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Let $M$ be a smooth manifold and let $N$ be a manifold with boundary, both with the same dimension $n$. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}N $.

I am trying to prove this theorem to prove a result about smooth embeddings. Here is how I am thinking the problem could be solved. Assume thet $F(p)$ is a boundary point of $N$. Then there exists a chart $(V,\psi)$ at $F(p)$ such that $\psi(V) $ is an open subset of the upper half space $\mathbb{H}^{n}$. I guess we have to use some fact about $M$ having no boundary and the the fact that $dF_p$ is an isomorphism to show that there is a contradiction, but I cannot figure that out.

Kelvin Lois
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Eigenfield
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1 Answers1

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By passing to charts near $p$ and $F(p)$, it suffices to prove this in the case $M = \mathbb R^n,N = H^n$. Since $dF_p$ is an isomorphism, by the inverse function theorem (for $\mathbb R^n$, thinking of $H^n$ as a subset of $\mathbb R^n$) we can find an open set $U \ni p$ such that $F(U)$ is open (in $\mathbb R^n$!) and $F : U \to F(U)$ is a diffeomorphism. Since $F(p)\in F(U)$, this rules out $F(p) \in \partial H^n$, because no $\mathbb R^n$-neighbourhood of a boundary point is contained in $H^n$.

Anthony Carapetis
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  • Now I completely understand. Thank you. – Eigenfield May 11 '17 at 10:46
  • Please, to apply the Inverse Function Theorem, don't we need that that the image be open? – Danilo Gregorin Afonso Apr 17 '21 at 17:42
  • I agree with @DaniloGregorinAfonso -- I think this answer is incorrect. The image of $F$ must be open in $\mathbb{R}^n$ to apply the IFT, and $\mathbb{H}^n$ is not. – Danny May 20 '23 at 19:19
  • @Danny The IFT does not require the image to be open - what source are you using that requires this? – Anthony Carapetis May 22 '23 at 05:46
  • @AnthonyCarapetis The problem stated in this question is taken from Lee's Introduction to Smooth Manifolds. In his book, he states the Inverse Function Theorem as follows:

    Suppose $U$ and $V$ are open subsets of $\mathbb{R}^n$ and $F: U \to V$ a smooth function. If $DF(a)$ is invertible at some point $a \in U$, then there exist connected neighborhoods $U_0 \subseteq U$ of $a$ and $V_0 \subseteq V$ of $F(a)$ such that $F: U_0 \to V_0$ is a diffeomorphism.

     – Danny May 22 '23 at 17:35
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    @Danny: $F:U \to V$ does not imply that $F(U) = V.$ You can apply the theorem to our $F$ with $U = V = \mathbb R^n.$ – Anthony Carapetis May 23 '23 at 00:25
  • @AnthonyCarapetis I see what you mean. Thanks for the clarification! – Danny May 24 '23 at 20:15