Let $A$ and $B$ unital $C^\ast$-algebras, $f:A\to B$ a linear, bounded map such that $f(a^\*)=f(a)^\*$ for all $a\in A$, $f(1\_A)=1\_B$ and $f(a)f(b)=0$ for all $a,b\in A\_{sa}$ with $ab=0$. Follows $f(a^2)=f(a)^2$ for all $a\in A$?


I have tried to proof this first for self-adjoint elements $a\in A$ using the continuous functional calculus but I'm stuck. Do you know a proof or a reference? Greetings



map between $C^*$-algebras with special properties, why is $f(a^2)=f(a)^2 $ for all $a\in A$?
Yemon Choi pointed out in a comment to [this related question](https://mathoverflow.net/questions/91700/properties-of-orthogonality-preserving-c-p-maps-between-c-algebras) that we can conclude $f$ is a \*-homomorphism under the stronger assumption that $ab = 0$ implies $f(a)f(b) = 0$ for arbitrary (not just self-adjoint) $a,b \in A$. This folows from a result stated in the [abstract](http://journals.impan.pl/cgi-bin/doi?sm193-2-3) of the paper Alaminos, Brešar, Extremera, and Villena, Maps preserving zero products, *Studia Math*. **193** (2009), 131-159 by taking $\phi(a,b) = f(a)f(b)$ and setting $b = 1$ in the conclusion.


