If $f\colon X\to Y$ is continuous and $X$ is compact Hausdorff and connected, $Y$ is given an order topology. Is $f$ necessarily onto?

Question

I want to ask you for following statement

$f$ is continuous from $X$ to $Y$, where $X$ is compact Hausdorff, connected and $Y$ is ordered set in order topology, then $f$ is onto?

My attempt

For given $y \in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff, $B=f^{-1}(\bar N_y$) is compact in X , Where $\bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* \in B$ such that $f(x*)=y$ . Thus $f$ is onto

First, I want to ask whether this proof is correct or not.

Second, Is there any counter example if X has only connect condition ?

Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$

Thank you!

Answer 1

Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.

A very simple counterexample to your statement is the inclusion $[0, 1] \to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $\{0\} \to \{0, 1\}$, assuming that $Y$ need not be connected).

Answer 2

Your proof is not correct as pointed out by Mees de Vries.

However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:X\rightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.