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

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 CalgebraHomomorphismsHaveClosedImages 2013-03-22 17:44:37 2013-03-22 17:44:37 asteroid (17536) asteroid (17536) 9 asteroid (17536) Theorem msc 46L05 image of $C^{*}$-homomorphism is a $C^{*}$-algebra