# A doubt about a proof of the dimension of tangent space in Guillemin and Pollack’s book

differential-geometryproof-explanationsmooth-manifolds

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.

This is one of many typos in Guillemin and Pollack. They intended the domain of $$\Phi'$$ to be $$W$$, an open set containing $$\phi(x)$$.