Let $X=\mathbb{S}^2$ and $Y=\mathbb{R}^2$ and $f:X\to Y$ a continuous mapping. Is it true that there must exist a nonempty set $V\subset f(X)$, open in $f(X)$ (in the subspace topology), such that each point of $V$ has at least two preimages under $f$?
Two followup questions:
- If the answer is negative, what additional assumptions can we place on $f$ to make the answer positive? For example, is it enough to assume that $f$ sends open sets in $X$ to sets with nonempty interior in $Y$?
- Is dimension 2 important here?
