I am confused with the underlined equation in the following picture. enter image description here I know that a *-isomorphism commutes with continuous functional calculus since every continuous functions on the compact subset of $\mathbb{R}$ can be approximated uniformly by polynomials.

To prove the underlined equation, we have to prove that $f\to \Phi(f(j_A(a)))$ is the Borel functional calculus on $\Phi(j_A(a))$. Given a net $(f_\lambda)$ of continuous functions converging to $0$ respect to $\sigma(C(\sigma(j_A(a)))^{**},C(\sigma(j_A(a)))^*)$ topology , we have $f_\lambda(j_A(a))\xrightarrow{\sigma(A^{**},A^*)}0$, but how to obtain $\Phi(f_\lambda(j_A(a)))\xrightarrow{\sigma(B^{**},B^*)}0$?

Notes: 1.The whole article is here;
2. My interest is the unital case, you can simply take $M(A)=A$;
3. $j_A$ in the picture is the canonical injection from $A$ to $A^{**}$.


1 Answer 1


Any $*$-isomorphism between von Neumann algebra is also an order isomorphism and hence is normal. So it is automatically weak* continuous.


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