Gelfand-Kolmogorov Theorem for the space $C(X)$ with compact $X$: ring vs algebra version.

Question

A theorem by Gelfand and Kolmogorov comes in two different guises, depending on which structure we consider on $C(X)$, namely if that of unital commutative ring or that of algebra.

Ring version Let X and Y be compact spaces. Then, $C(X)$ and $C(Y)$ are isomorphic as rings if, and only if, $X$ and $Y$ are homeomorphic.
Moreover, every rings isomorphism $T : C(Y)→ C(X)$ is of the form $Tf = f\circ h$ where $h:X →Y$ is a homeomorphism.

Identically, we have:

Algebra version Let X and Y be compact spaces. Then, $C(X)$ and $C(Y)$ are isomorphic as algebras if, and only if, $X$ and $Y$ are homeomorphic.
Moreover, every algebra isomorphism $T : C(Y)→ C(X)$ is of the form $Tf = f\circ h$ where $h:X →Y$ is a homeomorphism.

Now these two versions are claimed equivalent, but I cannot find an explicit proof of the fact, nonetheless I think the equivalence is based on the following

Lemma:
Each nonzero ring homomorphims $\omega:C(Y)\rightarrow \mathbb{R}$ is surjective and for each constant $c\in\mathbb{R}$, letting $\delta\equiv 1$, we have $\omega(c\delta)=c$.

• Is this sufficient to prove that any ring homomorphism $\lambda: C(X)\rightarrow C(Y)$ is also an algebra homorphism? If not how to prove it?

Morover, I think the two equivalent versions of the theorem have a categorical rephrasing: basically we have two contravariant functors:
$C(−):Top→ComRing$
$C(−):Top→ComAlg $

• Can we say something more about the categorical interpretation of this theorem?
• Has the Lemma or whatsoever the result needed to prove the equivalence, or the equivalence itself, a categorical interpretation too?

Answer 1

I recalled a reference from a classic:

Exercise 1I in Gillman and Jerison's book Rings of continuous functions says:

let $\mathfrak{t}$ be a (ring) homomorphism from $C(Y)$ or $C^\ast(Y)$ (bounded continuous real-valued functions) into $C(X)$.

1. $\mathfrak{t}\mathbf{r} = \mathbf{r}\cdot \mathfrak{t}\mathbf{1}$ for each $ r \in \Bbb R$ (where $\mathbf{r}$ is the constant function with value $r$). With hint

For each $x \in X$, the mapping $x \to (\mathfrak{t}\mathbf{r})(x)$ is a homomorphism from $\Bbb R$ into $\Bbb R$, and hence is either the zero homomorphism or the identity (0.22). So $(\mathfrak{t}\mathbf{1})(x) = 0$ or $1$.

2. $\mathfrak{t}$ is an algebra homomorphism, i.e. $\mathfrak{t}(\mathbf{r}g) = \mathbf{r}\cdot \mathfrak{t}(g)$ for all $ r \in \Bbb R$ and $g \in C(Y)$.

I'll refrain from commenting on the categorial part.

Answer 2

It's not hard to prove two stronger statements (which I would call the commutative Gelfand-Naimark theorem) with clear categorical meanings, which are that

• the functor $C(-)$, regarded as taking values in either rings or algebras, is fully faithful. This means that it induces a bijection of sets $\text{Hom}(X, Y) \cong \text{Hom}(C(Y), C(X))$. Essentially this means we can replace "isomorphism" and "homeomorphism" in the second half of your statements with "homomorphism" and "continuous function." (This implies the "isomorphism" and "homeomorphism" versions of the statements, basically by functoriality.)
• the functor $C(-)$, regarded as taking values in commutative real C*-algebras, is a (contravariant) equivalence of categories. (This theorem is usually stated for complex C*-algebras but it's also true for real ones, the two categories are equivalent in the commutative case.)