# The set of all regular points of a smooth map is open

Let $M$ be an smooth manifold manifold of dimension $m$ and let $N$ be a smooth manifold of dimension $n$, and let $F:M\rightarrow N$ be a smooth map. Then the set $W=\left\{ p\in M:F\mbox{ has full rank at }p\right\}$ is open.

Here is my attempt at the proof.

Suppose without loss of generality that $m<n$. Let $\left(U,\varphi\right)$ be a chart at $p$ and let $\left(V,\psi\right)$ be a chart at $F\left(p\right)$ such that $F\left(U\right)\subseteq V$. By identifying linear maps with matrices (with respect to the standard bases for $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$), the map $G:\varphi\left(U\right)\rightarrow M\left(n\times m,\mathbb{R}\right)$ defined by $x\longmapsto d\left(\psi\circ F\circ\varphi^{-1}\right)_{x}$ is continuous. For each $n\times m$ matrix $A$, let $m_A$ be the set of all invertible $m\times m$ submatrices of $A$. It is a standard result in linear algebra that $A$ has full rank if and only if $m_A$ is nonempty. The map $H:M(n\times m,\mathbb{R})\rightarrow\mathbb{R}$ defined by $A\longmapsto\sum_{S\in m_{A}}\left|\mbox{det}S\right|$ is continuous. It follows that the composition $H\circ G\circ\varphi$ is continuous. Hence, $\left(H\circ G\circ\varphi\right)^{-1}\left(\mathbb{R}-\left\{ 0\right\} \right)$ is an open subset of $U$. Taking the union of all such sets will result in $W$. Therefore $W$ is open.

Is it correct?

• This looks like the same basic idea as this proof. – Michael Lee Jun 16 '17 at 0:15

The proof is correct, although I would suggest doing a few things slightly differently. Set $k = \min(m,n)$ and let $p \in W$. Choose a coordinate system around $p$ and consider your map $G$. At $\varphi(p)$, the matrix $G(\varphi(p))$ has rank $k$. Hence, there exists some $k \times k$ submatrix whose determinant is non-zero. Denote by $H \colon M_{n \times m}(\mathbb{R}) \rightarrow \mathbb{R}$ the function which returns the determinant of the specific $k \times k$ submatrix which works for $G(\varphi(p))$. Then $H$ is continuous and $(H \circ G \circ \varphi)^{-1}(\mathbb{R} \setminus \{ 0 \})$ is an open neighborhood around $p$ in which $\operatorname{rank} G = k$ is full.