Irreducible mapping between compact Hausdorff spaces with no singleton fibers

Question

A continuous surjection $f\colon X \to Y$ is called irreducible if for every proper closed subset $A \subset X$, $f(A) \neq Y$. I have to construct an irreducible mapping between compact Hausdorff spaces, which has no singleton fibers.

Here is my observation.

Let $f\colon X \to Y$ be such a mapping. Since it is a mapping from a compact space into a Hausdorff space, it is closed. Thus $f$ is a quotient mapping. The irreducibility of $f$ can be rephrased to this: For every nonempty open set $U$ (here we may fix a basis to pick from), there exists $y \in Y$ s.t. $f^{-1}(y) \subset U$ i.e. $U$ contains a fiber. $Y$ being Hausdorff is equivalent to saying that two distinct fibers can be separated by saturated open sets. To sum up, we should find a partition of a compact Hausdorff space containing no singletons such that every nonempty open set contains a cell(an element of the partition) and every cell can be separated by saturated open sets.

Starting from $X = [0, 1]$, let $x \sim y$ when $x, y$ are triadic rationals different than $0, 1$ and $x$ can be transformed to $y$ with changing the rightmost digit in ternary representation from $1$ to $2$ or $2$ to $1$. With this I was able to tuck two point cells in every nonempty open set and find saturated open sets separating them. The problem is that $\sim$ has too many singleton cells. And further identifying singleton cells modulo $1/3$ breaks the Hausdorff condition. I'm stuck here.

How can one construct such mapping?

Answer 1

Here's a hint for one way to construct an example. Try taking a product of two totally ordered sets $A\times B$ with the lexicographic order and the projection map $A\times B\to A$. The idea is that if you choose $A$ and $B$ appropriately, then any open interval in the lexicographic order will be forced to contain an entire copy of $B$. Getting all the details to work out is a little tricky, though, and will require some minor modification of this idea.

More details are hidden below.

Let us first consider $X=[0,1]\times\{0,1\}$ with the lexicographic order, and the projection map $f:X\to [0,1]$. Given a point $(a,1)\in X$, an open interval around it must contain both $(b,0)$ and $(b,1)$ for all sufficiently small $b>a$. Similarly, an open interval around $(a,0)$ must contain both $(b,0)$ and $(b,1)$ for all sufficiently large $b<a$. So, this would seem to show that $f$ is irreducible. However, there is an issue at the endpoints: the singletons $\{(0,0)\}$ and $\{(1,1)\}$ are open.

So now the trick is, let's have our interval $[0,1]$ wrap around so that it doesn't have any endpoints. That is, let $X'$ be the quotient of $X$ that identifies $(0,0)$ with $(1,0)$ and $(0,1)$ with $(1,1)$. You can think of $X'$ as $S^1\times \{0,1\}$ with the "lexicographic cyclic order". There is then still a continuous projection map $f:X'\to S^1$, which now really is irreducible and has no singleton fibers.