How to lift a $\ast$-homomorphism to $A^\infty\cap B'$ to a $\ast$-homomorphism to $\ell^\infty(\mathbb{N},A)$?

Question

Recently, I read Santiago's paper Crossed products by actions of finite groups with the Rokhlin property. In the proof of Corollary 1, the author defined a $\ast$-homomorphism $\phi:\mathbb{C}[G]\to A^\infty\cap B'$ as $\phi(\sum_{g\in G}z_gg)=\sum_{g\in G}z_gp_g$, where $z_g\in\mathbb{C}$, $A$ is a $C^\ast$-algebra, $G$ is a finite group acting on $A$, $B$ is a subalgebra of $A$. And then the author said using the $C^\ast$-algebra $\mathbb{C}[G]$ is semiprojective, we can lift $\phi$ to a unital map from $\mathbb{C}[G]$ to $\ell^\infty(\mathbb{N},A)$. But I can not see from the definition of simiprojectivity how to lift it. Can anyone give me a hint? Thanks!

Answer 1

By definition $$ A^{\infty} = \ell^{\infty}(\mathbb{N},A)/c_0(\mathbb{N},A) $$ Now note that $c_0(\mathbb{N},A) = \overline{\cup I_n}$ where $$ I_n = \{(x_j) \in \ell^{\infty}(\mathbb{N}, A): x_j = 0 \quad\forall j > n\} $$ Hence if $D$ is a semi-projective algebra, then any $\ast$-homomorphism and $\phi : D \to A^{\infty}$ lifts to a map $\psi : D \to \ell^{\infty}(\mathbb{N},A)/I_n$ for some $n\in \mathbb{N}$. However, the map $$ \ell^{\infty}(\mathbb{N},A) \to \ell^{\infty}(\mathbb{N},A) \text{ given by } (x_j) \mapsto (x_{n+1}, x_{n+2},\ldots) $$ induces an isomorphism $$ \ell^{\infty}(\mathbb{N},A)/I_n \cong \ell^{\infty}(\mathbb{N},A) $$ which gives you the result you want.