Topological manifolds (dimension) 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. 

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.
