Show that a smooth map $F : M \rightarrow N$ has Constant rank if $F$ has a linear coordinate representation.

Solution 1:

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 $M$.