Proving manifold dimension theory using homology The text I am using stated that any manifold with different dimension cannot be homeomorphic, but did not prove it (only proved the case when $n = 2$) until now, and it is given as an exercise in the homology chapter, but I am not sure how to approach this problem. Can anyone provide a solution or hint?
 A: Just check that being locally homeomorphic to $\mathbb{R}^n$ is a property that is invariant under homeomorphism; the conclusion follows from the fact that $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic if and only if $m=n$ (and to prove that you need homology).
A: Let $h : M \to N$ be a homeomorphism and $p \in M$. Then $h$ restricts to a homeomorphism $h'  : M \setminus \{p\} \to N \setminus \{h(p)\}$. Thus long exact sequences of $(M,M \setminus \{p\})$ and $(N, N \setminus \{h(p)\})$ are connected by the map $h$ and its restrictuion $h'$. But $h_* : H_k(M) \to H_k(N)$ and $h'_* : H_k(M \setminus \{p\}) \to H_k(N \setminus \{h(p)\})$ are isomorphisms, therefore the five lemma gives us an isomorphism $H_k(M,M \setminus \{p\}) \to H_k(N, N \setminus \{h(p)\})$.
By excision we see that $H_k(M,M \setminus \{p\}) \approx H_k(U,U \setminus \{p\})$, where $U$ is an open neigbhorhood of $p$ which is homeomorphic to an open ball in $\mathbb R^n$,  $n = \dim M$. Thus $H_k(M,M \setminus \{p\}) \approx \mathbb Z$ for $k = 0,n$ and $H_k(M,M \setminus \{p\}) = 0$ for all other $k$.
This shows that we must have $\dim M = \dim N$.
