In induced von Neumann algebra, do we have $P(PMP) =P\cdot P(M)P$ and $U(PMP) =P\cdot U(M)P$? Let $M$ be a von Neumann algebra and $P\in M'$ be a projection. Then, $PMP$ is an induced von Neumann algebra. I know that $ P Z(M) P = Z(PMP)$ and $ (PMP)' = PM'P $ (here, $Z(M)$ is the center and $M'$ is the commutant). I would like to ask, do we have 
$$P(PMP) =P\cdot P(M)P$$
and 
$$U(PMP) =P\cdot U(M)P,$$
where $P(PMP)$ denotes the set of all projections in $PMP$ and $U(PMP)$ denotes all unitary elements of $PMP$.
It looks that Kadison-Ringrose I, Proposition 5.5.5 implies what I want above.

Edit: *This argument works when $P\in Z(M)$. Together with Proposition 5.5.5 in Kadison-Ringrose, as mentioned by @user92646, this gives the desired proof. 
If $Q\in M$ is a projection, then $QP$ is also a projection, since $PQ=QP$. Conversely, if $PX\in PMP$ is a projection, we have $(PX)^2=PX$. Let $Q=PX+I-P$. Then $PQ=PX$, and $Q^2=(PX)^2+I-P=PX+I-P=Q$. 
Similarly, if $U\in M$ is a unitary, $(UP)^*UP=PU^*UP=P$, $(UP)(UP)^*=PUU^*P=P$, so $UP$ is a unitary in $PM$. If $XP$ is a unitary in $PM$, this means that $P=(XP)^*XP=X^*XP$; let $U=PX+I-P$. Then $U^*U=PX^*XP+I-P=P+I-P=I$, and $UU^*=I$. 
The fact that $P\in M'$ was essential above. When $P$ is arbitrary, neither the compression of a unitary nor a projection has to be such in the compressed algebra. You can always do the following: let $X\in B(H)$ be any contraction. Then 
$$
U=\begin{bmatrix} X& (I-XX^*)^{1/2} \\ (I-X^*X)^{1/2} &-X^*\end{bmatrix}
$$
is a unitary in $B(H\oplus H)$ with $PUP=\begin{bmatrix} X&0\\0&0\end{bmatrix}$, where $P=\begin{bmatrix} I&0\\0&0\end{bmatrix}$.
If $X$ is positive, $$
Q=\begin{bmatrix} X& (X-X^2)^{1/2} \\ (X-X^2)^{1/2} &I-X\end{bmatrix}
$$
is a projection in $B(H\oplus H)$ with $PQP=\begin{bmatrix} X&0\\0&0\end{bmatrix}$.
