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In Do Carmo's Differential Geometry of Curves and Surfaces there's an excercise in section 2-3 that says:

  1. Prove that the definition of a differentiable map between surfaces does not depend on the parametrization chosen.

According to the book, the definition of a differentiable map is, if $\varphi:S_1\to S_2$ is a continuous map, where both $S_1,S_2$ are regular surfaces. then $\varphi$ is differentiable at $p\in S_1$ if given parametrizations: $f:U \subset\Bbb R^2\to S_1$ and $g:W \subset\Bbb R^2\to S_2$, with $p\in f(U)$ and $\varphi(f(U))\subset g(W)$, the map $g^{-1}\circ\varphi\circ f: U\to W$ is differentiable at $q=f^{-1}(p)$.

I'm getting a little bit confused on how to prove this without making a mess. I'm relying on the following diagram: enter image description here

Where $f_1,f_2$ are two diferent parametrizations for $S_1$, and $g_1,g_2$ are two diferent parametrizations for $S_2$. And my guess is that we have to prove that $g_1^{-1}\circ\varphi\circ f_1=g_2^{-1}\circ\varphi\circ f_2$, althought would it be also necesary to prove $g_1^{-1}\circ\varphi\circ f_2=g_2^{-1}\circ\varphi\circ f_1$? Am I considering this wrong?

An attempt to do this:
We know that $f_1=f_2\circ f_2^{-1}\circ f_1$ and $g_1^{-1}=(g_2\circ g_2^{-1}\circ g_1)^{-1}=g_1^{-1}\circ g_2\circ g_2^{-1}$, hence $$g_1^{-1}\circ\varphi\circ f_1=(g_1^{-1}\circ g_2\circ g_2^{-1}) \circ\varphi\circ (f_2\circ f_2^{-1}\circ f_1) \\=g_1^{-1}\circ g_2\circ (g_2^{-1} \circ\varphi\circ f_2)\circ f_2^{-1}\circ f_1 \\=G \circ (g_2^{-1} \circ\varphi\circ f_2)\circ F$$

because $S_1,S_2$ are regular, then $F,G$ are diff. but I'm not sure what to do from here, can we conclude that $g_1^{-1}\circ\varphi\circ f_1\subseteq g_2^{-1}\circ\varphi\circ f_2$?

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    $\begingroup$ I wish the notion of diffeomorphism was used, it would clarify a lot of things. $\endgroup$
    – orangeskid
    Oct 6, 2014 at 20:53
  • $\begingroup$ @orangeskid How would that be? $\endgroup$
    – Ana Galois
    Oct 6, 2014 at 20:57

1 Answer 1

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$$g_1^{-1}\circ\varphi\circ f_1=g_1^{-1}\circ g_2\circ (g_2^{-1} \circ\varphi\circ f_2)\circ f_2^{-1}\circ f_1= \\=G \circ (g_2^{-1} \circ\varphi\circ f_2)\circ F$$

and also

$$g_2^{-1}\circ\varphi\circ f_2=g_2^{-1}\circ g_1\circ (g_1^{-1} \circ\varphi\circ f_1)\circ f_1^{-1}\circ f_2= \\=G^{-1} \circ (g_1^{-1} \circ\varphi\circ f_1)\circ F^{-1}$$

The maps $F$ and $G$ are diffeomorphisms: they are differentiable, bijective and their inverses are also differentiable.

Now use: a composition of differentiable functions is also differentiable.

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    $\begingroup$ Thank you for your answer. However I fail to see how this solves the problem. What you get with this is that $g_1^{-1}\circ\varphi\circ f_1$ and $g_2^{-1}\circ\varphi\circ f_2$ are both differentiable, is it all it needs to be done? I can't say that they are the same because their domains are not necessarily the same open sets. $\endgroup$
    – Ana Galois
    Oct 7, 2014 at 17:58
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    $\begingroup$ No, no, what you really get is this : if $g_1^{-1} \circ\varphi\circ f_1$ is differentiable then $g_2^{-1} \circ\varphi\circ f_2$ is also differentiable. Also the converse. And the moral is: to check that the map $\phi$ is differentiable at a point $p \in S_1$ it's enough to choose any pair of parametrizations around $p$ on $S_1$ and around $\phi(p)$ on $S_2$. If it checks with $f_1$ and $g_1$ then it will surely check with $f_2$ and $g_2$, so the choice won't matter. $\endgroup$
    – orangeskid
    Oct 7, 2014 at 18:05
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    $\begingroup$ Oh! I see now, so its not about them being the same thing, but both carrying out the same thing. $\endgroup$
    – Ana Galois
    Oct 7, 2014 at 18:17

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