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 $\forall q\in U$, $\operatorname{rank}_{q}F\geqslant r$.
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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.