# A smooth map which has full rank at a point is locally a submersion/immersion.

Let $F:M\rightarrow N$ be a smooth map between manifolds and let $p\in M$ .

1. If $dF_{p}$ is surjective, then there exists a neighbourhood $U$ of p such that $F|_{U}$ is a submersion.

2. If $dF_{p}$ is injective, then there exists a neighbourhood $U$ of $p$ such that $F|_{U}$ is an immersion.

I saw this theorem in Lee's Introduction to Smooth Manifolds. This is his proof: What I don't understand is the part where says "by continuity". Is he talking about a function that is continuous? If so, what function is it?

• The function he refers could be the determinant of a minor of the matrix. May 9, 2017 at 14:00
• See math.stackexchange.com/questions/2299042/… for the same question. Dec 23, 2017 at 9:27

In coordinates, $\Bbb d F$ is represented by a matrix. Since $\Bbb d F _p$ has full rank, there exists a minor in it of maximal dimension that is non-zero at $p$. The determinant of this minor is a continuous (in fact polynomial, being a sum of products of entries in the matrix) function of the coordinates, so if it's non-zero at $p$, by continuity it will be non-zero on some small neighbourhood of $p$, so the rank will be maximal on this neighbourhood.
Suppose $U$ is a coordinate neighbourhood of $p$ in $M$, contained in the preimage under $F$ of a coordinate neighbourhood of $F(p)$ in $N$. We have a continuous map $$x \in U \subset \mathbb R^n \mapsto (dF)_x \in {\rm M}(n \times m , \mathbb R).$$ (Of course, I used the coordinate charts to identify $U$ with $\mathbb R^n$ and to identify the respective tangent spaces with $\mathbb R^n$ and $\mathbb R^m$.)
Since $F$ is a submersion at $x = p$, we know that the $n \times m$ matrix $(dF)_x$ is of maximal rank at $x = p$. Said another way, at least one of the $m \times m$ submatrices of $(dF)_p$ has non-vanishing determinant.
This suggests that we should consider the map $g : U \to \mathbb R^{\frac{n!}{m!(n-m)!}}$ that sends each point $x \in U$ to an $\frac{n!}{m!(n-m)!}$-component vector containing the determinants of all $m \times m$ submatrices of $(dF)_x$. Clearly $g$ is also continuous.
Finally, $g(p) \neq 0$. By continuity of $g$, there is an open neighbourhood $V \subset U$ of $p$ such that $g(x) \neq 0$ for all $x \in V$. This means that, for all $x \in V$, the matrix $(dF)_x$ has at least one $m \times m$ submatrix with non-vanishing determinant. So the original $F$ is a submersion on $V$.