$C^*$-algebra homomorphisms have closed images
CAlgebraHomomorphismsHaveClosedImages
2008-01-14 20:19:19
2008-04-21 23:18:59
Theorem
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{\bf Theorem -} Let $f: \mathcal{A} \longrightarrow \mathcal{B}$ be a *-homomorphism between the \PMlinkname{$C^*$-algebras}{CAlgebra} $\mathcal{A}$ and $\mathcal{B}$. Then $f$ has \PMlinkname{closed}{ClosedSet} \PMlinkname{image}{Function}, i.e. $f(\mathcal{A})$ is closed in $\mathcal{B}$.
Thus, the image $f(\mathcal{A})$ is a $C^*$-subalgebra of $\mathcal{B}$.
$\,$
{\bf \emph{Proof:}} The kernel of $f$, $\mathrm{Ker} f$, is a closed two-sided ideal of $\mathcal{A}$, since $f$ is continuous (see \PMlinkname{this entry}{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 \PMlinkname{this entry}{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 $\mathcal{B}$. $\square$