I have two questions regarding the proof of a proposition in Landsman's notes on $C^∗$-Algebras, Hilbert $C^∗$-modules, and Quantum Mechanics. (Note that $\Delta(\mathcal{A})$ is the structure space of a commutative Banach algebra $\mathcal{A}$, hence the set of all nonzero homomorphisms $\mathcal{A}\to\mathbb{C}$)
$\mathbf{Proposition 2.4.3}$ Let $X$ be a compact Hausdorff space and regard $C(X)$ as a commutative $C^*$-algebra. Then $\Delta(C(X))$ (equipped with the Gel'fand topology) is homeomorphic to $X$.
For all $x\in X$ we define the linear map $\omega_x:C(X)\to\mathbb{C}; f\mapsto f(x)$. With this map we define the Evaluation map $E:X\to\Delta(C(X))$ given by $E(x):f\to f(x)$.
After showing that $E$ is bijective, it remains to be shown that $E$ is an homeomorphism. We define $X_0$ as $X$ with the original topology and $X_G$ with the topology induced by $E^{-1}$ (the open sets are the sets of the form $E^{-1}(\mathcal{O})$ with $\mathcal{O}\subset\Delta(C(X))$ open). Moreover it holds that $\hat{f}\circ E=f$ ($\hat{f}$ is the Gel'fand transform of $f$) and the Gel'fand topology is by def the weakest topology s.t. all $\hat{f}$ are continuous. Then it was said that therefore $X_G$ is weaker than $X_0$ (since $f$, lying in $C(X_0)$, is continuous)
I don't understand why $X_G$ is weaker. The later following argumentation is clear to me.
Thanks for your help
Landsman's proof here is confusing. What I think he tries to say is that since $\hat{f} \circ E$ is continuous in the original topology, the original topology on $X$ makes $\hat{f}$ continuous (when you identify $X$ with $\Delta(C(X))$ through $E$). However, $X_G$ is the weakest topology on $X$ which does this, so $X_G$ is indeed weaker than $X_o$. What he does is to go back and forth between $X$ and $\Delta(C(X))$, having already identified them as sets via $E$.
I think the "classical" way of showing this is much easier to understand and so I'll add it here. Observe that since $X$ and $\Delta(C(X))$ are compact Hausdorff spaces, it suffices to show that $E$ is continuous. Indeed, consider a net $(x_i)_i$ in $X$ converging to $x \in X$. We have to show that $(E(x_i))_i$ converges to $E(x)$ in the weak topology on $\Delta(C(X))$. However, that is immediate since that is the topology of pointwise convergence, i.e. for every $f \in C(X)$ we have $$E(x_i)f = f(x_i) \xrightarrow{i} f(x) = E(x)f.$$