Let $H$ be a Hilbert space and $A$ be a strongly closed $*$-subalgebra of $B(H)$ contains identity, that is, a von Neumann algebra. Let $B$ be a maximal abelian $*$-subalgebra of $A$. Let $Z(A)$ be the center of $A$. Let $f:A \to Z(A)$ be a linear map which satisfies the following properties:
$(1)~f$ maps positive elements of $A$ to the positive elements of $Z(A)$
$(2)~f(x)=x$ for all $x \in Z(A)$
$(3)~ f(xay)=xf(a)y,$ for all $x,y \in Z(A)$ and all $a \in A$
$(4) ~f(a_1a_2)=f(a_2a_1),$ for all $a_1,a_2 \in A$.
More generally $f$ is called center valued trace on $A$. Now I want to prove the followings:
$(a)$ Prove that $Z(A)$ is a proper proper subalgebra of $B$.
$(b)$ Since $Z(A) \subsetneq B$, let $p \in B$ a projection which does not belong to $Z(A)$. Then prove that $f(p)$ is not a projection.
To prove $(a)$, I already proved that $Z(A)$ is a subalgebra of $A$, but I am unable to show that $Z(A)$ properly included in $B$. I goes for contradiction, that if not let $Z(A)=B$, but failed to reach at the contradiction.
And for $(b)$, I goes for contradiction, that let $f(p)$ be a projection. And also note that $f(p) \in Z(A).$ So, $pf(p)=f(p)p.$ So the closed subspace $f(p)(H)$ of $H$ is invariant for $p$. But again I do not know how to proceed. Somehow if I show that $p=f(p)$, then we get a contradiction. But failed again.
Please help me to solve this two. Thanks for your time and help.
a) For any $a\in Z(A)$, $b\in B$ you have $ab=ba$ by definition of center. Then $W^*(B,Z(A))$ is an abelian von Neumann subalgebra of $A$ that contains $B$. As $B$ is maximal, $W^*(B,Z(A))\subset B$ and so $Z(A)\subset B$. You cannot prove that $Z(A)$ is a proper subalgebra of $B$ because it need not be, if $A$ is abelian. If $A$ is not abelian, you cannot have $Z(A)=B$ because given any $a\in A\setminus Z(A)$ you have $W^*(Z(A),a)$ is an abelian algebra larger than $B$, contradicting maximality.
b) The center-valued trace is unique and faithful. Suppose that $f(p)$ is a projection. As $f(p)\in Z(A)$, we have $f(f(p))=f(p)$. Then $$ f(p)=f(p)f(p)=f(pf(p)). $$ So $$ 0=f(p-pf(p))=f(p(1-f(p))=f(p(1-f(p))p). $$ By the faithfulness of $f$, we get that $p(1-f(p))p=0$, and so $(1-f(p))p=0$. That is, $p=f(p)p$. This means that $p\leq f(p)$. Now $$ 0\leq f(f(p)-p)=f(f(p))-f(p)=f(p)-f(p)=0, $$ and the faithfulness of $f$ gives us that $p=f(p)$, a contradiction.