I have a problem with the following lemma (Takesaki Vol 1-page 298):
Lemma 1.25. If $e$ is an abelian projection in a von Neumann algebra $\mathcal M$, then for any projection $f\in\mathcal M$ with $z(f)\geq e$ we have $e\succsim f$.
Proof. By Theorem 1.8, we have only to show that $e\succsim f$ implies $e\sim f$. In this case, we have $z(e)=z(f)$. Let $\mathcal Z$ denote the center of $\mathcal M$. By assumption, we have $e\mathcal M e=\mathcal Z e$; see Proposition II.3.10. Choose a partial isometry $u\in\mathcal M$ with $u^*u=f$ and $uu^*=e_1\leq e$. There exists a central projection $z$ in $\mathcal M$ with $e_1=ez$. Since equivalent projections have the same central support, we have $z(f)=z(e_1)\leq z(e)$; hence we have $z(e_1)=z(e)$. Therefore $z(e)\leq z$, so that $e=e_1$. Thus $f\sim e_1=e$.
Let us consider $M=B(H)$. Let $e$ be a minimal projection and $f$ be the identity operator. The assertion works just when $H$ is one dimensional.
I think there are some misprints. Could you please check the notations in this lemma?
There is a typo in the statement: it should say $e\precsim f$. The rest looks ok to me.
As the proof says, one uses Theorem 1.8: there exists central $z$ with $ze\precsim zf$ and $(1-z)e\succsim (1-z)f$. The proof shows that $(1-z)e\sim (1-z)f$. Then $$ e=ze+(1-z)e\precsim zf+(1-z)f=f. $$