I stumbled over a useful consequence, that is apparently wrong for only continuous maps. Imagine $A \subset \mathbb{R}^{n-1}$ is a compact set and $F : \mathbb{R}^{n-1} \rightarrow S^{n}$ a smooth map, then we cannot have $F(A) = S^{n}.$ In other words: Smooth maps must preserve the dimension somehow.
Does anybody know how to show this?
This is a consequence of Sard's Theorem. Since $\dim\Bbb R^{n-1}<\dim S^n$, a map $F:\Bbb R^{n-1}\to S^{n}$ has critical values everywhere and hence its image $F(\Bbb R^{n-1})$ has measure zero in $S^n$.
This is not true in general for continuous functions, for example because of the existence of space filling curves $\Bbb R\to S^2$.