$*$-homomorphism of the tensor product

Question

Let $A,~B,~C,~D$ be the C*-algebras and the "$\odot$" denotes the algebraic tensor product.

Proposition 3.1.16 (Tensor product morphisms). Given $*$-homomorphisms $\phi: A\rightarrow C$ and $\psi: B\rightarrow D$, the tensor product map $\phi \odot \psi : A\odot B \rightarrow C\odot D$ is also a *-homomorphism.

Proposition 3.1.17 (Product morphisms). Given two $*$-homomorphsims $\pi_{A}: A\rightarrow C$ and $\pi_{B}: B\rightarrow C$ with commuting ranges, the product map $\pi_{A}\times\pi_{B}: A\odot B\rightarrow C$ is also a $*$-homomorphsim.

Exercise 3.1.1. Observe that if $C$ and $D$ are both unital, then Proposition 3.1.16 is a special case of Proposition 3.1.17. How about in the nonunital case?

But, in the Exercise 3.1.1, if $C$ and $D$ are both unital, how to explain Proposition 3.1.16 is a special case of Proposition 3.1.17?

Answer 1

If $C$ and $D$ are unital, given $\phi : A\to C$ and $\psi : B\to D$, define $$ \pi_A : A\to C\otimes D \quad\text{ and }\quad \pi_B :B\to C\otimes D $$ by $$ \pi_A(a) := \phi(a)\otimes 1_D\quad\text{ and }\quad \pi_B(b) := 1_C\otimes \psi(b) $$ The ranges of these maps commutes and the product map $\pi_A\times \pi_B$ is nothing by $\phi\otimes \psi$.