If the inverse image of a set is open then is the set open?

Question

Let $f:M\to N$ be a continuous map between smooth manifolds and suppose that $U\subset N$. If $f^{-1}(U)$ is open then can we conclude that $U$ is open?

Answer 1

No, we cannot. Take, for instance$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\displaystyle\frac1{1+x^2}.\end{array}$$Then $U=\left(\frac12,1\right]$ is not open, but $f^{-1}(U)=(-1,1)$, which is open.

Answer 2

A very simple counterexample: $$ f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto 0. $$ Then $f^{-1}(\{0\})=\mathbb{R}$ is open, but $\{0\}$ is not.

Answer 3

You can see why this need not be true in general by noting that what you're asking for is essentially that for your continuous mapping, you're asking whether open subsets of the domain are mapped to open images. But this is not required for continuity. Hence it need not hold.