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Let $X$ denote a finite dimensional normed space.

A non empty set $M \subset X$ is called $d$-dimensional differentiable submanifold of $X$, if for all $a \in M$ there exists an open neighborhood $U$ of $a$ and a diffeomorphism $\varphi \colon U \to V$ with open $V \subset \mathbb{R}^n$, such that $$\varphi(M \cap U) = \mathbb{R}_0^d \cap V,$$ where $\mathbb{R}_0^d := \{x \in \mathbb{R^n} \mid x_{d+1} = \dots = x_n = 0\}$.

Why is $d$ well defined?

I know that if there exists a diffeomorphism between to normed spaces, they have to have the same dimension, since its differential induces an isomorphism. Hence, $\dim X = n$. But how does this help me in showing the uniqueness of $d$ (the book I am citing says it would)?

el_tenedor
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    Consider two such charts $\varphi_k \colon U \to V_k \subset \mathbb{R}^n$ such that $\varphi_k(U\cap M) = \mathbb{R}_0^{d_k} \cap V_k$. Then look at the map $\varphi_2 \circ \varphi_1^{-1}$ which is a diffeomorphism $V_1 \to V_2$. What does it do with $\mathbb{R}_0^{d_1} \cap V_1$? – Daniel Fischer Mar 06 '16 at 21:50
  • @DanielFischer Thanks, this shows that $\varphi_2 \circ \varphi_1^{-1}(\mathbb{R}_0^{d_1} \cap V_1) = \mathbb{R}_2^{d_2} \cap V_2$. But I am not sure how this should imply $d_1 = d_2$. – el_tenedor Mar 06 '16 at 22:45
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    Consider also the map $\varphi_1 \circ \varphi_2^{-1}$. And consider the restrictions to $\mathbb{R}_0^{d_k}\cap V_k$. Identifying that with an open set in $\mathbb{R}^{d_k}$, we have a diffeomorphism between an open subset of $\mathbb{R}^{d_1}$ and an open subset of $\mathbb{R}^{d_2}$. – Daniel Fischer Mar 06 '16 at 22:51
  • @DanielFischer Thanks again, now I got it. – el_tenedor Mar 07 '16 at 08:28

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