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I am reading a paper by David Blecher, which contains the following:

" If $T: Y \to Z$ is a surjective isometric module map between $W^{*}$-modules over $M$, then $T$ is unitary. Also, $T$ is a $w^{*}$-homeomorphism, and the unique preduals of $Y$ and $Z$ are completely isometrically isomorphic via the module map $T_{*}$."

I have no problem with the unitary part. Also, I know that if $T$ is $w^{*}$-continuous then it is a $w^{*}$-homeomorphism. What I don't get is why do we know $T$ is $w^{*}$-continuous?

Since this statement is given without proof, it makes me think that this is a well known fact. I"ll appreciate any explanation or reference to a proof of this.

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  • $\begingroup$ Could you please add a link to the paper or add its title? $\endgroup$
    – user82740
    Dec 4, 2015 at 4:24
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    $\begingroup$ The statement is at the end of the first paragraph of pg 68 dmitripavlov.org/scans/blecher.pdf $\endgroup$
    – epsilon
    Dec 4, 2015 at 5:02

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It is a weak* homeomorphism because the predual of a W*-module is unique. If $T$ weren't a weak* homeomorphism then it could be used to transfer the weak* topology on $Y$ to a new, different weak* topology on $Z$. But $Z$ only has one predual, and hence it only has one weak* topology.

For slightly more details on why the predual is unique, see Theorem 2.6 of this paper.

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  • $\begingroup$ Thank you Nick, I actually think I got it in a different way. It is easy to show that T is adjointable, which implies is weak$^{*}$ continuous. By the way, how is the map $T_{*}$ between the preduals defined? $\endgroup$
    – epsilon
    Dec 4, 2015 at 6:33
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    $\begingroup$ Let $f \in Z_*$, i.e., $f$ is a weak* continuous linear functional on $Z$. Then $f\circ T$ is a weak* continuous linear functional on $Y$, i.e., $f\circ T \in Y_*$. $T_*$ is the map $f \mapsto f\circ T$. $\endgroup$
    – Nik Weaver
    Dec 4, 2015 at 12:59
  • $\begingroup$ Thank you Nick. If it's not too much trouble, could you say a few words about the "getting a new , different weak $^{*}$ topology" part of your answer? How would you use $T$ to get that new topology? $\endgroup$
    – epsilon
    Dec 4, 2015 at 15:07
  • $\begingroup$ If you have a bijection between two sets and a topology on one of them, you can define a topology on the other set as the set of images of open sets. Also, it's spelled "Nik". $\endgroup$
    – Nik Weaver
    Dec 4, 2015 at 15:19

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