Trouble with the proof of Proposition 4.3.18 of Pedersen's Analysis Now

Recently, I wrote all this out in detail for myself, so here I share my notes with you. Note that the assumption that $X$ is Tychonoff can be ommitted. The construction works for every topological space. The Tychnoff assumption is only there to ensure that the canonical inclusion is injective.

Recall that if $A$ is a commutative $C^*$-algebra, then we can consider the space of characters $\Omega(A)$ . If $A$ is a unital $C^*$-algebra, then this becomes a compact Hausdorff space for the weak$^*$-topology. Note that we have a natural map $$i_X: X \to \Omega(C_b(X)): x \mapsto \text{ev}_x.$$ Clearly this is a continuous map, as an easy argument with nets shows.

Lemma: The map $i_X$ has dense image.

Proof: Assume to the contrary that $\overline{i_X(X)}\subsetneq \Omega(C_b(X))$. Then Urysohn's lemma applied to the compact Hausdorff space $\Omega(C_b(X))$ gives a non-zero continuous function $f: \Omega(C_b(X))\to \mathbb{C}$ that is zero on $i_X(X)$. Consider the canonical isomorphism $$\Psi: C_b(X) \to C(\Omega(C_b(X))): \omega \mapsto \text{ev}_\omega.$$ Choose $\omega \in C_b(X)$ with $\text{ev}_\omega = f$. Then for all $x \in X$, we have $$\omega(x) = \text{ev}_x(\omega) = \text{ev}_\omega(\text{ev}_x) = f(i_X(x)) = 0$$ so $\omega = 0$, which is a contradiction. $\quad \square$

Theorem: If $X$ is a topological space, then $(\Omega(C_b(X)), i_X)$ is a Stone-Čech compactification of $X$.

Proof: Let $K$ be a compact Hausdorff space and let $f: X \to K$ be a continuous map. This induces a $*$-morphism $$C(f): C(K) \to C_b(X): g \mapsto g \circ f$$ and this then induces a continuous map $$\Omega(C(f)): \Omega(C_b(X)) \to \Omega(C(K)): \chi \mapsto \chi \circ C(f)$$ Consider the homeomorphism $$i_K: K \to \Omega(C(K)): k \mapsto \text{ev}_k.$$

Then we define the continuous map $F:= i_K^{-1}\circ \Omega(C(f)): \Omega(C_b(X)) \to K$. Moreover, we have $F\circ i_X= f$. Indeed, if $x \in X$, then $$i_K(F \circ i_X(x)) = i_K (F(\text{ev}_x)) = \Omega(C(f))(\text{ev}_x) = \text{ev}_x \circ C(f)= \text{ev}_{f(x)}= i_K(f(x))$$ so that by injectivity of $i_K$ we obtain $F \circ i_X = f$.

The condition $F \circ i_X = f$ determines $F$ uniquely on $i_X(X)$, which is dense in $\Omega(C_b(X))$ by the preceding lemma. Thus $F$ is unique. $\quad \square$

Consider a special set of characters of $C_b(X)$, for each $x\in X$ define:

$$\delta_x: C_b(X)\to\Bbb C, \quad g\mapsto g(x)$$

Since the (non-zero) characters of $C_b(X)$ are the points of $\beta X$ this gives you a way of embedding $X$ into $\beta X$. Now if $f$ is some continuous function on $\beta X$ we may identify it also with an element $\tilde f\in C_b(X)$, namely $\tilde f = \delta^{-1}(f)$ using your notation. Remember that $$f(\delta_x) = \delta(\tilde f)\,(\delta_x) = [\phi \mapsto \phi(\tilde f)]\,(\delta_x)= \delta_x(\tilde f) = \tilde f(x) $$

Asking that $f$ vanishes on $X$ is asking that $f(\delta_x)=0$ for all $x\in X$, in particular looking at $\tilde f$ this becomes: $$\tilde f(x)=0\quad \forall x\in X$$ the only function in $C_b(X)$ satisfying this property is the zero function.