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.
 A: The proof I know is not super hard, but it's not an easy exercise either.
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 $\beta h=\Gamma$.
