# Show that if $dF_p$ is nonsingular, then $F(p)\in \text{Int}N$ : Lee's Smooth Manifolds

Suppose $M$ is a smooth manifold without boundary and $N$ is a smooth manifold with boundary and let $F : M \rightarrow N$ is a smooth map. Show that if $p \in M$ is a point such that $dF_p$ is nonsingular then $F(p) \in \text{Int} N$.

Recently i addressed this question and aware that this already been asked here but i'm not really satisfy or i'm feeling like it miss some steps (forgive me if that is trivial). So i think i can modify it a bit as follows :

The idea is same, that is try to applying IFT to the representation of the map $F$ and find a contradiction. Assume for the contrary that $F(p) \in \partial N$. Now choose charts $(U,\varphi)$ in $M$ centered at $p$ and boundary chart $(V,\psi)$ centered at $F(p)$. Because $\psi(V) = \hat{V}$ is open subset in $\mathbb{H}^n$ with $\psi(F(p)) \in \partial \mathbb{H}^n$, we can't apply IFT directly because the codomain is not open subset of $\mathbb{R}^n$. So we need to somehow extend its codomain to $\mathbb{R}^n$.

When we doing this we actually compose the map $\hat{F}$ with then inclusion map $\iota : \hat{V} \hookrightarrow \mathbb{H}^n$ and $\iota' : \mathbb{H}^n \hookrightarrow \mathbb{R}^n$ right ?

(So i'm feel unconfortable in the answer (the link above) that when he regard the codomain as $\mathbb{R}^n$ the map is still $\hat{F}$, not composed with the inclusions).

So therefore the map should be $$\hat{U} \xrightarrow{\hat{F}} \hat{V} \xrightarrow{\iota} \mathbb{H}^n \xrightarrow{\iota'} \mathbb{R}^n$$ We know that $d\iota_{\hat{F}(\hat{p})} : T_{\hat{F}(\hat{p})}\hat{V} \rightarrow T_{\hat{F}(\hat{p})}\mathbb{H}^n$ and $d\iota'_{\hat{F}(\hat{p})} : T_{\hat{F}(\hat{p})}\mathbb{H}^n \rightarrow T_{\hat{F}(\hat{p})}\mathbb{R}^n$ isomorphism (from Prop. 3.9 and Lemma 3.11 in Lee's smooth manifold) therefore the differential of $\iota' \circ \iota \circ \hat{F} : \hat{U} \rightarrow \mathbb{R}^n$, $$d\iota'_{\hat{F}(\hat{p})} \circ d\iota_{\hat{F}(\hat{p})} \circ d\hat{F}_{\hat{p}}$$ is also an isomorphism. By this we can apply Inverse Function Theorem in Euclidean space and deriving the contradiction.

Is this correct ? I hope someone can clarify this. Thank you.

• Yes, this is correct. But as I noted in the paragraphs following the proofs of Prop. 3.9 and Lemma 3.11, it's good to start getting used to thinking of the tanget space of an open subset of $\mathbb R^n$ or of $\mathbb H^n$ as being identified with the tangent space to $\mathbb R^n$, which in turn is identified with $\mathbb R^n$ itself by Prop. 3.13. This is possible because all of these isomorphisms are canonical (independent of any choices). Thus the two maps $d\iota_{\hat F(\hat p)}$ and $d\iota'_{\hat F(\hat p)}$ are essentially both the identity map of $\mathbb R^n$. ... – Jack Lee Dec 24 '17 at 20:35
• Don't get me wrong -- it's a good idea to go through the argument in detail the way you just did. But in order to make further progress in differential geometry, you'll probably need to get used to making these kinds of harmless abuses of notation. – Jack Lee Dec 24 '17 at 20:37
• @JackLee Thank you very much for the clarification and the advice Prof Lee. It seems bit awkward to me also adding this details, but next time i'll try to get used to it whenever it seems possible. Thank you for going through my details, i really appreciate it. – Sou Dec 25 '17 at 2:55
• – Sou Jan 11 '18 at 10:17