$*$-homomorphism induced by a c.p.c order zero map

Question

The $*$-homomorphism $\Phi:C_0((0,1])\otimes A\rightarrow B$ is defined in (3).

If $f$ is any nonzero element in $C_0((0,1])$, how to define $\Phi(f\otimes a)$? where $a\in A$

Answer 1

Given the formula, you can get an explicit calculation of $\Phi(p\otimes a)$ whenever $p$ is a polynomial with $p(0)=0$ and $a\in A$. Since such polynomials are dense in $C_0((0,1])$, we can approximate $\Phi(f\otimes a)$ for any $f\in C_0((0,1])$.

Let $p$ be such a polynomial, with $p(x)=\sum_{k=1}^n\lambda_kx^k$, and assume $a\in A$ is positive (since positive elements span $A$, we can find $\Phi(p\otimes a)$ for arbitrary $A$ by linearity). Then we have $$\Phi(p\otimes a)=\sum_{k=1}^n\lambda_k\Phi(x^k\otimes a)=\sum_{k=1}^n\lambda_k\left(\Phi(x\otimes a^{1/k})\right)^k =\sum_{k=1}^n\lambda_k\phi(a^{1/k})^k.$$

If $A$ is unital and $f\in C_0((0,1])$, choose a sequence of polynomials $(p_n)$ in $C_0((0,1])$ converging to $f$. Since $1^{1/k}=1$ for all $k$, the above formula implies that $$\Phi(f\otimes 1)=\lim_{n\to\infty}\Phi(p_n\otimes1)=\lim_{n\to\infty}p_n(\phi(1))=f(\phi(1)).$$