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

Theorem - Let $f:\mathcal{A}\u27f6\mathcal{B}$ 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 $\mathcal{B}$.

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*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 $\stackrel{~}{f}:\mathcal{A}/\mathrm{Ker}f\u27f6\mathcal{B}$.

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

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

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