Theorem - Let $f: \mathcal{A} \longrightarrow \mathcal{B}$ be a *-homomorphism between the $C^*$ -algebras $\mathcal{A}$ and $\mathcal{B}$ . Then $f$ has closed image, i.e. $f(\mathcal{A})$ is closed in $\mathcal{B}$ .

Thus, the image $f(\mathcal{A})$ is a $C^*$ -subalgebra of $\mathcal{B}$ .

$\,$

Proof: The kernel of $f$ , $\mathrm{Ker} f$ , is a closed two-sided ideal of $\mathcal{A}$ , since $f$ is continuous (see this entry). Factoring threw the quotient $C^*$ -algebra $\mathcal{A}/\mathrm{Ker} f$ we obtain an injective *-homomorphism $\widetilde{f}:\mathcal{A}/\mathrm{Ker} f \longrightarrow \mathcal{B}$ .

Injective *-homomorphisms between $C^*$ -algebras are known to be isometric (see this entry), hence the image $\widetilde{f}(\mathcal{A}/\mathrm{Ker} f)$ is closed in $\mathcal{B}$ .

Since the images $\widetilde{f}(\mathcal{A}/\mathrm{Ker} f)$ and $f(\mathcal{A})$ coincide we conclude that $f(\mathcal{A})$ is closed in $\mathcal{B}$ . $\square$