$C^{*}$ -algebra homomorphisms have closed images

Theorem - Let $f:\mathcal{A}\longrightarrow\mathcal{B}$ be a *-homomorphism between the $C^{*}$ -algebras (http://planetmath.org/CAlgebra) $\mathcal{A}$ and $\mathcal{B}$ . Then $f$ has closed (http://planetmath.org/ClosedSet) image (http://planetmath.org/Function), 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 (http://planetmath.org/HomomorphismsOfCAlgebrasAreContinuous)). 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 (http://planetmath.org/InjectiveCAlgebraHomomorphismIsIsometric)), 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 $\mathcal{B}$ . $\square$

Title	$C^{*}$ -algebra homomorphisms have closed images
Canonical name	CalgebraHomomorphismsHaveClosedImages
Date of creation	2013-03-22 17:44:37
Last modified on	2013-03-22 17:44:37
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	9
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46L05
Synonym	image of $C^{}$ -homomorphism is a $C^{}$ -algebra

C*-algebra homomorphisms have closed images

$C^{*}$ -algebra homomorphisms have closed images