Let $M\subseteq B(H)$ be a von Neumann algebra. Is it true that the mapping
$$\psi: M \to B(H \otimes H): m \mapsto m \otimes \text{id}\_H$$
is $\sigma$-weakly continuous? Here the $\sigma$-weak topology can be described in two ways:


(1) Let $M\_\*$ be any predual of $M$. Then the $\sigma$-weak topology is the weak$^ \*$-topology on $M$ coming from the isomorphism $M \cong (M\_\*)^\*$.


(2) It is the topology generated by the normal functionals. I.e. if $x \in M$ and $(x\_i)$ is a net in $M$, then $x\_i \to x$ if and only if $\text{Tr}((x\_i-x)y) \to 0$ for all trace class operators $y$ on $H$.


We have to show that if $m\_i \to m$ in the $\sigma$-weak topology, then also $m\_i \otimes \text{id}\_H \to m \otimes \text{id}\_H$. I guess we will need some result that connects trace class operators on $H \otimes H$ and trace class operators on $H$?



Is $x \mapsto x \otimes 1$ $\sigma$-weakly continuous?
There are many ways to do this. Maybe the quickest is to notice that $\psi$ is a $\*$-isomorphism between $M$ and $\psi(M)$, hence an order isomorphism, hence normal. (For this reason, $\*$-isomorphisms between von Neumann algebras are always weak\* continuous.)


