Injective Morphism of Commutative $C^*$-algebras

Question

Let $X$ and $Y$ be compact Hausdorff spaces, and let $\varphi : C(X) \to C(Y)$ an injective morphism of $C^*$-algebras. The book I'm reading through claims that there exists a continuous surjection $\alpha : Y \to X$ such that $$\varphi (f) = f \circ \alpha$$ for every $f \in C(X)$. But I am having trouble verifying this.

Answer 1

Here's how $\alpha$ is defined. If $y\in Y$, then $$\operatorname{ev}_y\circ\ \varphi:C(X)\to\mathbb C$$ is a non-zero $*$-homomorphism, hence there is some $\alpha(y)$ such that $\operatorname{ev}_{\alpha(y)}=\operatorname{ev}_y\circ\ \varphi$. Continuity of $\alpha$ is readily verified.

Now, suppose that $\alpha$ is not surjective. Then $\alpha(Y)$ is a proper closed subspace of $X$, hence there is a non-zero function $f\in C(X)$ which vanishes on $\alpha(Y)$ (by Urysohn's lemma). But then $$\|\varphi(f)\|_\infty=\sup_{y\in Y}|(\operatorname{ev}_y\circ\ \varphi)(f)|=\sup_{y\in Y}|\operatorname{ev}_{\alpha(y)}(f)|=0,$$ and thus $\varphi$ is not injective.