Given Hilbert spaces $H_1,H_2$ and a functional in the predual $\psi\in B(H_1)_*$ we may consider the slice map $S:B(H_1)\otimes B(H_2)\to B(H_2)$ defined on the spatial tensor product given by extension of the map $a\otimes b \mapsto \psi(a)b$ defined on the algebraic tensor product of $B(H_1)$ and $B(H_2)$.
I want to know why we can extend this slice map to the weak (also, strong) operator closure $B(H_1)\overline{\otimes}B(H_2)\subseteq B(H_1\otimes H_2)$ such that the extension is a $*$-homomorphism.
Thank you!
As mentioned in the comments, the slice map will only be a $*$-homomorphism if $\psi$ is multiplicative. This is already clear on the purely algebraic level, even before taking completions. However, it is true that the algebraic slice map extends to the von Neumann tensor product, provided that you slice with a normal functional (i.e. an element of the predual). This is actually part of a more general statement, namely that the algebraic tensor product of two normal c.p. linear maps between von Neumann algebras extends (uniquely) to a normal linear map on the related von Neumann algebra tensor products. This is for instance proven in Takesaki's book "Theory of operator algebras I" in the chapter on tensor product (chapter IV).