Proper smooth map has its regular values an open set

Question

If $f:M\rightarrow N$ is proper and smooth how can I conclude that its regular values form an open set? I know the regular points do but I cannot show that the map is open.

Answer 1

Let $y \in N$ be a regular value and assume that every open neighborhood $y \in U \subseteq N$ of $y$ in $N$ contains a non-regular value. Hence, we can choose a sequence $y_n \rightarrow y$ such that $y_n$ are non-regular values. This means that there are $x_n \in M$ with $f(x_n) = y_n$ such that $df|_{x_n}$ is not onto. The set $K = \{ y_n \}_{n \in \mathbb{N}} \cup \{ y \}$ is a compact subset of $N$ and so $f^{-1}(K)$ is a proper subset of $M$. Since $x_n \in f^{-1}(K)$, we can find a subsequence $x_{n_k} \rightarrow x$ with $x \in f^{-1}(K)$. By continuity, $y_{n_k} = f(x_{n_k}) \rightarrow f(x)$ and so $f(x) = y$. We have managed to find a sequence $x_{n_k} \rightarrow x$ such that $df|_{x_{n_k}}$ is not onto but $df|_{x}$ is. Since being onto is an open condition in $\operatorname{Hom}(\mathbb{R}^m,\mathbb{R}^n)$ we arrive a contradiction.