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!