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Math Help - confusion on proof in do carmo: regular surfaces

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    confusion on proof in do carmo: regular surfaces

    let  S \subset \mathbb{R}^3 be a regular surface with  p \in S . Then there exists a neighborhood V of p in S such that V is the graph of a differentiable function that is either z = f(x,y), y = g(x,z), or x = h(y, z).

    he starts off by letting  x: U \subset \mathbb{R}^2 \rightarrow S be a parametrization of S in p and that x(u,v) = (x(u,v), y(u,v), z(u,v)), (u,v) belonging to U. he supposes that  \frac{\partial(x, y)}{\partial(u,v)}(q) \neg 0 . Then he composes the projection map  \pi(x,y,z) = (x,y) with the function so that we have  /pi o x . Then the inverse function theorem applies so there exists neighborhoods V1 of q and V2 of  \pi o x(q) such that there is a diffeomorphic mapping from V2 to V2.

    Then he says that it follows that  \pi restricted to x(V1) = V is one to one and that there is a differentiable inverse  (\pi o x)^{-1}: V2 \rightarrow V1 . I always seem to get confused when domains become restricted in differential geometry. i don't know how it follows that  \pi is one to one and why that is even necessary in this proof.

    the rest of his proof just says: observe that x is a homeomorphism so that V is a neighborhood of p in S. now if we compose  (\pi o x)^{-1} : (x,y) \rightarrow (u(x,y), v(x,y)) with  (u,v) \rightarrow z(u,v) and we see that V is the graph of the differentiable function z = f(x,y). the remaining cases can be treated in the same way.

    i'm confused about the middle part of the proof on how did he know that  \pi is one to one and why did he even need that result in the first place. any help in clearing this up will be greatly appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: confusion on proof in do carmo: regular surfaces

    Quote Originally Posted by oblixps View Post
    let  S \subset \mathbb{R}^3 be a regular surface with  p \in S . Then there exists a neighborhood V of p in S such that V is the graph of a differentiable function that is either z = f(x,y), y = g(x,z), or x = h(y, z).

    he starts off by letting  x: U \subset \mathbb{R}^2 \rightarrow S be a parametrization of S in p and that x(u,v) = (x(u,v), y(u,v), z(u,v)), (u,v) belonging to U. he supposes that  \frac{\partial(x, y)}{\partial(u,v)}(q) \neg 0 . Then he composes the projection map  \pi(x,y,z) = (x,y) with the function so that we have  /pi o x . Then the inverse function theorem applies so there exists neighborhoods V1 of q and V2 of  \pi o x(q) such that there is a diffeomorphic mapping from V2 to V2.

    Then he says that it follows that  \pi restricted to x(V1) = V is one to one and that there is a differentiable inverse  (\pi o x)^{-1}: V2 \rightarrow V1 . I always seem to get confused when domains become restricted in differential geometry. i don't know how it follows that  \pi is one to one and why that is even necessary in this proof.

    the rest of his proof just says: observe that x is a homeomorphism so that V is a neighborhood of p in S. now if we compose  (\pi o x)^{-1} : (x,y) \rightarrow (u(x,y), v(x,y)) with  (u,v) \rightarrow z(u,v) and we see that V is the graph of the differentiable function z = f(x,y). the remaining cases can be treated in the same way.

    i'm confused about the middle part of the proof on how did he know that  \pi is one to one and why did he even need that result in the first place. any help in clearing this up will be greatly appreciated.
    You know \pi\circ x is injective on V_1 because this is a conclusion of the inverse function theorem, clearly then \pi is injective on x(V_1) since taking two elements, x(t),x(t') if \pi sent them to the same thing then \pi\circ x wouldn't be injective. I don't quite follow the proof, at least as you wrote it, beyond that point. The idea then is that \psi=x\circ (\pi\circ x)^{-1} will be chart at p, but then \pi\circ\psi=\text{id} and this implies that \psi is of the form \psi(x,y)=(x,y,f(x,y)) for some smooth function f.
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