Mrthdrb

It is well-known that a compact Hausdorff $X$ space is scattered if and only if admits no continuous maps onto the unit interval $[0,1]$.

Surprisingly, but I cannot find a good reference to this well-known fact (desirably some textbook).

In the survey paper "Scattered spaces" in Encyclopedia of General Topology this fact is not mentioned, unfortunately.

The proof in the direction that there is no continuous surjection from a compact scattered $X$ onto $[0,1]$ may be found here (Theorem 1):

W. Rudin, Continuous functions on compact spaces without perfect subsets, Proc. Amer. Math. Soc. 8 (1957), 39-42.

And the full characterisation may be found in (Theorem 8.5.4, p. 148):

Z. Semadeni, Banach spaces of continuous functions, vol. 1, PWN - Polish Scientific Publishers, Warsaw 1971.

I'm not properly answering since you're asking for a reference and I don't know any; however here's a hopefully reasonably concise proof.

First implication:

(1) Let $X,Y$ be compact Hausdorff topological spaces, such that there $X$ is scattered and such that there exists a continuous surjective map $Xto Y$. Then $Y$ is scattered.

Proof: otherwise, we can reduce to the case when $Y$ is perfect (nonempty); let $f$ be the map. By compactness, let $Z$ be a minimal compact subset of $X$ on which $f$ is surjective. So $Z$ has an isolated point $z$, and since $Y$ is perfect, $f$ is still surjective on $Zsmallsetminusz$, contradiction.

Reverse implication:

(2) Let $X$ be compact Hausdorff and not scattered. Then there exists a continuous surjective map $Xto [0,1]$.

Proof. If $X$ is not totally disconnected, choose $xneq x'$ in the same connected component and directly apply Urysohn's lemma (which ensures the existence of a continuous map $Xto [0,1]$ mapping $x$ to $0$ and $x'$ to 1; connectedness ensures surjectivity.

Otherwise, suppose that $X$ is totally disconnected and non-scattered; in this case it's enough to prove that $X$ has a continuous surjection onto the Cantor set. By Stone duality, it's enough to embed a free BA of countable rank in the Boolean algebra of $X$. Since we can lift free BA's, we can assume that $X$ is perfect (nonempty, by assumption). In this case, it's immediate by an induction to produce a countable non-atomic subalgebra.

Edit: now Damian Sobota has provided a complete reference. Actually the above proof of (1) is the same argument as Rudin's, which is the same as the one given in Semadeni's book. For (2), the dichotomy between the totally disconnected case and the other case also appears in Semadeni's proof; my proof is essentially the same as Semadeni's, except that I used a formulation in terms of Boolean algebras while Semadeni's one is directly formulated in terms of subdivisions of clopen subsets.