# proving that this function is a diffeomorphism

i have a question on section 2-3

I have a question in the book do Carmo Differential Geometry, Section 2-3, Proposition 1, the book says is a composition of $$h\left| N \right. = F^{ - 1} \left( {y\left| N \right.} \right)$$ differential maps, then by the chain rule h is differentiable at r. Since r is arbitrary, h is differentiable on $$y^{ - 1} \left( W \right)$$

Ok here I agree on everything, except the last, and I see that this implies that h is a local diffeomorphism, that thing I'm not seeing?

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Can you include more detail? Some of the objects in your question is not defined, and not everybody has a copy of do Carmo sitting around handy. Please try to make all the notation in your question as self-contained as possible. – Willie Wong Jul 30 '11 at 22:26
I already which can be but I'm not quite sure, if f is a homeomorphism and it is a local diffeomorphism, it is necessarily a diffeomorphism? – Daniel Jul 30 '11 at 23:22
@Daniel Yes, there are two conditions for a map to be a diffeomorphism, it must be smooth, and it must have a smooth inverse. If it is a homeomorphism, we know that an inverse exists, and if it is a local diffeomorphism, we know that it is smooth. However, smoothness is a local property, and being a local diffeomorphism tells us that the inverse is also smooth. – Aaron Jul 30 '11 at 23:54
Thanks Aaron, I have one last question related to the topic In the same section of the book, wants to prove that the inverse of a parametrization is differentiable, but using the last lemma proves that $$x^{ - 1} \circ y$$ is differentiable, this implies that x ^ -1 is differentiable? is the reciprocal of the chain rule, but this? – Daniel Jul 30 '11 at 23:59
What is $y$? Or $x$? I don't have the book, and I can't make any sense of your original question (you aren't just missing context, you are missing words which would turn your individual sentences into things which make sense, mathematics completely aside). Is $y$ an arbitrary smooth function from your manifold to $\mathbb{R}$? There is a theorem that $f:M\to N$ is smooth if and only if $\rho\circ f$ is smooth for every $\rho:N\to \mathbb{R}$. If $y$ is not arbitrary, then it is impossible for me to say what is going on. – Aaron Jul 31 '11 at 0:20

Having just obtained a copy of the book, you are leaving out very key things about the proposition and its proof. In particular, you do not yet have abstract smooth manifolds, and you have no definition of smooth map from one manifold to another. At a higher level, the proposition you refer to is essentially a definition, but here you only have the notion of smooth maps from $\mathbb{R}^m$ to $\mathbb{R}^n$.
The problem is that $x^{-1}$ is only defined on (part of) the surface, and not on (a neighborhood of) $\mathbb{R}^3$, and so you can't directly use the chain rule or other basic calculus facts. Instead, you have to fatten up $x$ to a new map $\widetilde{x}:U\times \mathbb{R}\to S\times \mathbb{R} \to \mathbb{R}^3$. In slightly fancy language, you change perspectives from looking at the surface $S$ to looking at a tubular neighborhood of $S$. Because $\widetilde{x}:U\to V$ where $U$ and $V$ are open subsets of $\mathbb{R}^3$, and $\widetilde{x}$ is invertible, we can use the chain rule and the inverse function theorem to conclude that $x^{-1}\circ y = p_{12}\circ \widetilde{x}^{-1}\circ y$ is a local diffeomorphism and a homeomorphism between two open sets in $\mathbb{R}^2$ (here $p_{12}:\mathbb{R}^3\to \mathbb{R}^2$ is the projection onto the first two coordinates).
Finally, the inverse function theorem implies that a map which is both a homeomorphism and a local diffeomorphism between open subsets of $\mathbb{R}^n$ must be a diffeomorphism.