# A group isomorphism $\text{Homeo}(X)\to\text{Aut}(C(X))$?

Let $$X$$ be a compact Hausdorff space and consider the associated commutative C*-algebra $$C(X)$$ of complex values continuous functions on $$X$$. For a homeomorphism $$\phi\colon X\to X$$ we have a *-isomorphism $$T_{\phi}\colon C(X)\to C(X)$$ given by $$T_{\phi}f:=f\circ\phi^{-1}$$. This yields a group homomorphism $$\beta\colon\text{Homeo}(X)\to\text{Aut}(C(X))$$ given by $$\beta(\phi):=T_{\phi}$$. I suspect that $$\beta$$ is a group isomorphism, but I cannot prove surjectivity. Injectivity should follow fairly quickly from Urysohn's lemma. Any suggestions would be greatly appreciated.

• By Gelfand duality $X\mapsto C(X)$ is an equivalence of categories between compact Hausdorff spaces and commutative unital C*-algebras, so it respects automorphism groups and anything else May 12, 2021 at 21:47

Let $$\Gamma:C(X)\to C(X)$$ be a C$$^*$$-algebra isomorphism. For each $$x\in X$$, the map $$\Gamma_X:C(X)\to\mathbb C$$ given by $$\Gamma_x(g)=(\Gamma g)(x)$$ is a nonzero multiplicative linear functional (that is, a character). It is a well-known fact (that requires several steps to be proven) that the characters of $$C(X)$$ are the point evaluations; so there exists a point $$h(x)\in X$$ with $$(\Gamma g)(x)=g(h(x)),\qquad g\in C(X).$$ Since $$\Gamma g$$ is continuous, if $$x_j\to x$$ then $$g(h(x_j))=(\Gamma g)(x_j)\to (\Gamma g)(x)=g(h(x)).$$ This can be done for any $$g\in C(X)$$, so we get that $$h(x_j)\to h(x)$$. Thus $$h$$ is continuous. Repeating the process but now for $$\Gamma^{-1}$$, there exists $$h':X\to X$$, continuous, with $$(\Gamma^{-1}g)(x)=g(h'(x))$$. Then, for any $$f\in C(X)$$, $$f(x)=(\Gamma\Gamma^{-1}f)(x)=(\Gamma^{-1}f)(h(x))=f(h'\circ h(x)).$$ As this works for all $$f\in C(X)$$, we get that $$h(h'(x))=h'(h(x))=x$$, so $$h$$ is a homeomorphism with $$\beta h=\Gamma$$.