Let $f:U\rightarrow f(U)$ and $g:V\rightarrow g(V)$ be two charts on $M$ such that $f(x_{0})=g(y_{o})=p$. I want to prove that $Df(x_{0})(\mathbb{R}^{m})=Dg(y_{0})(\mathbb{R}^{m})$.
I believe the following proof is correct:
Proof: $g^{-1}\circ f:f^{\dashv}\left(f(U)\cap g(V)\right)\rightarrow g^{\dashv}\left(f(U)\cap g(V)\right)$
is smooth and its inverse is also smooth. (Assume this has already been proved).
Hence, $D\left(g^{-1}\circ f\right)(x_{0})$
is an automorphism of $\mathbb{R}^{m}$. ($\star$)
Now, $$Df(x_{0})(\mathbb{R}^{m})=D(g\circ g^{-1}\circ f)(x_{0})(\mathbb{R}^{m})=Dg(g^{-1}\circ f(x_{0}))\circ D(g^{-1}\circ f)(x_{0})(\mathbb{R}^{m})=Dg(g^{-1}\circ f(x_{0}))(\mathbb{R}^{m})=Dg(y_{0})(\mathbb{R}^{m})$$ $\square$
My Question: Why does ($\star$) hold? That is, what is the reasoning behing the "Hence"?
We know that $T = D(g^{-1}\circ f)(x_0): \mathbb{R}^m\to\mathbb{R}^m$ is a linear map. Any linear endomorphism over a finite dimensional vector space is an automorphism if and only if it is surjective. The fact that $g^{-1}\circ f$ is a local diffeomorphism implies that $T$ is of full rank, i.e., surjective, and therefore $T$ is invertible, as desired.