Homeomorphic manifolds have the same dimension

Question

So I want to prove: If two manifolds $M$ and $N$ are homeomorphic then $dim(M) = m = n = dim(N)$.

My idea was to use the property of the manifolds that they are locally homeomorphic to the $\mathbb{R}^m$ and $\mathbb{R}^n$ respectively. So I want to take an open subset $U \subset M$ which is homeomorphic to the $\mathbb{R}^m$, but my problem is how do I know that the image $f(U) = V \subset N$ is homeomorphic to the $\mathbb{R}^n$. I probably don't, but do we even have the existence of such $U$ and $V$? If so, why? Is this even the right approach to the proof?

Answer 1

Let $M,N$ be manifolds with dimension m, n respectively. Suppose $\Phi:M \to N$ is homeomorphism. Let $\psi_1: M \supset D \to U \subset \mathbb R^m$ be a map (that is homeomorphism of some open subset of $M$ and open subset of Euclidean space). $U$ is open so we can choose open some ball in it and restrict $\psi_1$ to subset that is mapped onto this ball. So without loss of generality you can assume some subset of $M$ is homeomorphic to open ball in $\mathbb R^m$. Analogously we construct homeomorphism $\psi _2$ from open subset of $N$ to open ball in $\mathbb R^n$. Therefore $\psi_1 \circ \Phi \circ \psi_2 ^{-1}$ maps ball in $\mathbb R^n$ to ball in $\mathbb R^m$. That is what I meant before editting. Open ball in $\mathbb R ^n$ is homeomorphic to $\mathbb R^n$ so final result is that $\mathbb R^n $ and $\mathbb R^m$ are homeomorphic. I assume you know that this implies $n=m$.

Answer 2

Consider a neighbourhood $O_M$ in $M$. Since $M$ is a manifold, it's locally homeomorphic to $\mathbb{R}^m$ for some integer $m$. Let $f_1$ be the homeomorphic mapping from $\mathbb{R}^m$ to $O_M$. Now consider the homeomorphism $f_2$ from $M$ to $N$. Since $f_2$ is a homeomorphism, $f_2(O_M)$ is a neighbourhood in $N$, call it $O_N$. And finally, $O_N$ is homeomorphic to $\mathbb{R}^n$ for some integer $n$ and call that homeomorphism $f_3$.

Now composing the three homeomorphisms, we get $f_3 \circ f_2 \circ f_1$, which is a homeomorphism from $\mathbb{R}^m$ to $\mathbb{R}^n$. Now if $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^m$, then $n=m$, by some theorem whose name I've forgotten.

Answer 3

Choose a homomorphism $h:M\rightarrow N$.

Let $x\in M$. There exists an open nighborhood $U\subseteq M$ of $x$ and a homomorphism $\psi_1:U\rightarrow B^m$, (where $B^m$ denotes an open ball in $\mathbb{R}^m$). Now, consider $h(x)\in N$. There exists an open set $V\subseteq N$ s.t. $h(x)\in V$ and a homomorphism $\psi_2:V\rightarrow B^n$. By restricting $\psi_2$ to $h(U)\cap V$ you obtain a homomorphism $$\phi_2:=\psi_2\mid_{h(U)\cap V}:h(U)\cap V\rightarrow \psi_2(h(U)\cap V)$$ (note that $\psi_2(h(U)\cap V)$ is open in $\mathbb{R}^n$).

Now, $h^{-1}(h(U)\cap V)\subseteq U$ and it's open, thus you obtain a homomorphism $$\phi_1:=\psi_1\mid_{h^{-1}(h(U)\cap V)}:h^{-1}(h(U)\cap V)\rightarrow \psi_1(h^{-1}(h(U)\cap V))$$ (note that $\psi_2(h^{-1}(h(U)\cap V))$ is open in $\mathbb{R}^m$). By composing $\phi_2\circ h\circ \phi_1^{-1}$ ($h$ has to be appropriately restricted) you obtain a homomorphism from $\psi_1(h^{-1}(h(U)\cap V))$ (open in $\mathbb{R}^m$) to $\psi_2(h(U)\cap V)$ (open in $\mathbb{R}^n$), and this forces $n=m$.