I am finding difficulties in finding a counterexample that if $f:A\to B$ is a homomorphism of $C^*$algebras A and B (which means: f is linear and multiplicative) and let f be isometric, this implies that f preserves involution.
Maybe we could take A and B =$\mathbb{C}$ or the tensorproduct of $\mathbb{C}$, endow A and B with a suitable involution and take f=identity to obtain a counterexample. But I don't know what to take exaktly. Or if someone knows an other example, I'm interested in it (often you obtain counterexamples if you consider matrices). Do you have an idea? Maybe.. it could be that there is no counterexample. I would appreciate your help. Regards
At least if $A$ and $B$ are unital and $f(1)=1$, there is no counterexample. Suppose these extra assumptions on $A$, $B$, and $f$ hold.
Because $A$ is the linear span of its unitary elements, it suffices to show that $f(u^*)=f(u)^*$ for each unitary element $u\in A$. Let such $u$ be fixed and let $v=f(u)$. Then $v^{-1}=f(u^*)$, so $\|v\|=\|v^{-1}\|=1$. For an operator $V$ on a Hilbert space, it is plain that $\|V\|=\|V^{-1}\|=1$ implies that $V$ is a unitary operator, so in light of the Gelfand–Naimark theorem allowing $B$ to be embedded isometrically via a $*$-homomorphism into $B(H)$ for some Hilbert space $H$, we must have $v$ unitary in $B$, and thus $f(u)^*=v^*=v^{-1}=f(u^*)$.