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

Question

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.

Answer 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 $M$.

Answer 2

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