# Prove that the tangent space has the same dimension as the manifold

I asked this question a couple of days ago. And I thought that I totally understood the question. However it turned out that I didn't, since the argument I constructed was proved to be wrong just now: Diffeomorphism preserves dimension

The original text from Guillemin and Pollack's book was:

"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 he inverse $\phi^{-1}$. Choose an open set $W$ in $\mathbf{R}^N$ and a smooth map $\Phi': \mathbf{R}^N \rightarrow \mathbf{R}^k$ 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

$\mathbf{R}^k\xrightarrow{d\phi_0}T_x(X)\xrightarrow{d\Phi_x'}\mathbf{R}^k$

is the identity map of $\mathbf{R}^k$. It follows that $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism, so $\dim T_x(X)=k$."

Just for everyone's information, in GP's book, the convention is that $\phi$ is a diffeomorphism from $\mathbf{R}^k$ to a k-dimensional manifold $X\subset \mathbf{R}^N$ ($\phi :\mathbf{R}^k\rightarrow X$).

My question is here:

How does $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism follow from previous argument? From the link given, it was shown that the existence of one-sided inverse does not guarantee that a linear map is an isomorphism. Thus I think there is a loophole in arguing "it follows that $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism"...

Thanks a lot for your help!

That $d\phi_0$ has a left inverse implies that it is an injective linear map. The tangent space $T_x(X)$ is defined by Guillemin and Pollack to precisely be the image of $\mathbb{R}^k$ under the map $d\phi_0$, so it is also a surjection. Thus $d\phi_0$ is a linear isomorphism.