$ for the space of characters of $A$, viewed as a subspace of the unit ball of the dual $A^\*$, and endowed with the weak\*-topology (i.e., the topology of pointwise convergence). This is a compact Hausdorff space if $A$ is a unital $C^\*$-algebra (in general it is locally compact Hausdorff, if $A$ is not unital).


We claim that there exists a natural bijection
$$\Hom(A,C(S))\cong \Hom(S,\Omega(A)),$$
where the left hom-set is in the category of unital $C^\*$-algebras with unital $\*$-homomorphisms, and the right hom-set is in the category of compact Hausdorff spaces.


Given $f\in \Hom(A,C(S))$, we define $\tilde f\in \Hom(S,\Omega(A))$ by $\tilde f(s)(a):=f(a)(s)$ for all $s\in S$, $a\in A$. It is clear that
$$\Hom(A,C(S))\ni f\mapsto \tilde f\in \Hom(S,\Omega(A))$$
is a well-defined injective map. To see that it is surjective, take $g\in \Hom(S,\Omega(A))$ and define $f(a)(s):=g(s)(a)$. Then $\tilde f=g$, so that the map above is also a surjection, therefore a bijection, as desired.


