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I am trying to solve Problem 4-2 on Lee's Introduction to Smooth Manifold. The problem states the following:

Suppose $M$ is a smooth manifold (without boundary), $N$ is a smooth manifold with boundary and $F: M \to N$ is smooth. Show that if $p \in M$ is a point such that $dF_p$ is non-singular, then $F(p)\in$ Int$N$.

I know that there are two questions addressing this problem (this and this). I tried to solve in my own way, but it is likely wrong since my solution does not use the inverse function theorem which seems to be crucial in the solutions above. In this question, I would like to ask what went wrong with my proof.

My attempt

Since $F(p)$ belongs to either $\textrm{Int }N$ or $\partial N$, we suppose $F(p)\in \partial N$ to prove the contrapositive. In the simplest case, our problem is that a smooth function $F:\mathbb{R}\to \mathbb{H}^1=[0,\infty)$ satisfies that if $F(p)=0$, then $F'(p)=0$. This is true because $F$ attains the minimum at $p$. I (try to) extend this idea to the general case.

Take any chart $(U,\phi)$ containing $p$ and a boundary chart $(V,\psi)$ containing $F(p)$ such that $\psi(F(p))\in \partial \mathbb{H}^n$. Then $\hat{F}=\psi \circ F\circ \phi^{-1}$ is a smooth map from an open subset of $\mathbb{R}^n$ to an open subset of $\mathbb{H}^n$ such that $\hat{p}=\phi(p)$ maps to $\partial \mathbb{H}^n$. Thus, its $n$th component $\hat{F}^n$ is a smooth map to $[0,\infty)$ and $\hat{F}^n(\hat{p})=0$.

Therefore, $\hat{F}^n$ attains minimum at $\hat{p}$, so its partial derivatives should be all zero. This implies the last row of the Jacobian is all zero at $\hat{p}$, so the Jacobian is singular. Therefore, $dF_p$ is singular.

Now we have that if $F(p) \in \partial N$, $dF_p$ is singular. By taking the contrapositive, we have the proposition.

Kaira
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  • I may be missing something, but my first thought is that the claim is wrong. Let $M$ be the unit circle on the plane (a 1-dimensional manifold without boundary). Let $N$ be the half plane defined by $x\le 1$ (a 2-dimensional manifold with boundary). Let $F:M\to N$ be the identity mapping (hopefully smooth everywhere). Is the point $P=(1,0)$ not a counterexample? – Jyrki Lahtonen Jul 17 '23 at 08:42
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    @JyrkiLahtonen: I think that since $dF_p$ is assumed to be non-singular, it is implicit that $\dim M = \dim N$ (otherwise, you could take your example or even the inclusion $\mathbb{R}\to\mathbb{H}^2$ given by $x\mapsto (x,0)$). – Balloon Jul 17 '23 at 08:45
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    @Balloon That is probably the intention. I seem to have used full rank as a substitute for non-singular :-) – Jyrki Lahtonen Jul 17 '23 at 08:49

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