Nonunital C*-Algebras: Morphism Contractive

Question

Problem

Given C*-algebras $\mathcal{A}$ and $\mathcal{B}$.

Consider a *-morphism $\pi:\mathcal{A}\to\mathcal{B}$.

Then it is contractive: $\|\pi[\mathcal{A}]\|\leq\|A\|$

The proof I know critically uses: $\pi[1]=1\in\mathcal{B}$.

But what if either $1\notin\mathcal{A}$ or $1\notin\mathcal{B}$ or $\pi[1]\neq1$?

Answer 1

Case: $1\in\mathcal{A}$

Consider the C*-subalgebra $\overline{\mathrm{im}\pi}$ and the restricted morphism $\pi:\mathcal{A}\to\overline{\mathrm{im}\pi}$.

Case: $1\notin\mathcal{A}$

Adjoin a unit if necessary $\mathcal{B}\oplus\mathbb{C}$.

Adjoin a unit $\mathcal{A}\oplus\mathbb{C}$ and extend the morphism by $\pi[A+a1]:=\pi[A]+a1$.

Answer 2

As pointed out to you in the (at the moment) deleted answer: calculate the norm of $\pi'[A + a1]$. You don't need to assume that $A$ is a $C^\ast$ algebra you can show it for any Banach $\ast$-algebra $A$.

In other words, the following theorem holds:

Let $A$ be a Banach $\ast$-algebra and $B$ be a $C^\ast$-algebra and $\varphi: A \to B$ a $\ast$-homomorphism. Then $\varphi$ is norm decreasing.