A corollary of commutation theorem for tensor product of von Neumann algebras. Please help me to understand the proof of the following corollary from the book 'Lectures on von Neumann algebras' by Stratila and Zsido.
Corollary: Let $\mathscr{M}_1$ and $\mathscr{M}_2$ be von Neumann algebras, $\mathscr{Z}_1,\,\mathscr{Z}_2$ their centers, respectively. Then the center of $\mathscr{M}_1\overline{\otimes}\mathscr{M}_2$ is $\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2$.
To prove the corollary, they have used the commutation theorem for tensor products (i.e. for two von Neumann algebras $\mathscr{M}_1$ and $\mathscr{M}_2$, we have $(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)'=\mathscr{M}_1'\overline{\otimes}\mathscr{M}_2'$). Here is my attempt to understand the proof.
Let $\mathscr{Z}$ be the center of $\mathscr{M}_1\overline{\otimes}\mathscr{M}_2$. Then $$\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2=(\mathscr{M}_1\cap\mathscr{M}_1')\overline{\otimes}(\mathscr{M}_2\cap\mathscr{M}_2')\subseteq (\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)\cap(\mathscr{M}_1'\overline{\otimes}\mathscr{M}_2')=(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)\cap(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)'=\mathscr{Z}.$$
Now conversely, we have to prove $\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2\supseteq\mathscr{Z}$, i.e. $(\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2)'\subseteq\mathscr{Z}'$, i.e. $\mathscr{Z}_1'\overline{\otimes}\mathscr{Z}_2'\subseteq\mathscr{Z}'$, i.e. $\mathfrak{A}(\mathscr{M}_1\cup\mathscr{M}_1')\overline{\otimes}\mathfrak{A}(\mathscr{M}_2\cup\mathscr{M}_2')\subseteq\mathfrak{A}((\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)\cup(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)')$, where $\mathfrak{A}(A)$ stands for the von Neumann algebra generated by the set $A$. But I   did not get the way the authors prove it. For the proof given in the book see here.
Thanks in advance for any help.
 A: In your attempt for the easy inclusion, I'm not comfortable enough with tensor products to assess your exchange of tensor and intersection. But that's not needed. Elements of $Z_1\otimes Z_2$ commute with every elementary tensor in $M_1\otimes M_2$, so commute with all elements of the algebraic tensor product, and then commute with limits of these. This shows that
$$
Z_1\otimes Z_2\subset Z.
$$
For the reverse inclusion, what the authors do is to get, from the definition of centre, that
$$\tag1
M_1\otimes M_2\subset Z'.
$$
Then, since every element of $M_1'\otimes M_2'$ commutes with every elements of $M_1\otimes M_2$,
$$\tag2
M_1'\otimes M_2'\subset (M_1\otimes M_2)'\subset Z',
$$
the last inclusion being simply $Z\subset M_1\otimes M_2$. Then,
$$\tag3
Z_1'\otimes Z_2'=(M_1\cap M_1')'\otimes (M_2\cap M_2')'
=(M_1'\cup M_1)''\otimes (M_2'\cup M_2)''\subset Z'.
$$
The last inclusion follows from $(1)$ and $(2)$, because if you have $a\otimes b$ with $a\in M_1'$, $b\in M_2$, then
$$
a\otimes b=(a\otimes 1)(1\otimes b)\in Z'
$$
since $a\otimes 1\in Z'$ by $(2)$ and $1\otimes b\in Z'$ by $(1)$. The other cases are treated similarly, to get $(M_1'\cup M_1)\otimes (M_2'\cup M_2)\subset Z'$. Taking sums and limits in the first coordinate we get $(M_1'\cup M_1)''\otimes (M_2'\cup M_2)\subset Z'$. And then taking sums and limits in the second coordinate, $(3)$ follows.
Finally, from $(3)$ one gets
$$
Z\subset (Z_1'\otimes Z_2')'=Z_1''\otimes Z_2''=Z_1\otimes Z_2. 
$$
