TITLE: Continuous open surjection on Reals
QUESTION [7 upvotes]: If $f:\mathbb{R}\rightarrow \mathbb{R}$ is a continuous, open surjection, must it be a homeomorphism?

REPLY [6 votes]: It suffices to show that $f$ is injective. (Indeed, then $f$ is a bijection, and since $f$ is open $f^{-1}$ must be continuous.)
Suppose $f(x)=f(y)$ where $x<y$. I claim that $f((x,y))$ is not open. (Indeed, if $f$ is contant on $[x,y]$ then $f((x,y))$ is a point. Otherwise, $f$ achieves either a maximum or a minimum on $[x,y]$ which is not $f(x)$, and this must be an extreme point of $f((x,y))$.)