If $f \colon M \to N$ is a smooth map of smooth manifolds, for any point $p \in M$, is there an open neighbourhood $V$ of $p$ such that $\forall q \in V \colon \mathrm{rnk}_q (f) \geq \mathrm{rnk}_p (f)$?
is a non-falling rank of smooth maps an open condition?
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differential-geometry
1 Answers
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Yes. Note that if $f$ has rank $k$ at $p$, then $Df(p)$ has rank $k$. Therefore in coordinates, there is a non-vanishing $k \times k$-minor of $Df(p)$. As having a non-vanishing determinant is an open condition, the same minor will not vanish in a whole neighbourhood $V$ of $p$, giving $\operatorname{rank}_q f \ge k$, all $q \in V$: