Suppose that $A,C$ are $C^*$-algebras and $\phi:A \to C$ is a completely positive, orthogonality-preserving linear map. (Orthogonality preserving means: if $a,b \in A$ satisfy $ab=0$ then $\phi(a)\phi(b) = 0$.) Then:

(i) For any $a,b,c \in A$, $$ \phi\left(ab\right)\phi\left(c\right) = \phi\left(a\right)\phi\left(bc\right) $$ (in the special case that $A$ is unital, this is equivalent to $\phi\left(a\right)\phi\left(b\right) = \phi\left(1\right)\phi\left(ab\right)$ for any $a,b \in A$);

(ii) For any $a,b \in A$, $$ \left\| \phi\left(ab\right) \right\| \leq \|a\|\cdot\left\|\phi\left(b\right)\right\|; $$

(iii) If $A$ is unital and simple, then for any $a \in A$, $$ \left\| \phi\left(a\right) \right\| = \|a\|\cdot\left\|\phi\left(1\right)\right\|. $$

In fact, there is a rich structure theorem about completely positive, orthogonality-preserving maps (in the literature, they are called "order zero" instead of "orthogonality-preserving" ), Theorem 2.3 of Winter, Zacharias, "Completely positive maps of order zero," Münster J. Math, 2009 (see also Corollary 3.1); and I can prove these statements easily using the structure theorem. But, my question is: can we prove any of the facts above directly (without appealing to this structure theorem)?

(I am intentionally not restating the structure theorem here because my question is about not using it.)

Maps preserving zero products, J. Alaminos, M. Brešar, J. Extremera, A. R. Villena. Studia Math. 193 (2009), 131-159. dx.doi.org/10.4064/sm193-2-3 (To apply the result mentioned there, note that any unital C*algebra is linearly spanned by the unitaries). IIRC, their proof rests on the fact that the diagonal subgroup of ${\mathbb T}\times{\mathbb T}$ is a set of synthesis $\endgroup$ – Yemon Choi Mar 20 '12 at 21:523more comments