# If an open set $U \subseteq\mathbb{R}^n$ is diffeomorphic to an open set $V\subseteq \mathbb{R}^m$ then…

analysissmooth-manifolds

Consider the problem asked in my smooth manifolds assignment:

If an open set $$U \subseteq\mathbb{R}^n$$ is diffeomorphic to an open set $$V\subseteq \mathbb{R}^m$$ then prove that n=m.

I think the question means that there exits a diffeomorphism f (say) from $$\mathbb{R}^n \to \mathbb{R}^m$$. Now, I thought of proving that such map f cannot be both one -one or onto.But I am unable to argue because f is not given.So, how should I try proving that if such map is both 1-1 and onto (assuming that it exists), m=n.

If there is some other way of proving that m=n, that is also welcome.

Let $$F : U \rightarrow V$$ be a diffeomorphism and let $$p \in U$$. Then the derivative $$DF_p : T_pU \rightarrow T_{F(p)}V$$ is an isomorphism of vector spaces. Since $$T_pU$$ is isomorphic to $$\mathbb{R}^n$$ and $$T_{F(p)}V$$ is isomorphic to $$\mathbb{R}^m$$, we must have that $$n = m$$.