# diffuseness of a corner of a von Neumann algebra

Let $$M$$ be a finite von Neumann algebra, $$B$$ a von Neumann subalgebra of $$M$$, and $$p$$ be a projection in $$B$$. Given that $$B'\cap M$$ is diffuse, is it true that the relative commutant in the corner $$(pBp)'\cap pMp=B'p \cap pMp$$ is also diffuse?

I am reading Chapter 19.2 of the book "An introduction to $$II_1$$ factors" by Anantharaman-Delaroche and Popa, and though they wrote this as if it is trivial, I couldn't come up with a proof.

We prove $$(B'\cap M)p =B'p \cap pMp$$, from which the result immediately follows.
$$LHS\subset RHS$$ is trivial. We prove the converse. Let $$x\in B'p \cap pMp$$. Then $$x=yp$$ for some $$y\in B'$$. If $$y\in B'\cap M$$ then we are done. Take $$z\in Z(B)=Z(B')$$ as the minimal projection in $$Z(B)$$ satisfying $$zp=p$$. Then, the kernel of the $$\ast$$-hom $$B'\to B'p:x\mapsto xp$$ equals $$B'(1-z)$$, and $$x=yp=yzp$$.
Then, for every $$w\in M'\subset B'$$, one gets $$ywzp=yzwp=yzpw=xw=wx=wyzp$$.
Then we get $$(yw-wy)z\in B'(1-z)$$, therefore $$yzw=wyz$$ and $$yz \in (M')'=M$$.