why the regular value of a smooth map is everywhere dense?

Question

Let $ f:M\to N$ be a smooth mapping between two smooth manifold(dim M>dimN). A.B.Brown's theorem told me :the set of regular values is everywhere dense in N. so ,what about $f:U \to R^n$ where U is an open subset of $R^n$enter image description here

(I have known Sard theorem which describes the critical set is measure zero.)

Answer 1

Note that each $q \in N \setminus f(M)$ is a regular value because then $f^{-1}(q) = \emptyset$ and thus trivially each $p \in f^{-1}(q)$ is a regular point.