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 $C^\infty$?
 A: See Banach-Stone theorem for desription of isometric isomorphisms of $C_0$ spaces.
A: 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.


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*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 $M$.

*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.)

*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).
A: 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.)
