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Suppose $F:M\to N$ is a smooth map, with $M,N$ smooth manifolds and $M$ connected. I want to show that if for each $p$ in $M$ there exists smooth charts near $p$ and $F(p)$ in which the coordinate representation of $F$ is linear, then $F$ has constant rank. The first step is to show that the rank of $F$ is constant in a neighborhood of each point because $F$ is linear. The next is to show $F$ has constant rank in all $M$ by connectedness hypothesis.

How can I use the connectedness of $M$ to say that the rank is constant on all of $M$? This comes from the book of "Introduction to Smooth Manifolds" by Lee.

Kelvin Lois
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sgshsmsdsdos
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  • Fix an integer $r\leq\min(\dim M,\dim N)$. What can you say about the subset of points of $M$ where the rank of $F$ is $r$? – Olivier Bégassat Dec 06 '12 at 06:32
  • Note that the rank is constant locally. – lee Dec 06 '12 at 08:49
  • Hmm? Maybe I got your question wrong, but rank is NOT a local property. That a map is IMMERSION or SUBMERSION is. For example, consider $f(x) = x^3$, which has rank 1 everywhere but at zero. – chriseur Jan 03 '15 at 05:19

2 Answers2

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Note that for any connected topological space $M$ with open cover $\mathscr{U}$, any two points $a,b \in M$ there is a collections of open sets $\{U_1,\dots,U_n\ | U_i \in \mathscr{U} \}$ such that $a\in U_1$ only , $b \in U_n$ only, and $U_i \cap U_j \neq \emptyset$ for $|i-j|\leq 1$ (Willard's Topology).

By hypothesis, for any point $p \in M$ there are neighbourhood $U_p$ such that the rank of $F$ constant there. Take $\mathscr{U} = \{U_p : \forall p\in M\}$ as the open cover for $M$. By above theorem, for any point $a,b \in M$, there are $\{U_1,\dots U_n\ : U_i \in \mathscr{U}\}$ such that $a \in U_1$, $b \in U_n$ and $U_i \cap U_j \neq \emptyset$ for $|i-j|\leq 1$. Lets say that the rank of $F$ at $U_1$ is $r$. For any $p \in U_1 \cap U_2$, rank $F$ at $p$ is $r$, hence rank $F$ on $U_2$ is $r$. Do this inductively we have rank $F$ on $U_n \ni b$ is $r$. Since $a,b \in M$ arbitrary then rank $F$ constant on $M$.

Kelvin Lois
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By the first step the rank is locally constant. Since locally constant functions on connected spaces are constant, the rank is constant.

Claire
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