Let $\tau,\phi:A \rightarrow \mathbb{C}$ be faithful states on a C* algebra A, with $\tau \geqslant \phi$.
Write $H_{\tau}, H_{\phi}$ for the GNS representation with respect to $\tau,\phi$ respectively, i.e. $H_{\tau}$ is the completion of $A$ with respect to the inner product $(a,b) \mapsto \tau(b^{*}a)$, and similarly for $H_{\phi}$.
The condition $\tau \geqslant \phi$ implies that the identity on $A$ extends to a map $p: H_{\tau} \rightarrow H_{\phi}$.
Is the map $p$ injective or surjective? If it is not surjective, are there any conditions that guarantee that it is? When is $pp^{*} = 1$?
Left multiplication in $A$ extends to a representation of $A$ on both $H_{\tau}$ and $H_{\phi}$. Write $A_{\tau}$ and $A_{\phi}$ for the image of $A$ in $B(H_{\tau})$ and $B(H_{\phi})$ respectively. Write $A_{\tau}''$, and $A_{\phi}''$ for the von Neumann completions of $A_{\tau}$ and $A_{\phi}$ respectively.
Is it true that the map $\hat{p}:B(H_{\tau}) \rightarrow B(H_{\phi}), x \mapsto pxp^{*}$ restricts to a surjective homomorphism $A_{\tau}'' \rightarrow A_{\phi}''$?
Here are some things that I've tried so far:
Let's suppose that $pp^{*} = 1$. This implies that $\hat{p}$ is a homomorphism.
Moreover, in this case $\hat{p}$ restricts to a homomorphism $A_{\tau}'' \rightarrow A_{\phi}''$:
For all $a \in A$, and all $v \in A \subset H_{\tau}$ we have \begin{equation*} p(av) = av = ap(v), \end{equation*} which then implies that the relation $p(av) = ap(v)$ holds for all $a \in A$ and all $v \in H_{\tau}$. This means that the map $\hat{p}$ restricts to the identity on $A$, which thus extends to a homomorphism $\hat{p}: A_{\tau}'' \rightarrow A_{\phi}''$.
Clearly, the map $B(H_{\phi}) \rightarrow B(H_{\tau}), x \mapsto p^{*}xp$ is a right inverse to $\hat{p}$. Does this map restrict to an algebra homomorphism $A_{\phi}'' \rightarrow A_{\tau}''$?
A sufficient condition is $p^{*}p \in A_{\tau}''$. In that case we can proceed as follows: Suppose that $a \in A_{\phi}$, and $x \in A_{\tau}'$ and $v \in H_{\tau}$ then we compute \begin{align*} p^{*}apxv = p^{*}paxv = p^{*}pxav = xp^{*}pav = xp^{*}apv, \end{align*} whence $p^{*}apx = xp^{*}ap$, and hence $p^{*}ap \in A_{\tau}''$.
I don't have a general answer, but here are some ideas. Given a state $\tau$, let $h\in Z(A)_+$ with $\|h\|=1$, and define $\phi(a)=\tau(ha)$; then $\phi$ is a state with $\phi\leq\tau$.
You can check that in this case $p^*=p^*p=pp^*=M_h$ (multiplication by $h$). So $p$ is injective/surjective if and only if $h$, and you can construct examples with all three possible situations.
For $\hat p:x\longmapsto pxp^*$ to be a homomorphism you need, in particular, that $pxp^*px^*p^*=pxx^*p^*$ for all $x\in A_\tau''$. This equality is $$ px(I-p^*p)x^*p=0, $$ so $(I-p^*p)xp=0$ for all $x\in A_\tau''$. As the range of $p$ is dense, you get $(I-p^*p)x=0$. And by using $x\in A$ you obtain that $I-p^*p=0$. So $\hat p$ is a homomorphism if and only if $p^*p=I$.