Does $C_0(X)$ completely determine $X$?

Let $X$ and $Y$ be compact metric spaces. Let $C_0(X)$ and $C_0(Y)$ be the Banach spaces of continuous real-valued functions over $X$ and $Y$, respectively. If $F : X \rightarrow Y$ is a homeomorphism, then $f\mapsto f\circ F$ is an isomorphism from $C_0(Y)$ to $C_0(X)$. Now suppose conversely that $C_0(X)$ and $C_0(Y)$ are isomorphic as Banach spaces. Are $X$ and $Y$ homeomorphic?

What if we replace metric spaces by manifolds and $C_0$ by $C^\infty$?

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You only need compact Hausdorff. This is also true for second-countable smooth manifolds, I think. –  Qiaochu Yuan Nov 14 '12 at 1:24
@Jason: sorry, by "this" I meant the version of the question in the last sentence with $C$ replaced by $C^{\infty}$. –  Qiaochu Yuan Nov 14 '12 at 2:51
What exactly do you mean by "isomorphic as Banach spaces"? Should there be a bijective isometry between them, or only a linear homeomorphism? –  Nate Eldredge Dec 4 '12 at 17:41

In the smooth case, you definitely need to know $C^\infty$, as there are examples of compact smooth manifolds which are homeomorphic but not diffeomorphic.

On the other hand $C^\infty(M)$ does determine $M$.

Here's the sketch of the proof, which is essentially outlined in the early exercises to Milnor and Stasheff's Characteristic Classes book.

1. For each $x\in M$, define $ev_x:C^\infty(M)\rightarrow\mathbb{R}$ by $ev_x(f) = f(x)$ - it evaluates $f$ at the point $x$. Prove this is a surjective ring homomorphism onto a field, hence the kernel is a maximal ideal of $M$.

2. Show that, in fact, every maximal ideal of $M$ is of the form $\ker(ev_x)$ for some $x\in M$. (This part is no longer true in the noncompact setting. For example, the collection of smooth compactly supported functions forms a maximal ideal which is not of this form.)

3. Given a ring isomorphism $\phi:C^\infty(M)\rightarrow C^\infty(N)$, note that $\phi$ must take maximal ideals in $C^\infty (M)$ to maximal ideals in $C^\infty (N)$. Use this to define $f: M\rightarrow N$ by $ev_{f(x)} \circ \phi = ev_x$. To see $f$ is smooth, notice that this equation gives that for any $g\in C^\infty(M)$, we have $g(x) = (\phi(g)\circ f)(x)$. It follows that the composition of $f$ with any function in $C^\infty(N)$ is smooth, in particular with local coordinate functions (extended to the whole of $N$ using a partition of unity).

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I'm not familiar with any proofs of the Banach-Stone theorem (indeed, I'm only vaguely aware of the theorem itself.) As such, I'm not sure how close this proofs models one which proves Banach-Stone. –  Jason DeVito Nov 14 '12 at 3:05
My understanding is that in the noncompact case step 2) should be replaced by "show that every maximal ideal $M$ with residue field $\mathbb{R}$..." although I haven't worked through the details. The proof of the compact Hausdorff version has essentially the same outline (the main technical result needed is Urysohn's lemma). –  Qiaochu Yuan Nov 14 '12 at 3:11

See Banach-Stone theorem for desription of isometric isomorphisms of $C_0$ spaces.

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Isomorphism of $C(X)$ and $C(Y)$ does not mean that $X$ and $Y$ are homeomorphic. Easy example: $X = [0,1]$, $Y= X \cup \{2\}$. The Banach Stone theorems says that $X$ and $Y$ are homeomorphic if $C(X)$ and $C(Y)$ are isometric. ($X, Y$ compact Hausdorff.)

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What's an example of a isomorphism between $C(X)$ and $C(Y)$? –  Nate Eldredge Dec 4 '12 at 17:42
$C[0,1]$ contains a complemented copy of $c_0$. Write $C[0,1] = Z\oplus c_0 = Z\oplus (c_0 \oplus \mathbb{R}) = (Z\oplus c_0) \oplus \mathbb{R} = C[0,1]\oplus \mathbb{R} = C([0,1]\cup \{2\})$. [$=$ in the sense of isomorphism.] –  Denny Leung Dec 7 '12 at 8:52