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Let $ X $ be a compact Hausdorff space. Let $ \psi $ be a homeomorphism on $ X $. Let $ \text{Aut}(C(X)) $ be the group of automorphisms of $ C(X) $, and $ \text{Homeo}(X) $ be the group of homeomorphisms on $ X $. Show that the mapping $ \Psi: \text{Homeo}(X) \to \text{Aut}(C(X)) $ defined by $ \Psi(\phi) := \Phi $, where $ [\Phi(f)](x) := f(\phi(x)) $ for $ x \in X $ and $ f \in C(X) $, is a group isomorphism.

I was able to show the map $ \Psi $ is bijective. I am having trouble with showing $ \Psi $ is a homomorphism. For let $ \phi_{1},\phi_{2} \in \text{Homeo}(X) $, $ \Phi_{1} = \Psi(\phi_{1}) $ and $ \Psi(\phi_{2}) = \Phi_{2} $. Then $ {\Phi_{1}}(f) = f \circ \phi_{1} $ and $ {\Phi_{2}}(f) = f \circ \phi_{2} $. Hence, $ \Psi(\phi_{1} \circ \phi_{2})(f) = f \circ (\phi_{1} \circ \phi_{2}) $ but $ (\Psi(\phi_{1}) \circ \Psi(\phi_{2}))(f) = (\Phi_{1} \circ \Phi_{2})(f) = f \circ \phi_{2} \circ \phi_{1} $, which does not equal $ \Psi(\phi_{1} \circ \phi_{2}) $.

Am I doing something wrong or is the problem wrong?

Haskell Curry
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john
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    There is a problem of variance: the definition of $\Psi$ makes it an anti-isomorphism, i.e., $\Psi(\phi_1) \circ \Psi(\phi_2) = \Psi(\phi_2 \circ \phi_1)$. If you want an honest isomorphism you should define $\Phi(f)(x) = f(\phi^{-1}(x))$ and things work out the way they should. – Martin Jan 14 '13 at 21:07
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    Martin, I think you can turn your comment into an answer. – Norbert Jan 14 '13 at 21:40

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