Order isomorphic order intervals Let $M$ be a von Neumann algebra. If $x$ is positive, then Lemma 2.1(3) of the paper "Order Isomorphisms of Operator Intervals in von Neumann Algebras" (Mori, Integral Equations and Operator Theory, 2019, https://arxiv.org/abs/1811.01647) states that if $p$ is the support projection of $x$, then the order interval $[0,p]$ is order isomorphic to $[0,x]$ through the map $y \mapsto x^{1/2} y x^{1/2}$. The proof here is claimed to be "easy to see".
But in the earlier paper "Order isomorphisms of operator intervals" (Semrl, Integral Equations and Operator Theory, 2017, https://www.fmf.uni-lj.si/~semrl/preprints/orderoperatorintervals.pdf), Semrl proves this in the special case where $M = B(H)$ and $x$ is injective, i.e., its support projection is $I$. His proof takes 1.5 pages (pages 38-39). I do not understand why Mori claims that his more general result is "easy to see". I need this in one of my papers and so I proved this myself based on characterising the inverse as the map $y \mapsto \lim_n f_n(x) y f_n(x)$ (weak*-limit) where $f_n(t) := t^{-1/2}$ for $t \geq 1/n$ and 0 elsewhere. But this proof is not trivial, one has to show, amongst other things, that this limit always exists if $0 \leq y \leq x$.
Am I missing a simple argument showing that $y \mapsto x^{1/2} y x^{1/2}$ is an order isomorphism between $[0,p]$ and $[0,x]$?
 A: (It is always hard to know what a "simple argument" is.  One Mathematician's "simple" can be another Mathematician's 3 page argument.  However, I think the follow is fairly "simple").
We can suppose our von Neumann algebra $M$ acts non-degenerately on $H$.
I think it's not so hard to reduce (by cutting down by the support) to the case when $x$ is positive and injective.
Then, the non-trivial thing to prove is that if $0\leq y\leq x$ then there is a positive contraction $z\in M$ (so $0\leq z\leq 1$) with $y = x^{1/2} z x^{1/2}$.  It suffices to find $w\in M$ a contraction with $y^{1/2} = wx^{1/2}$ (as then set $z=w^*w$).  This is just Douglas's Lemma which I'll now show.
(A little pre-lemma: $x^{1/2}$ has dense range.  Indeed, for any $a\in B(H)$ notice that $a\xi=0 \implies a^\ast a\xi=0 \implies (a^\ast a\xi|\xi)=0 \implies \|a|xi\|^2=0 \implies a\xi=0$ so $\ker(a) = \ker(a^\ast a)$.  Further, $\ker(a) = \textrm{Im}(a^*)^\perp$.
Thus $\ker(x^{1/2}) = \ker(x) = \{0\}$ it follows that $x^{1/2}(H)^\perp=\{0\}$ so $x^{1/2}$ has dense range.)
We construct $w\in B(H)$ as follows.  Firstly define on $x^{1/2}(H)$ by
$$ w x^{1/2} \xi = y^{1/2}\xi \qquad (\xi\in H). $$
Then $\|y^{1/2}\xi\|^2 = (y\xi|\xi) \leq (x\xi|\xi) = \|x^{1/2}\xi\|$ so $w$ is contractive.  As $w$ is densely-defined it extends by continuity.
It remains to show that $w\in M = M''$.  For $a\in M'$ we have that $wa x^{1/2}\xi =  wx^{1/2}a\xi = y^{1/2}a\xi = ay^{1/2}\xi = awx^{1/2}\xi$ so again as $x^{1/2}(H)$ is dense, it follows that $wa = aw$ as required.
