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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?

Cosmare
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    Use invariance of domain. – anomaly Jul 09 '15 at 16:27
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    You can try to show that $\mathbb{R}^m$ and $\mathbb{R}^n$ are not homeomorphic if $m \neq n$. – aGer Jul 09 '15 at 16:28
  • I already have that, that's not my question. I want the existence of two homeomorphic subsets of the manifolds respectively which are each homeomorphic to the $\mathbb{R}^m$ and $\mathbb{R}^n$. – Cosmare Jul 09 '15 at 16:31
  • Invariance of domain states that any continuous, injective map $f: U\to \mathbb{R}^n$ for $U\subset \mathbb{R}^n$ open is a homeomorphism onto its image. That's exactly what you want (modulo some minor issues with the dimension). – anomaly Jul 09 '15 at 17:17
  • I think the proof of this theorem is harder than my proof and since we didn't cover this theorem I can't use it except I proof it myself. – Cosmare Jul 09 '15 at 17:21
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    Hint to prove that two open subsets $U\subseteq\mathbb{R}^n,V\subseteq\mathbb{R}^m$ can only be homeomorphic for $n=m$: Prove that the homeomorphism restricts to a homeomorphism between pairs $(U,U\backslash{x})$ and $(V,V\backslash{y})$ for some points $x\in U$ and $y\in V$. Use excision to calculate the homology of these pairs and you'll get the result. – archipelago Jul 09 '15 at 20:52

3 Answers3

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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 $n=m$.

John
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  • Why does it force $n = m$? I only know the proof for $\mathbb{R}^m \cong \mathbb{R}^n$. Let's say $m > n$, then $\mathbb{R}^n$ is an open subset of $\mathbb{R}^m$ but obviously they are not homeomorphic. – Cosmare Jul 09 '15 at 17:18
  • Sorry, but I can't follow, first of all it's an algebraic topology problem. And surely if you have a subspace $V$ of a space $X$ with dimension $n$ then $V$ doesn't have dimension $n$ but dimension lower than $n$. Maybe your last detail can be proven via homology but I don't see how. – Cosmare Jul 09 '15 at 17:39
  • yes you are right sorry (I had to cancel the comment as I realized I was not right). It's quite a well known result anyway (it is also cited in this question), but I can't think of an easy proof right now. Sorry! – John Jul 09 '15 at 17:39
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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 $n=m$.

Blazej
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  • I don't understand your answer, sorry. – Cosmare Jul 09 '15 at 15:45
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    I edited it to make it more readable – Blazej Jul 09 '15 at 16:13
  • Yes, you can choose a subset $D \subset M$ which is homeomorphic to the $\mathbb{R}^m$ and we can choose a subset $E \subset N$ which is homeomorphic to the $\mathbb{R}^n$ but there is no reason why $D$ and $E$ should be homeomorphic. That's my question. – Cosmare Jul 09 '15 at 16:22
  • You have shown that $M$ and $N$ are homeomorphic. This is not the question of the OP. – aGer Jul 09 '15 at 16:22
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    Composition of homeomorphisms is homeomorphism. So I explicitly wrote down homeomorphism between R^n and R^m. – Blazej Jul 09 '15 at 16:25
  • No, because you didn't proof that $D$ and $E$ are homeomorphic. – Cosmare Jul 09 '15 at 16:33
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    I have made a mistake. Here is a solution. I use symbols as in post before. $f=\Phi \circ \psi _1 ^{-1}$ is homeomorphic map of $\mathbb R ^m$ in open subset of $N$. Choose (open) subset $V$ of its range such that it is homeomorphic with $\mathbb R ^n$. By restricting f to $f^{-1}(V)$ you get homeomorphism of open subset of $\mathbb R^m$ with $\mathbb R^n$ which is enough. – Blazej Jul 09 '15 at 17:11
  • Why is it enough? – Cosmare Jul 09 '15 at 20:14
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    There is a theorem that says that if two open subsets of $\mathbb R ^n$ and $\mathbb R ^m$ respectively are homeomorphic then $n=m$. – Blazej Jul 09 '15 at 22:21
  • It is consequence of so-called Domain Invariance Theorem, which can be prove using Poincaré-Miranda theorem. – mikis Jul 10 '15 at 09:15
  • It's immediate to prove if one knows the very basics of homology, see my comment to the question. – archipelago Jul 10 '15 at 16:36
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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.

sayantankhan
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    Why is $f_2(O_M)$ homeomorphic to $\mathbb{R}^n$? We just know there is a neighborhood for every point in N that is homeomorphic to the $\mathbb{R}^n$ but not that every neighborhood is homeomorphic to the $\mathbb{R}^n$. – Cosmare Jul 09 '15 at 16:32
  • $f_2(O_M)$ is an open neighbourhood in $N$, and since $N$ is a manifold, every open neighbourhood in $N$ is homeomorphic to $\mathbb{R}^n$. – sayantankhan Jul 09 '15 at 16:34
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    No, I don't think so? – Cosmare Jul 09 '15 at 16:34
  • Yes, you're actually right. I'll try and correct my statement as soon as possible. – sayantankhan Jul 09 '15 at 16:37