# A theorem due to Gelfand and Kolmogorov

For any topological space $X$, we can define $C(X)$ to be the commutative ring of continuous functions $f\,:\,X\rightarrow \mathbb{R}$ under pointwise addition and multiplication. Then $C(-)$ becomes a contravariant functor $C(-)\,:\,\bf{Top}\rightarrow \text{ComRing}$.

A theorem due to Gelfand and Kolmogorov states the following:

Let $X$ and $Y$ be compact Hausdorff spaces. If $C(X)$ and $C(Y)$ are isomorphic as rings, then $X$ and $Y$ are homeomorphic.

I encountered this theorem as an example in a book on homological algebra, without proof. I have searched for the proof, but have been unable to find it.

If anyone has an idea of how to prove this, or a reference to a proof, I would appreciate it greatly.

• You might already know this, but there is a similar statement which is known as Gelfand duality. It deals with functions to $\mathbb{C}$ and commutative $C^*$-algebras though. Nov 1 '12 at 12:37
• Hist. ref.: И. М. Гельфанд, А. Н. Колмогоров, “О кольцах непрерывных функций на топологических пространствах”, Докл. АН СССР, 22:1 (1939), 11–15 Nov 1 '12 at 12:39
• An english translation of (part of) the paper referenced by @Grigory's (I hope...). Nov 1 '12 at 12:47

I found a proof in these lecture notes by Garrido and Jaramillo. See Theorem 18. They also have historical references.

• Note that the theorem in those notes uses the $\mathbb{R}$-algebra structure on $C(X)$, not just the ring structure. Aug 6 '19 at 5:54
• @EricWofsey: The R-algebra structure can be canonically reconstructed from the ring structure in this case. Nov 24 '19 at 16:22
• @DmitriPavlov: I'm aware of that; just pointing out that the reference given in this answer does not fully answer the question. Nov 24 '19 at 16:27

Dugundji's Topology has a very short, readable proof.

you can find the proof on page 289. its very readable.

• Can you provide a more specific reference? Jan 24 '14 at 12:29

The key to the proof is the following fact.

Lemma: Let $$X$$ be a compact space and let $$\varphi:C(X)\to\mathbb{R}$$ be a ring-homomorphism. Then there exists $$x\in X$$ such that $$\varphi(f)=f(x)$$ for all $$f\in C(X)$$.

Proof: Note that if $$f\in C(X)$$ is such that $$f\geq 0$$ everywhere, then $$\varphi(f)=\varphi(\sqrt{f})^2\geq 0$$. I claim furthermore that $$\varphi$$ is an $$\mathbb{R}$$-algebra homomorphism, so $$\varphi(r)=r$$ for any $$r\in\mathbb{R}$$ (thinking of it in $$C(X)$$ as a constant function on $$X$$). Indeed, we know this must be true if $$r\in\mathbb{Q}$$; for arbitrary $$r\in\mathbb{R}$$, now use the fact that $$\varphi(r-q)\geq0$$ if $$q\leq r$$ and $$q\in\mathbb{Q}$$ and $$\varphi(q-r)\geq 0$$ if $$q\geq r$$ and $$q\in\mathbb{Q}$$.

Now suppose that $$\varphi$$ is not given by evaluation at any point. Then for each $$x\in X$$, there is a function $$f_x\in C(X)$$ such that $$\varphi(f_x)\neq f_x(x)$$. Letting $$r=\varphi(f_x)$$ and replacing $$f_x$$ with $$f_x-r$$, we may assume $$f_x(x)\neq 0$$ and $$\varphi(f_x)=0$$. Replacing $$f_x$$ with its square, we may further assume that $$f_x\geq 0$$ everywhere. By compactness of $$X$$, finitely many of the sets $$\{y:f_x(y)>0\}$$ cover $$X$$, and so adding together the corresponding $$f_x$$'s, we get a function $$f\in C(X)$$ such that $$f>0$$ everywhere and $$\varphi(f)=0$$. But then $$1/f$$ is continuous so $$f$$ is a unit and so $$\varphi(f)$$ cannot be $$0$$, so this is a contradiction.

Given this fact, the result you ask for follows easily. If $$X$$ is compact Hausdorff, then we can recover the set $$X$$ from $$C(X)$$ (up to canonical bijection) as the set of ring-homomorphisms $$C(X)\to\mathbb{R}$$. We can moreover recover the topology on $$X$$ since it is the coarsest topology that makes each element of $$C(X)$$ continuous, by Urysohn's lemma. (Here if we are identifying $$X$$ with homomorphisms $$C(X)\to\mathbb{R}$$, we can think of an element of $$C(X)$$ as a function on $$X$$ by evaluation.) So we can recover the space $$X$$ up to homeomorphism from the ring $$C(X)$$.

(In fact, it similarly follows from the Lemma that if $$X$$ and $$Y$$ are compact Hausdorff, then ring-homomorphisms $$C(X)\to C(Y)$$ are naturally in bijection with continuous maps $$Y\to X$$, and this preserves composition. So this gives a contravariant equivalence of categories between compact Hausdorff spaces and rings of the form $$C(X)$$.)

• Hey Eric, can this approach be made to work when one considers complex-valued functions instead? Nov 17 '20 at 1:55
• To prove what result, exactly? That if $C(X,\mathbb{C})\cong C(Y,\mathbb{C})$ as rings then $X\cong Y$ (for compact Hausdorff spaces $X,Y$)? Nov 17 '20 at 2:08
• The issue with that is that the direct analogue of my Lemma is horribly false over $\mathbb{C}$ (for instance, even in the case where $X$ is a single point). But you can replace the Lemma with a different Lemma that says every maximal ideal is the set of functions that vanish at some point, and the proof is very similar. Nov 17 '20 at 2:16
• The remainder of the proof using the Lemma is then similar, except that you have to determine the topology on $X$ (identified with the maximal ideals of $C(X,\mathbb{C})$) differently, since you don't know what the evaluation maps are. However, the closed sets of a compact Hausdorff space are generated by the zero sets of elements of $C(X,\mathbb{C})$, so you can recover $X$ topologically as the maximal spectrum of $C(X,\mathbb{C})$ with the Zariski topology. Nov 17 '20 at 2:18
• That's great, thanks! Nov 17 '20 at 2:23

Gillman-Jerison, Rings of continuous functions, Theorem 7.3.