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Suppose that $X$ and $Y$ are two compact Hausdorff spaces and $F\colon C(X) → C(Y)$ is a continuous isomorphism of algebras. Can I say $X$ and $Y$ are homeomorphic?

The key words always lead me to other questions. Can anyone give me some reference? Thanks a lot!

Tomasz Kania
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yoyo
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2 Answers2

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This is the so-called Gelfand-Kolmogorov theorem. It says:

Let $X$ and $Y$ be compact, Hausdorff spaces. Suppose that there exists a ring isomorphism $T\colon C(X)\to C(Y)$. Then there exists a homeomorphism $h\colon Y\to X$ such that $$Tf = f\circ h\text{ for all }f\in C(X).$$ In particular, $T$ is a continuous algebra isomorphism.

Click here for some references. Actually there is a more general result due to Milgram which asserts that two compact Hausdorff spaces $X$ and $Y$ are homeomorphic if and only if there exists a multiplicative bijection between $C(X)$ and $C(Y)$.

Tomasz Kania
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An algebra isomorphism of $C(X)$ to $C(Y)$ is automatically continuous in the norm, because $\|f\| = \sup \{|\lambda|: \lambda \in \sigma(f)\}$. $F$ induces a homeomorphism of the maximal ideal spaces, and these are in turn homeomorphic to $X$ and $Y$ respectively.

Robert Israel
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  • Thanks for answer the question, but I still a little bit confused on how to induce a homeomorphism. – yoyo Aug 06 '16 at 03:37
  • Is your "automatically continuous" means "give any algebra isomorphism $\phi :C(X) → C(Y)$, then $\phi$ is continuous."? – yoyo Aug 06 '16 at 03:41
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    @yoyo, yes. It is actually automatically of the form $\phi(f)=f\circ h$, where $h$ is a homeomorphism. This is the content of the Gelfand-Kolmogorov theorem. See my post below. – Tomasz Kania Aug 06 '16 at 07:57