$\pi^{\ast \ast}: A^{\ast \ast} \to B^{\ast \ast}$ is a $C^{\ast}$-homomorphism. 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.
 A: 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 $A\to A^{\ast\ast}$.
A: Here is a second answer that works directly with the definition of the Arens product without invoking any density or continuity properties.
I will write $\langle\cdot,\cdot\rangle$ for the various dual pairings in this setting. First note that if $a,a'\in A$ and $\omega\in B^\ast$, then
$$
\langle \pi^\ast(\omega\pi(a)),a'\rangle=\langle \omega\pi(a),\pi(a')\rangle=\langle \omega,\pi(a)\pi(a')\rangle=\langle \omega,\pi(aa')\rangle=\langle \pi^\ast(\omega)a,a'\rangle.
$$
Thus $\pi^\ast(\omega\pi(a))=\pi^\ast(\omega)a$. An analogous computation show $\pi^\ast(\pi(a)\omega)=a\pi^\ast(\omega)$.
Now let $a\in A$, $\omega\in B^\ast$ and $f\in A^{\ast\ast}$. Using the identity from above, we get
\begin{align*}
\langle \pi^\ast( \pi^{\ast\ast}(f)\omega),a\rangle&=\langle  \pi^{\ast\ast}(f)\omega,\pi(a)\rangle\\
&=\langle \pi^{\ast\ast}(f),\omega\pi(a)\rangle\\
&=\langle f,\pi^\ast(\omega\pi(a))\rangle\\
&=\langle f,\pi^\ast(\omega)a\rangle\\
&=\langle f\pi^\ast(\omega),a\rangle.
\end{align*}
Thus $\pi^\ast(\pi^{\ast\ast}(f)\omega)=f\pi^\ast(\omega)$. Again, $\pi^\ast(\omega \pi^{\ast\ast}(f))=\pi^\ast(\omega)f$ is analogous.
Finally, if $f,f'\in A^{\ast\ast}$ and $\omega\in B^\ast$, then
\begin{align*}
\langle \pi^{\ast\ast}(ff'),\omega\rangle&=\langle ff',\pi^\ast(\omega)\rangle\\
&=\langle f,f'\pi^\ast(\omega)\rangle\\
&=\langle f,\pi^\ast(\pi^{\ast\ast}(f')\omega)\rangle\\
&=\langle \pi^{\ast\ast}(f),\pi^{\ast\ast}(f')\omega\rangle\\
&=\langle \pi^{\ast\ast}(f)\pi^{\ast\ast}(f'),\omega\rangle.
\end{align*}
This settles $\pi^{\ast\ast}(ff')=\pi^{\ast\ast}(f)\pi^{\ast\ast}(f')$.
