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If $X$ and $Y$ are compact Hausdorff spaces, show that for any algebra homomorphism $$ F:C(Y) \to C(X) $$ there exists a continuous function $f:X\to Y$ such that $$ F(\phi)=\phi \circ f, \forall \phi \in C(Y) $$

The spaces are compact Hausdorff, so presumably one should use the Gelfand-Naimark theorem and construct a continuous function from the spectra of $C(X)$ and $C(Y)$. To be honest I'm pretty confused. I don't know exactly what I should try to do; using characters and spectra instead of points in spaces doesn't seem to get me closer to the answer. Any help would be appreciated.

Please note that this is not homework.

EDIT: I've found the solution.

Proof. Let $A=C(X)$ and $B=C(Y)$ be two commutative algebras over $\mathbb{R}$. If $X$ and $Y$ are compact Hausdorff spaces, show that for any algebra homomorphism $$ F:B \to A $$ there exists a continuous function $f:X\to Y$ such that $$ F(\phi)=\phi \circ f, \forall \phi \in C(Y) $$ We first define all the relevant maps. Let $X_A$ and $Y_B$ be the spectra of $A$ and $B$ respectively. Let $\sigma: Y_B \to Y$ and $\tau: X_A \to X$ be the homeomorphisms induced by the Gelfand Naimark theorem, since $X$ and $Y$ are compact Hausdorff. Let $\lambda: A \to \mathbb{R}$ and $\chi: B\to \mathbb{R}$ be two characters in $X_A$ and $Y_B$ respectively. Notice that $\lambda \circ F$ is a function from $B \to \mathbb{R}$ that lies in $Y_B$ since $F$ respects the $\mathbb{R}$-linearity and multiplicative identities of the algebra structure on $\mathbb{B}$ and $\lambda$ is a character and therefore respects the same identities by definition. So there exists an injective function $h: X_A \to Y_B$.

We prove that $h$ is continuous. Apply Exercise 4: we have that $ f_a \circ h = f_a(h(\lambda))= f_a(\lambda \circ F)=\lambda(F(a))= g_{F(a)}$ with $g_a: X_A \to \mathbb{R}$ and $g_a(\lambda)=\lambda(a)$ and $f_a: Y_B \to \mathbb{R}$ and $f_a(\chi)=\chi(a)$.

Define $f$ as follows: $f(x):= (\sigma \circ h \circ \tau)(x)$. The composition of continuous function is continuous so f is continuous.

We now prove the identity $\phi \circ f=F(\phi)$. \begin{align*} (\phi \circ f)(x)&= (\phi \circ \tau \circ h \circ \sigma)(x) \\ &= (\phi \circ \tau\circ h) (\chi_x)\\ &= (\phi \circ \tau \circ \chi_x \circ F)\\ &= (\phi \circ \tau \circ \chi_y)\\ &=\phi(y)\\ &= \chi_y(\phi)\\ &=(\chi_x \circ F)(\phi)\\ &=F(\phi)(x) \end{align*}

Lee Wang
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  • Dear Lee, You are correct that one can apply GN and construct a continuous function on spectra (as the answer below shows). Since you seemed to anticipate this, why exactly are you confused? Regards, – Matt E Jun 17 '13 at 12:27
  • Dear Matt, I simply don't have a lot of experience dealing with these kinds of more abstract questions, being far removed from the geometric or analytic notions of pointset topology I'm familiar with. Also the proof below seems to use only the algebras not the spectra? – Lee Wang Jun 17 '13 at 12:44
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    Dear Lee, The spectrum of $C(X)$ is the set of maximal ideals of $C(X)$, and this is what the answer is using. The basic point is that the spectrum is naturally in bijection with $X$ (via $x \mapsto I_x$, as in the answer), and so this gives a way to get back from $C(X)$ to $X$. Regards, – Matt E Jun 17 '13 at 13:10

1 Answers1

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Any point $x\in X$ gives rise to a maximal ideal $I_x=\{f\in C(X):f(x)=0\}$. Then $F^{-1}(I_x)$ is a maximal ideal of $C(Y)$, so, thanks to Gelfand-Naimark (and compactness and Hausdorffness), there is a unique $y\in Y$ such that $F^{-1}(I_x)=\{g\in C(Y):g(y)=0\}$. Define $f(x)$ to be $y$. There's a lot that needs to be checked --- continuity of $f$ and the formula $F(\phi)=\phi\circ f$, but I believe all the relevant work is already implicit in the proof (if not the statement) of the Gelfand-Naimark theorem.

(My computer's spell-checker tells me that "Hausdorffness" isn't a word, and I have to agree with that, but you know what it means anyway. My computer also tells me that "Gelfand" isn't a word, but apparently "Naimark" is OK.)

Andreas Blass
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  • Thanks for this answer. I have one small question though, how do we know that $I_y$ is maximal? Apparently it fails in general rings, though perhaps not algebras: http://math.stackexchange.com/questions/31776/inverse-image-of-maximal-ideals – Lee Wang Jun 17 '13 at 16:31
  • @LeeWang You're right; I should have included the following in my answer. Compose $F$ with the quotient map $C(X)\to C(X)/I_x$ to get a homomorphism whose kernel is $F^{-1}(I_x)$. So it induces an embedding $C(Y)/F^{-1}(I_x)\to C(X)/I_x$. But $C(X)/I_x$ is just the field $\mathbb C$. So $C(Y)/F^{-1}(I_x)$ is a unital $\mathbb C$-algebra embedded in $\mathbb C$. The only such algebra is $\mathbb C$, so $C(Y)/F^{-1}(I_x)$ is isomorphic to $\mathbb C$. In particular, it's a field, and so $F^{-1}(I_x)$ is a maximal ideal. – Andreas Blass Jun 18 '13 at 20:46
  • Thank you, I've found the solution. Took a slightly different road. Solution in the original post. – Lee Wang Jun 22 '13 at 11:49