Show $\mathbb{R}^n$ is not diffeomorphic to $\mathbb{R}^m$ for $m\neq n$

Question

Here's how I've approached the problem so far.

Without loss of generality, assume $n<m$. Now we assume there exists some diffeomorphism $f:\mathbb{R}^n\to\mathbb{R}^m$. A diffeomorphism satisfies the following:

• $f$ is bijective
• $f$ is differentiable
• $f^{-1}=g$ is differentiable

Now there exists a bijection from $\mathbb{R}^n$ to $\mathbb{R}^m$ so I don't believe showing a contradiction there is the way to go.

We have that $g$ is the inverse of $f$ so

• $g\circ f = \text{id}_{\mathbb{R}^n}$
• $f\circ g = \text{id}_{\mathbb{R}^m}$

I want to take the derivative of these with the chain rule and show a contradiction but can't find one. Is there one or is there another approach?

Answer 1

If you take the derivatives of these diffeomorphisms and apply the chain rule, then you will find that $d_0f$ is an isomorphism (of vector spaces) from the tangent space $T_0\mathbb{R}^n$ to the tangent space $T_{f(0)}\mathbb{R}^m$. But the tangent space $T_0\mathbb{R}^n$ is of dimension $n$, while the tangent space $T_{f(0)}\mathbb{R}^m$ is of dimension $m$, so they cannot possibly be isomorphic as vector spaces. This is the contradiction.