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I am taking an introductory course to topology and the professor defined a topological manifold of dimension $n$ if it is hausdorff and if for every point $x$ there exists an open set $U$ around $x$ such that $U$ is homeomorphic to $\mathbb{R}^n$. My question is: By the way she defined it, one could have (apriori) a topological space $X$ being a manifold of dimensions $n$ and $m$? Meaning I could find open sets $U$ and $V$ around every $x$ and $U$ is homeomorphic to $\mathbb{R}^n$ and $V$ is homeomorphic to $\mathbb{R}^m$, and this would make the notion of dimension not well defined. My guess is that since it is called "topological manifold of dimension $n$" what I described probably cant happen, but I dont see why not.

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Indeed, how does one know that $\mathbb R^n$ is not homeomorphic to $\mathbb R^m$ unless $n=m$? IIRC, one must look at homotophy groups, or at least at homology groups. – Michael Hardy May 19 '13 at 20:14
You are right. This is called invariance of domain. – 23rd May 19 '13 at 20:14
It's also possible to prove that the Lebesgue covering dimension of an $n$-manifold is $n$. But this too requires the tools of algebraic topology. – kahen May 19 '13 at 20:41
It's a somewhat frustrating fact that while basic topology is great at proving things the same, you need to work pretty hard to prove that things are different from each other. To prove that $A$ and $B$ are homeomorphic, just provide a homeomorphism, but proving they are not homeomorphic needs some kind of invariant on which they differ. – Ben Millwood May 19 '13 at 21:28
up vote 8 down vote accepted

You are correct.

There are several ways to show that it is impossible, and it basically boils down to the fact that no open subset of ${\bf R}^n$ is homeomorphic to an open subset of ${\bf R}^m$ if $n<m$ (as $U\cap V$ would be in your case). You can assume without loss of generality that the sets in question are connected.

One way to show it is to use the fact that you can find an $(n-1)$-dimensional sphere which disconnects any open subset of ${\bf R}^n$, but it is a general fact that if a compact set disconnects ${\bf R}^m$, then it must admit a non-nullhomotopic map onto $S^{m-1}$ (in fact, it is an equivalent condition), while on the other hand, no proper compact subset of $S^{m-1}$ admits such a map.

So, summing it all up, a set homeomorphic to $S^{n-1}$ can't disconnect an open subset of ${\bf R}^m$ if $m>n$ and we're done.

(The results used can be shown by combinatorial or algebraic topology, but are not simple enough to show here.)

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Thanks for noting me that point, Tomasz. :-) and +1 – S. Snape May 22 '13 at 11:28

You can use local homology groups to solve this problem. If $X$ is a topological space and if $x\in X$, then the local homology groups of $X$ at $x$ are defined by the rule $H_i^{\text{local}}(X,x)=H_i(X,X-x)$, where the right hand side denotes the relative homology groups of $X$ relative to $X-x$.

Exercise 1: Prove that if $X$ is an $n$-manifold, then the local homology groups of $X$ agree with the homology of an $n$-sphere. (Hint: use excision to reduce to computing the homology of an $n$-ball relative to its boundary, an $(n-1)$-sphere.)

Exercise 2: Prove that the dimension of a connected topological manifold is an invariant.

Exercise 3: What can you say about manifolds with boundary?

Hope this helps! Of course, tomasz's answer and the comments are excellent and address the question in other ways.

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