Prove that if $f$ is a diffeomorphism than its differential $D_{f}$ is an isomorphism

Question

I want to prove that if $f : U \to V$ , when $U,V \subset R^n$, is a diffeomorphism than its differential $D_{f}$ is an isomorphism.

Since $D_{f}$ is a linear transformation, isomorphism means that it's both injective and surjective. Moreover, $D_{f}$ is between two vector spaces with the same dimension (both are $R^{n*n}$) then we can use the dimension theorem for vector spaces. So we really need to prove only that $D_{f}$ is injective (or surjective).

However, I don't really see how to do it. I know that $f$ is $C^1$ (continuously differential) and also $f^{-1}$ is $C^1$.

Maybe, because the inverse of $D_f$ is $D_{f^{-1}}$. So if $D_f$ is not injective then it doesn't have an inverse - contradiction. I'm not really sure if I'm right here.

Help would be appreciated.

Answer 1

Since $f^{-1}\circ f=\operatorname{Id}$, $D_{f^{-1}}\circ D_f=D_{f^{-1}\circ f}=\operatorname{Id}$. But $D_f$ and $D_{f^{-1}}$ are linear maps from $\mathbb{R}^n$ into itself. So, each one is the inverse of the other one.

Answer 2

Well, if $f$ is a $\mathscr{C}^r$ diffeomorphism, then it has $\mathscr{C}^r$ inverse called $g$. We know that the differential $D$ has the property $D(f\circ g)=Df\circ Dg$ by the chain rule. So, since $f\circ g= \mathbb{1}_V$ and $g\circ f=\mathbb{1}_U$, we get that $Df$ and $Dg$ are mutually inverse, hence isomorphisms.