In this paper, specifically Theorem 4.1 on page 10, one of the last steps in the proof involves saying that a particular right-inverse exists. I'll try to restate what I think are the relevant pieces of info here, instead of just rewriting the whole proof.
Let $\mathfrak{A}$ be a von Neuman algebra with a faithful normal state $\rho$. In the last paragraph, the author looks at a self-adjoint, positive element $y \in \mathfrak{A}$, and then defines $q = \chi_{(1/2, 1]}(y)$, where this latter expression is meant in the sense of the Borel functional calculus on a self-adjoint $y \in \mathfrak{A}$. This $q$ satisfies $$q \leq 2y \leq q + 1 \; (\dagger).$$ He then claims that there must exist $z \in \mathfrak{A}$ such that $\| z \|\leq 2$ and $$1 - q = (1- y) z \; (\dagger \dagger) .$$
I'm confused by how he gets the right-hand inequality in $(\dagger)$, and how he conjures up the $z$ in $(\dagger \dagger)$. If anyone can clear this up, I'd greatly appreciate it!
The first inequality is most likely a typo, it should be $$ q\leq 2yq\leq 1+q. $$ This is simply functional calculus on the inequalities $$ 1_{\big(\tfrac12,1\big]}(t)\leq 2t\,1_{\big(\tfrac12,1\big]}(t)\leq 2\,1_{\big(\tfrac12,1\big]}(t)\leq 1+1_{\big(\tfrac12,1\big]}(t). $$ For $z$, take $$ h(t)=\frac1{1-t}\,1_{\big[0,\tfrac12\big]}(t) $$ and define $z=h(y)$. Since $|h(t)|\leq2$, you get that $\|z\|\leq2$.