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I am studying Guillemin and Pollack's book Differential Topology, and there is a proof in the book that:
The dimension of the Tangent space $T_x(X)$ is the dimension of manifold $X$.
Please find the proof that I found in the book in the following link.

Prove that the dimension of the tangent space $T_x(X)$ of a k-dimensional manifold is k

Assume that $X \subseteq \mathbb{R}^N$, and it is a $k$-dimension manifold. Consider a single point $x \in X$.
We know that there is an open set $U \subseteq \mathbb{R}^k$, and there is an open set $V \subseteq X$ that: $$\begin{equation} \begin{cases} \phi &: U \rightarrow V\\ \phi^{-1}&: V \rightarrow U \end{cases} \end{equation}\tag{1}\label{eq1} $$ Moreover, $\phi$ is a diffeomorphism. Therefore, both $\phi$ and $\phi^{-1}$ are bijective and smooth. The idea in the proof is to show that the $d\phi_0:\mathbb{R}^k \rightarrow \mathbb{R}^N$ is the isomorphism we need to show the dimension of the $T_x(X)$ as a vector space is $k$. In the book, there is an expansion of $\phi^{-1}$ for this issue. At first, I had problem why we need something like $\phi'$. Thanks to @ Christopher A. Wong post in the mentioned link, this question is solved now.
Now I do not know why this smooth expansion of $\phi^{-1}$ exists?
I know we we can have a smooth function on an open set of $\mathbb{R}^N$ to $\mathbb{R}^k$. Even though, I do not know how we should expand it on whole $\mathbb{R}^N$.

Additionally, I think that if we use the expansion on the open set of $\mathbb{R}^N$ that contains x works as well! Am I missing something?

Thanks a lot for everyone's help!

peek-a-boo
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Janbazif
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    Hint: What is your definition of smoothness of $\phi^{-1}$? – Moishe Kohan Sep 12 '21 at 18:20
  • If $f:X \rightarrow Y$ ($X \subseteq \mathbb{R}^n$ and $Y \subseteq \mathbb{R}^m$) is smooth. Then for each $x \in X$ there is an open neighborhood $x \in V \subseteq \mathbb{R}^n$, and a smooth function $F:V \rightarrow \mathbb{R}^m$. Additionaly, $F|_{V \cap X}=f$. – Janbazif Sep 12 '21 at 18:51
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    Now you should be able to answer your own question. – Moishe Kohan Sep 12 '21 at 18:52
  • I do not get it! How should I make the domain to be the whole $\mathbb{R}^n$, and have the same value on $X$ with $f$. I know I can do it locally, but these functions that we have for each point of $X$ may not have the same value in other parts. Are you saying we can merge these functions? – Janbazif Sep 12 '21 at 18:56
  • The proof is local, so the domain of the extension of $\phi^{-1}$ needs to be simply an open subset of $R^N$ which is a neighborhood of $x$ there. No need to extend to the entire $R^N$. – Moishe Kohan Sep 12 '21 at 19:04
  • Yeah I think the local is enough for the proof, but in the book, there is $\phi': \mathbb{R}^N \rightarrow \mathbb{R}^k$. Moreover, it is a smooth function which is the same with $\phi^{-1}$ on the whole $X$! – Janbazif Sep 12 '21 at 19:18
  • There is no need to extend the map to the whole space. The extension is supposed to be $\Phi^{\prime}\colon W\rightarrow\mathbb{R}^k$. Feel free to make use of this list of Errata by Ted Shifrin. – Thorgott Sep 12 '21 at 20:23
  • Thanks a lot., so it is an error. I looked up and see it did not use $W$ at all. :) – Janbazif Sep 13 '21 at 12:28

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