Let $H$ be a separable Hilbert space. Write $B(H)$ for the set of linear bounded operator on $H$ and $H \otimes H$ the tensor product of Hilbert space. For every $A,B \in B(H)$ we can define an operator $A \otimes B$ on $H \otimes H$ by $(A \otimes B)(x \otimes y) = Ax \otimes Ay$. Thus, we can define $\omega : T \mapsto T \otimes 1$ from $B(H)$ to $B(H \otimes H)$. Given how simple $\omega$ is, I guessed that it should be continuous with respect to both $\sigma$-weak topology. That's what I want to show. Here is what I've done so far :
Let $(A_i)_i$ be a $\sigma$-weak converging net in $B(H)$ and let $A$ be it limit. Let $(\xi_n)_n, (\eta_n)_n$ be sequences in $H \otimes H$ such that $\sum_n ||\xi_n||^2, \sum_n ||\eta_n||^2 < +\infty$. We have to show that $\sum_n (\omega(A_i)\xi_n \mid \eta_n) \xrightarrow[i \to +\infty]{}\sum_n (\omega(A)\xi_n \mid \eta_n)$. For every $n$, we can write $$\eta_n = \sum_r x_r^n \otimes y_r^n , \hspace{2mm} \xi_n = \sum_s z_s^n \otimes t_s^n$$
First I'll do things formally and then try to justify every step.
We have
\begin{align*} \sum_n (\omega(A_i)\xi_n \mid \eta_n) &= \sum_n \sum_r \sum_s (A_i z_s^n \otimes t_s^n \mid x_r^n \otimes y_r^n) \\ &= \sum_n \sum_r \sum_s (A_iz_s^n \mid x_r^n) (t_s^n \mid y_r^n) \\ &= \sum_n \sum_s (A_i z_s^n \mid \psi_n^s) \end{align*} Where $\psi_n^s = \sum_r (t_s^n \mid y_r^n)x_r^n$. I then want to use the $\sigma$-weak convergence of $(A_i)_i$ by re-indexing both sequences $(z_s^n)$ and $(\psi_n^s)$ but I'm not quite sure if I can control the series of norm of those sequences. For instance, we could have that $||z_1^s|| = s$ and $||t_1^s|| = \frac{1}{s^3}$ so that $||\xi_1|| = \sum_s ||z_1^s|| ||t_1^s||$ converges. This issue makes me think that my approach won't work. What I'm asking for is either an hint that would help me fix the issue I have or another proof idea. Thanks in advance.
Here is something useful to keep in mind.
A linear map $\pi: M\to N$ between von Neumann algebras is normal if its adjoint $\pi^*$ maps $N_*$ into $M_*$. In your case, you have $$\pi: B(H)\to B(H\otimes H): x \mapsto x\otimes 1.$$ Consider vectors $\xi, \eta, \xi', \eta'\in H$. Then we have for $x\in B(H)$ that
$$\pi^*(\omega_{\xi, \eta}\otimes \omega_{\xi', \eta'})(x) = \omega_{\xi, \eta}(x) \langle \xi', \eta'\rangle$$ so that $$\pi^*(\omega_{\xi, \eta}\otimes \omega_{\xi', \eta'}) = \langle \xi', \eta'\rangle \omega_{\xi, \eta} \in B(H)_*.$$
Since the functionals $\omega_{\xi, \eta}\otimes \omega_{\xi', \eta'}$ are linearly dense in $B(H\otimes H)_*$, we conclude that $$\pi^*(B(H\otimes H)_*))\subseteq B(H)_*$$ whence $\pi$ is normal.
Remark: No separability of $H$ was used, and you can replace the second factor of the tensor product by an arbitrary Hilbert space.
Alternatively, you can use that a $*$-isomorphism between two von Neumann algebras is automatically $\sigma$-weakly continuous (although using this is a sledgehammer argument in this case). Your map $\pi$ defines a $*$-isomorphism $B(H)\cong B(H)\bar{\otimes}\mathbb{C}$ and is therefore normal.