Is $x \mapsto x \otimes 1$ $\sigma$-weakly continuous? 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 $H$?
 A: 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.)
A: I want to note that the final question in the OP has a simple answer.  For a Hilbert space $H$ and $\xi,\eta\in H$ let $\omega_{\xi,\eta}$ be the trace-class operator, which gives the normal function $\newcommand{\mc}{\mathcal}\mc B(H) \rightarrow\mathbb C; T\mapsto (T\xi|\eta)$.  Any trace-class operator $\omega\in\mc T(H)$ can be written as a normal convergent sum
$$ \omega = \sum_{n=1}^\infty \omega_{\xi_n, \eta_n} 
\qquad\text{with}\qquad
\sum_n \|\xi_n\| \|\eta_n\| < \infty. $$
So for example Takesaki, Chapter II, Theorem 1.6 (which gives a stronger statement).
Now consider a Hilbert space $K$ and the Hilbert space tensor product $H\otimes K$.  If $(e_i)$ is any orthonormal basis of $K$ then any $\xi\in H\otimes K$ can be (uniquely) written as
$$ \xi = \sum_i \xi_i \otimes e_i
\qquad\text{with}\qquad
\|\xi\| = \sum_i \|\xi_i\|^2. $$
Thus, given $T\in\mc B(H)$ and $\xi,\eta\in H\otimes K$, then
$$ \langle T\otimes 1 , \omega_{\xi,\eta} \rangle
= \sum_{i,j} \big( (T\otimes 1)(\xi_i\otimes e_i) \big| \eta_j\otimes e_j \big)
= \sum_i (T(\xi_i)|\eta_i). $$
Notice that this sum converges absolutely, as
$$ \sum_i \big| (T(\xi_i)|\eta_i) \big|
\leq \|T\| \sum_i \|\xi_i\| \|\eta_i\|
\leq \|T\| \Big(\sum_i \|\xi_i\|^2\Big)^{1/2} \Big(\sum_i \|\eta_i\|^2\Big)^{1/2},
$$
by Cauchy-Schwarz.  Notice that we need something like this argument to show that $T\otimes 1$ is even defined!  Thus $\omega_{\xi,\eta}$ induces the functional
$$ \sum_i \omega_{\xi_i, \eta_i} \in \mc T(H). $$
We extend to $\mc T(H\otimes K)$ by absolute convergence.
