0

Problem 4.2, pg 98, John Lee's smooth manifold: Suppose $M$ is a smooth manifold (without boundary), $N$ is a smooth manifold with boundary, $F:M \rightarrow N$ is smooth. Show that if $p \in M$ is a point suchthat $dF_p$ is nonsingular, then $F(p) \in Int N$.

I could not really find a contradiction if $F(p) \in \partial N$. Let $(U,(x^i))$ be chart of $p$, and $(V, (y^j))$ chart of $F(p)$. Then we have the coordinate representation $$ v^i \frac{\partial}{\partial x^i} \Big|_p \mapsto \Big(\frac{\partial F^j}{\partial x^i} v^i \Big) \frac{\partial}{\partial y^j}\Big|_{F(p)}$$ being nonsingular doesn't really imply any thing (?)

Also, what happens when $p \in \partial M$?

Hints would be appreciated!

Bryan Shih
  • 9,934
  • 3
    First, $M$ has no boundary, so you can't have $p\in\partial M$. Next, you didn't state it explicitly, but you have $\dim M=\dim N$ (or else it makes no sense for $dF_p$ to be nonsingular). The main point is that the inverse function theorem tells you that when you translate into local coordinates, near $p$, for open sets $U\subset\Bbb R^n$ and $V\subset\Bbb H^n\subset\Bbb R^n$, $F|_U\colon U\to \Bbb R^n$ is an open map. – Ted Shifrin Sep 29 '18 at 16:57
  • So I suppose the argument goes like this: By definition, $F:U \rightarrow \Bbb H^n \hookrightarrow \Bbb R^n$. As $dF_p$ is nonsingular at $p$, by IFT, exists open subsets, $F:U_0 \rightarrow V_0$ is a diffeomorphism. If $F(p) \in \partial \Bbb H^n$, then the image is not contained $\Bbb H^n$. – Bryan Shih Sep 30 '18 at 07:26

0 Answers0