This is a problem in baby Rudin. Chapter 4, problem 15. I'm just looking at examples at the moment. It says that any continuous open map is monotone. Recall, an open map $f: X \to Y$ is such that any open set $U$ in $X$ will haves its image, $f(U)$ open in $Y$.
So, Let $$ f: \mathbb{R} \to [-1, 1] $$ $$f: x \mapsto \sin(x) $$ and consider the open interval $$ (\frac{\pi}{3}, \frac{7\pi}{12}) $$ The image of this interval in $[-1, 1]$ is $$ (\frac{\sqrt{3}}{2}, 1] $$ which is open in $[-1, 1]$ (its complement is $[-1, \frac{\sqrt3}{2}]$, certainly closed). The problem is, the function is not monotonic on this interval. It is continuous, so it must not be an open map. I'm pretty sure I have the image right, so, where am I making a mistake?
I really hope I'm not making a dumb error!