Is This Proof Involving the Sub-matrix of $D_pF$ Correct?

Question

The Problem to Solve

Prove that for any smooth map between manifolds $F\in C^{\infty}(M,N)$ and any given point $p\in M$, there is an open neighborhood $U$ of $p$ such that $rank_p(F)\leq rank_q(F)$ for all $q\in U$.

Note: The hint provided is to first work out why it is enough to show this fact holds for maps $\mathbb{R}^m \rightarrow \mathbb{R}^n$, then to show it is true for these maps, choose an invertible sub-matrix of $D_pF$ of size $rank_p(F)\times rank_p(F)$ and think about how the determinant of the sub-matrix varies as p changes.

Attempt

A smooth map between manifolds locally resembles the map $\mathbb{R}^m \rightarrow \mathbb{R}^n$. Hence, it is sufficient to show that this fact holds for $\mathbb{R}^m \rightarrow \mathbb{R}^n$.

Consider the smooth map $F: \mathbb{R}^m \rightarrow \mathbb{R}^n$. Let $p\in \mathbb{R}^m$, and $rank_pF = r$. We want to prove that in an open neighborhood $U$ of $p$, the rank of any other point $q$ is at least $r$.

First, choose an $r\times r$ sub-matrix of $D_pF$ whose determinant is non-zero at $p$. This sub-matrix has the largest linearly independent partial derivatives of $F$ at $p$. Also, the determinant of the sub-matrix is a continuous function of the entries of the sub-matrix. Since the determinant is non-zero, then it is invertible at $p$. Now, by continuity, there exists an open neighbourhood $U$ around $p$ in which this determinant is non-zero. This means the chosen sub-matrix remains invertible in $U$, and hence, the rank of any point in $U$ is at least $r$. Hence, in the open neighborhood $U$, $rank_pF\leq rank_qF$ for all $q$ in $U$.

Question:

Is there any area of this proof I need to improve/strengthen?

Answer 1

It is correct. You can improve it a little bit as follows:

1. Mention explicitly that an $(n \times m)$-matrix $A$ has rank $\ge r$ if and only if it has an $(r \times r)$-submatrix $\tilde A$ with $\det \tilde A \ne 0$. Note, however, that $\tilde A$ is in general not unique.

2. Apply this to the Jacobian matrix $D_pF = DF(p)$ to get an $(r \times r)$-submatrix $\tilde DF(p)$ with $\det \tilde DF(p) \ne 0$. Avoid to say "This sub-matrix has the largest linearly independent partial derivatives of $F$ at $p$". This sentence does not make much sense. Also there is no need to mention that the submatrix is invertible (although this is correct).

3. The determinant of the sub-matrix $\tilde DF(p)$ is a continuous function of the entries of the sub-matrix. Since the determinant is non-zero at $p$, then, by continuity, there exists an open neighbourhood $U$ around $p$ such that $\det \tilde DF(q) \ne 0$ for $q \in U$. Hence, for $q \in U$, the matrix $DF(q)$ contains the $(r \times r)$-submatrix $\tilde DF(q)$ with $\det \tilde DF(q) \ne 0$. By 1. one gets $rank_qF = rank({DF(q)} \ge r$ for all $q \in U$.