Rank of a smooth map is lower semicontinuous? Let $F:M\rightarrow N$ be a smooth map between manifolds, $p\in M$. Prove that if $\operatorname{rank}_{p}F=r$, then there exists a neighborhood $U$ of $p$, such that for $\forall q\in U$, $\operatorname{rank}_{q}F\geqslant r$.
Could anyone help me? Thanks in advance.
 A: Let us choose coordinate charts $(x_{1},\ldots,x_{n})$ around $p$ and $(y_{1},\ldots,y_{m})$ around $F(p)$. Writing $F=\big(F_{1}(x_{1},\ldots,x_{n}),\ldots,F_{m}(x_{1},\ldots,x_{n})\big)$, we can express the derivative $d_{p}F$ as a matrix
$$
\begin{pmatrix}
\frac{\partial F_{1}}{\partial x_{1}}(p) & \cdots & \frac{\partial F_{1}}{\partial x_{n}}(p)\\
\vdots & & \vdots\\
\frac{\partial F_{m}}{\partial x_{1}}(p) & \cdots & \frac{\partial F_{m}}{\partial x_{n}}(p)
\end{pmatrix}.
$$
By assumption, this matrix has rank $r$, so there is a nonzero minor of size $r\times r$:
$$
\begin{vmatrix}
\frac{\partial F_{i_{1}}}{\partial x_{j_{1}}}(p) & \cdots & \frac{\partial F_{i_{1}}}{\partial x_{j_{r}}}(p)\\
\vdots & & \vdots\\
\frac{\partial F_{i_{r}}}{\partial x_{j_{1}}}(p) & \cdots & \frac{\partial F_{i_{r}}}{\partial x_{j_{r}}}(p)
\end{vmatrix}\neq 0.
$$
The map
$$
G:q\mapsto \begin{vmatrix}
\frac{\partial F_{i_{1}}}{\partial x_{j_{1}}}(q) & \cdots & \frac{\partial F_{i_{1}}}{\partial x_{j_{r}}}(q)\\
\vdots & & \vdots\\
\frac{\partial F_{i_{r}}}{\partial x_{j_{1}}}(q) & \cdots & \frac{\partial F_{i_{r}}}{\partial x_{j_{r}}}(q)
\end{vmatrix},
$$
is continuous and nonzero at $p$, so there exists a neighborhood $U$ of $p$ such that $G(q)\neq 0$ for all $q\in U$. This implies that for all $q\in U$
$$
d_{q}F=\begin{pmatrix}
\frac{\partial F_{1}}{\partial x_{1}}(q) & \cdots & \frac{\partial F_{1}}{\partial x_{n}}(q)\\
\vdots & & \vdots\\
\frac{\partial F_{m}}{\partial x_{1}}(q) & \cdots & \frac{\partial F_{m}}{\partial x_{n}}(q)
\end{pmatrix}
$$
has rank at least $r$, since it has a nonzero $r\times r$ minor.
