Given C*-Algebras $\mathcal{A}$ and $\mathcal{B}$.
(Possibly unital!)
Morphisms are contractive: $$\varphi:\mathcal{A}\to\mathcal{B}:\quad\|\varphi\|\leq1$$ (Possibly nonunital!)
How to apply unitization?
Given C*-Algebras $\mathcal{A}$ and $\mathcal{B}$.
(Possibly unital!)
Morphisms are contractive: $$\varphi:\mathcal{A}\to\mathcal{B}:\quad\|\varphi\|\leq1$$ (Possibly nonunital!)
How to apply unitization?
On
Extending s.sharp's answer..
Algebra
Given a C*-Algebra $1_\mathcal{Z}\notin\mathcal{Z}$.
Unitization: $$\iota_\mathcal{Z}:\mathcal{Z}\hookrightarrow\mathcal{Z}\oplus\mathbb{C}:\quad\iota_\mathcal{Z}Z:=Z\oplus0$$ $$\pi_\mathcal{Z}:\mathcal{Z}\oplus\mathbb{C}\twoheadrightarrow\mathcal{Z}:\quad\pi_\mathcal{Z}(Z\oplus\lambda):=Z$$
Isometricity: $$\|\iota_\mathcal{Z}Z\|=\|Z\|\quad\|\pi_\mathcal{Z}(Z\oplus0)\|=\|Z\|$$
(Nonunital crucial!)
Morphism
Extension: $$\Phi:\mathcal{A}(\oplus\mathbb{C})\to\mathcal{B}(\oplus\mathbb{C}):\quad\Phi A(\oplus\lambda):=\varphi A\left(+\lambda1_{\mathcal{B}(\oplus\mathbb{C})}\right)$$
Range-Unit: $$1_\mathcal{R}:=\Phi1_{\mathcal{A}(\oplus\mathbb{C})}:\quad 1_\mathcal{R}R\equiv R\equiv R1_\mathcal{R}\quad(R\in\overline{\mathcal{R}\Phi})$$
Restriction: $$\Phi_\mathcal{R}:\mathcal{A}(\oplus\mathbb{C})\to\overline{\mathcal{R}\Phi}:\quad\Phi_\mathcal{R}A(\oplus\lambda):=\Phi A(\oplus\lambda)$$
Unit-Preserving: $$\Phi_\mathcal{R}1_{\mathcal{A}(\oplus\mathbb{C})}=\Phi1_{\mathcal{A}(\oplus\mathbb{C})}=1_\mathcal{R}$$
Action-Identic: $$\varphi=(\pi_\mathcal{B}\circ)\Phi(\circ\iota_\mathcal{A}) =(\pi_\mathcal{B}\circ)\Phi_\mathcal{R}(\circ\iota_\mathcal{A})$$ (Modulo Ranges!)
The way to show contractivity usually works like this:
If $\Phi : \mathcal A \to \mathcal B$ is a $*$-morphism between $C^*$ Algebras, you can assume in general $\mathcal A$ to be unital (otherwise append unit to $\mathcal A$ and extend $\Phi$ so that $\Phi(\mathbb{1}_{\mathcal A})=\mathbb{1}_{\mathcal B}$, if $\mathcal B$ is not unital then extend it also to make this work). In that case if you define $\pi= \Phi(\mathbb{1}_{\mathcal A})$ you have $\pi^2=\pi=\pi^*$, and either $\pi=0$ or $\pi$ is a projection of norm $1$.
You can then see that the image of $\mathcal A$ under $\Phi$ is a subset of $\mathcal B':=\pi \mathcal B \pi$ (because $\Phi(A)=\Phi(\mathbb{1}_{\mathcal A} A \mathbb{1}_{\mathcal A})=\pi\Phi(A)\pi$). $\pi$ is a unit in $\mathcal B'$ and $\mathcal B'$ is a $C^*$ algebra, completeness in the metric sense follows from:
$$\pi A_n \pi -B \to 0 \implies \pi (\pi A_n \pi - B)\pi = \pi A_n \pi - \pi B \pi \to 0$$
And you have reduced to the case of a unital $*$-morphism. I'll add the proof of that case for completeness:
If $A \in \mathcal A$ then $\sigma_\mathcal{A}(A)\supset \sigma_\mathcal{B'} (\Phi(A))$, because unitary morphisms conserve invertibility.
In the case of a self-adjoint $A$ you then have $\|A\|=\sup \sigma_{\mathcal A}(A)$ and $\|\Phi(A)\|=\sup \sigma_\mathcal{B'} (\Phi(A))≤\|A\|$.
If $A$ is not self adjoint, $A^*A$ is so and $\|A\|^2=\|A^*A\|≥\|\Phi(A^*A)\|=\|\Phi(A)^*\Phi(A)\|=\|\Phi(A)\|^2$.