Return to Answer

Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital C*-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

Sketch of proof: $\Rightarrow$ is trivial, and $\Leftarrow$ follows from Tietze extension theorem $\square$. By the way, we also have:

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, C*-subalgebras of $C(X)$ (i.e. closed unital -subalgebras) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of C-subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.

Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital $C^{\ast}$-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

Sketch of proof: $\Rightarrow$ is trivial, and $\Leftarrow$ follows from Tietze extension theorem $\square$. By the way, we also have:

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, $C^{\ast}$-subalgebras of $C(X)$ (i.e. closed unital $\ast$-subalgebras) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of $C^{\ast}$-subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.

Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital C*-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, C*-subalgebras of $C(X)$ (i.e. closed unital *-subalgebras) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.

Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital C*-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

Sketch of proof: $\Rightarrow$ is trivial, and $\Leftarrow$ follows from Tietze extension theorem $\square$. By the way, we also have:

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, C*-subalgebras of $C(X)$ (i.e. closed unital -subalgebras) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of C-subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.

Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital C*-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, subalgebras of $C(X)$ (in the same category, hence we assume them to be closed and unital) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.

Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital C*-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, C*-subalgebras of $C(X)$ (i.e. closed unital *-subalgebras) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.