Why is tangent space independent of choice of coordinates? This is a question about Spviak's Calculus on Manifolds, page 115.
Let $M$ be a $k$-dimension manifold in $\mathbb{R}^n$ and let $f:W\rightarrow \mathbb{R}^n$ be a coordinate system around $x=f(a)$, where $W$ is some open subset of $\mathbb{R}^k$. We know $f'(a)$ has rank $k$, by definition), so the linear transformation $f_*:\mathbb{R}^k_a\rightarrow \mathbb{R}^n_x$ is 1-1, and the image $f(\mathbb{R}^k_a)$ is $k$-dimensional. If $g:V\rightarrow \mathbb{R}^n$ is another coordinate system, with $x=g(b)$. Spviak claims:
$$g_*(\mathbb{R}^K_b) = f_*(f^{-1}\circ g)_*(\mathbb{R}^K_b) = f_*(\mathbb{R}^K_a)$$
I realize the chain rule is being used, but I am having trouble following the implications. In particular:
1) I think the chain rule gives the first implication, but I don't see why we can take the derivative of $f^{-1}$, because it is a map between spaces of different dimensions (so the inverse function theorem doesn't apply).
2) The last equality implies that $(f^{-1}\circ g)'$ is the identity. I do not see why this is true (but I do see that $(f^{-1}\circ g)(b)=a$, so the subscript works out).
 A: By shrinking $V$ and $W$, we can assume that $f^{-1} \circ g\colon V \to W$ is defined and is a bijection.


*

*You don't have to take the derivative of $f^{-1}$. The map $f^{-1} \circ g$ is a smooth map defined on an open subset of $\mathbf R^K$, so you can use the chain rule on the following diagram.


*You're right in that it doesn't have to be the identity—as a silly example, we could parametrize $M = \mathbf R$ using $f(x) = x$ and $g(x) = 2x$. Then $(f^{-1} \circ g)_*$ would be multiplication by $2$. However, all we need is for $(f^{-1} \circ g)_*(\mathbf R_b^K) = \mathbf R_a^K$ to be true. Luckily, at each point the linear map $(f^{-1} \circ g)_*$ is an isomorphism because $f^{-1} \circ g$ is a diffeomorphism, having a smooth inverse $g^{-1} \circ f$.
A: Indeed, derivative is the same as Jocobi Matrix in coordinate system, there are $m\times n$ Jacobi matrix, so you can expect taking derivative of a map $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ even if $m\neq n$.
The Chain rule says if $f:\mathbb{R}^m\rightarrow\mathbb{R}^n, g:\mathbb{R}^n\rightarrow\mathbb{R}^k$ are two mapping of class $C^1$, then we have
$(g\circ f)'=g'\circ f'$
The reason Spivak wrote $f_\star$ instead of $f'$ is that the tangential map $f_\star$ is a generalization of $f'$, to the case of $f$ being a $C^1$ map between two $C^1$ manifolds. (Note $\mathbb{R}^m, \mathbb{R}^n$ are both $C^1$ manifolds, in this case $f_\star$ is just $f'$.)
The suitable intrinsic definition of tangential mapping can make the Chain rule an obvious thing. For example, you can define in the following way
Let $\gamma$ be a curve passing the point $p$ on $M$, then $f\circ\gamma$ is a curve passing $f(p)=q$ on $N$, which determines an element in $T_qN$. So there is a map
$[\gamma]\mapsto[f\circ\gamma]$
where $[\gamma]$ is the element in $T_pM$ determined by the curve $\gamma$.
This map is called the tangential map of $f$, denoted as $f_\star$.
You can show that in the case $M=\mathbb{R}^m, N=\mathbb{R}^n$, it's just the derivative $f'$ defined in Calculus.
Now the Chain rule is easy,
$(g\circ f)_\star([\gamma])=[(g\circ f)\circ\gamma]=[g\circ(f\circ\gamma)]=
g_\star[f\circ\gamma]=g_\star(f_\star[\gamma])=(g_\star\circ f_\star)[\gamma]$
where $[\gamma]$ is an arbitrary element in $T_pM$, $f:M\rightarrow N, g:N\rightarrow K$ are both $C^1$ maps, $M,N,K$ are $C^1$ manifolds.
So that $(g\circ f)_\star=g_\star\circ f_\star$, the Chain rule follows.
