So I've started studying differentiable manifolds, and I came across the following problem:
Problem: Assume that $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic iff $m=n$. Let $M$ be a connected differential manifold and let $\mathcal{A}_1$ and $\mathcal{A}_2$ be atlases M with dimensions $m$ and $n$, respectively. Show that $m=n$.
First thing I noticed is that I can't use differentiability in order to solve this, since we're dealing with different atlases which might not be compatible.
We'll assume $\mathcal{A}_1$ and $\mathcal{A}_2$ are maximal atlases, so that any open neighborhood has an associated chart.
Now, fix $p \in M$ and let $(U, \phi) \in \mathcal{A}_1$ and $(U, \psi) \in \mathcal{A}_2$ be two charts such that $p \in U$ and $\phi: U \to \mathbb{R}^m$ is a homeomorphism.
The map $\psi \circ \phi^{-1}: \mathbb{R}^m \to \psi(U) \subset \mathbb{R}^n$ is, thus, also a homeomorphism.
We thus, have an open set $\psi(U) \subset \mathbb{R}^n$ which is homeomorphic to $\mathbb{R}^m$. For me this obviously implies that $m$ and $n$ should be equal or, at least, $m<n$ or something similar. However, I have no clue how I should proceed or if I'm even on the right track.
