I'm trying to prove that the dimension of the tangent space of a $k$-dimensional manifold $X$ is actually $k$. I came across this thread in which the proof from Guillemin and Pollack's book is quoted.
The dimension of the vector space $T_x(X)$ is, as you expect, the dimension $k$ of $X$. To prove this, we use the smoothness of the inverse $\phi^{-1}$. Choose an open set $W$ in $\mathbb{R}^N$ and a smooth map $\Phi': \mathbb{R}^N \rightarrow \mathbb{R}^k$ that extends $\phi^{-1}$. Then $\Phi'\circ\phi$ is the identity map of $U$, so the chain rule implies that the sequence of linear transformations $$\mathbb{R}^k \xrightarrow{\mathrm d\phi_0} T_x (X) \xrightarrow{\mathrm d \Phi_x'} \mathbb{R}^k$$ is the identity map of $\mathbb{R}^k$. It follows that $\mathrm d \phi_0 :\mathbb{R}^k \rightarrow T_x (X)$ is an isomorphism, so $\dim T_x(X)=k$.
Previously, the author said that
A map $f: X \rightarrow \mathbb{R}^{m}$ defined on an arbitrary subset $X$ in $\mathbb{R}^{n}$ is called smooth if it may be locally extended to a smooth map on open sets; that is, if around each point $x \in X$ there is an open set $U \subset \mathbb{R}^{n}$ and a smooth map $F: U \rightarrow \mathbb{R}^{m}$ such that $F$ equals $f$ on $U \cap X$.
So the extension map is defined on some open subset of $\mathbb R^N$. We don't know if such an extension to the whole $\mathbb R^N$ exists. As such, how can the extension map $\Phi'$ above be define on the whole $\mathbb R^N$. If this is not the case, then the composition $\Phi'\circ\phi$ is not well-defined.
Could you elaborate on my confusion?
Update: In this errata, $\Phi': W \rightarrow \mathbb{R}^k$. Assume that $\phi:U \to V$ and thus $\phi^{-1}: V \to U$. Here $W$ is open in $\mathbb R^N$, $U$ is open in $\mathbb R^k$, and $V$ is open in $X$. For $x \in U$, $\phi(x) \in V$. How do we know $\phi(x) \in W$, i.e., $V \subseteq W$? If not, the map $\Phi'\circ\phi$ is not well-defined and we can not apply the chain rule.
Best Answer
This is one of many typos in Guillemin and Pollack. They intended the domain of $\Phi'$ to be $W$, an open set containing $\phi(x)$.
(A list of the typos of which I am aware in the book can be downloaded from my webpage, linked in my profile.)