$\pi^{\ast \ast}: A^{\ast \ast} \to B^{\ast \ast}$ is a $C^{\ast}$-homomorphism.

Question

Let $A$ and $B$ be $C^{\ast}$-algebras and $\pi: A \to B$ be a $C^{\ast}$-homomorphism. I have always believed that $\pi^{\ast \ast}: A^{\ast \ast} \to B^{\ast \ast}$ is also a $C^{\ast}$-homomorphism. Since yesterday I am trying to write a proof of this using Arens multiplication but after multiple attempts I'm unable to write it clearly. Can someone please help me to show that $\pi^{\ast \ast}$ is multiplicative or provide a reference.

Answer 1

Instead of going through the definition of the Arens product and verifying this by hand, you can use a few useful facts:

• The Arens product is separately weak$^\ast$ continuous.
• The canonical image of $A$ inside $A^{\ast\ast}$ is weak$^\ast$ dense.
• On the image of $A$ inside $A^{\ast\ast}$, the Arens product coincides with the product of $A$.

Then you can argue as follows: As an adjoint, $\pi^{\ast\ast}$ is weak$^\ast$ continuous. Moroever, it coincides with $\pi$ on the image of $A$ inside $A^{\ast\ast}$. Thus, if $f,g\in A^{\ast\ast}$ and $(a_\lambda)$, $(b_\mu)$ are nets in $A^{\ast\ast}$ such that $j(a_\lambda)\to f$, $j(b_\mu)\to g$ in the weak$^\ast$ topology, then \begin{align*} \pi^{\ast\ast}(fg)&=\lim_\lambda \lim_\mu \pi^{\ast\ast}(j(a_\lambda)j(b_\mu))\\ &=\lim_\lambda\lim_\mu \pi^{\ast\ast}(j(a_\lambda b_\mu))\\ &=\lim_\lambda\lim_\mu j(\pi(a_\lambda b_\mu))\\ &=\lim_\lambda\lim_\mu j(\pi(a_\lambda)\pi(b_\mu))\\ &=\lim_\lambda\lim_\mu j(\pi(a_\lambda))j(\pi(b_\mu))\\ &=\lim_\lambda\lim_\mu\pi^{\ast\ast}(j(a_\lambda))\pi^{\ast\ast}(j(b_\mu))\\ &=\pi^{\ast\ast}(f)\pi^{\ast\ast}(g). \end{align*} Here I wrote $j$ for the canonical map $A\to A^{\ast\ast}$.