First of all, here is the setting for my problem:
Definition: Let $A$ be a $C^*$-Algebra, $a,b\in A$. We say, a and b are orthogonal, if $ab=ba=a^*b=ab^*=0$.
Proposition: Let $A$ be a $C^*$-Algebra, $a,b\in A$. Equivalent: a and b are orthogonal iff 1.$a^*a, b^*b$ 2. $a^*a, bb^*$, 3.$aa^*, b^*b$ and 4. $aa^*, bb^*$ are orthogonal.
I have some questions about the proof of the following
claim: Let $A$ and $B$ be $C^*-$algebras and $\phi:A\to B$ be a completely positive map. Equivalent:
(i) for all positive elements $a,b\in A$ such that a and b are orthogonal, are $\phi(a),\phi(b)$ orthogonal in $B$.
(ii) for all $a,b\in A$ such that a and b are orthogonal, are $\phi(a),\phi(b)$ orthogonal in $B$.
Proof:
(ii)=>(i) is trivially true.
(i)=>(ii): Let $a,b$ are arbitrary such that a and b are orthogonal. By the proposition above are 1.$a^*a, b^*b$ 2. $a^*a, bb^*$, 3.$aa^*, b^*b$ and 4.$aa^*, bb^*$ are orthogonal and therefore 1.$\phi(a^*a), \phi(b^*b)$ 2. $\phi(a^*a), \phi(bb^*)$, 3.$\phi(aa^*), \phi(b^*b)$ and 4.$\phi(aa^*), \phi(bb^*)$ are orthogonal, because the elements $a^*a$ and so on are postive. But since $\phi$ is completely positive, we have $0\le\phi(a^*)\phi(a)\le \phi(a^*a)$ for all $a\in A$ (similary for the other elements). Now, orthogonality is a hereditary property, $\phi(a^*)\phi(a)$ are orthogonal and so on. By proposition above $\phi(a),\phi(b)$ are orthogonal.
My questions are: 1) I only know the corollary of Stinespring's theorem "Let $A$ be a $C^*$a-legrba, $\phi:A\to B$ completely positive. It is $0\le \phi(a)^*\phi(a)\le \|\phi\|\phi(a^*a)$ for all $a\in A$," and I know the proof of it. But in the proof above it is $0\le\phi(a^*)\phi(a)\le \phi(a^*a)$ for all $a\in A$, which is a different statement. And I dont understand why $0\le\phi(a^*)\phi(a)\le \phi(a^*a)$ for all $a\in A$ is true.
2)Which is the cosidered hereditary $C^*$-subalgebra of B here?
You can repeat the proof with $\phi/\|\phi\|$, which does satisfy the inequality. If the authors didn't mention that their were sloppy, but the argument still works.
What they mean by "hereditary" is that if $a,b$ are orthogonal and positive and $a'\leq a$, $b'\leq b$, then $a'$ and $b'$ are orthogonal.
One way to see this last assertion is by representing $A$ faithfully in $B(H)$. From $ab=0$, we get $a^n b=0$ for all $n\geq 1$, so $p(a)b=0$ for all polynomials with $p(0)=0$, and by taking sot limits $f(a)b=0$ for all Borel functions $f$ on $\sigma(a)$. In particular $b$ is orthogonal to the range projection $p_a$ of $a$. By using the same idea on $b$, we get that $p_ap_b=0$. Then $$ a'b'=a'p_ap_bb'=0. $$ As were are assuming the elements are positive, there are no adjoints to deal with.