Question about two homeomorphic closed manifolds I was studying about algebraic topology with my study group.
So, there was a question that held all of the study members confused. 

If two closed manifolds are homeomorphic, they must have same
  dimension.

It sounds really trivial but we did not find any good way to prove this problem.
Any suggestions to start the proof?
Thanks in advance.
 A: As suggested, as the dimension is defined locally, the problem is also a local one:
(note that $dim$ is a locally constant function, hence it our manifolds are connected, the dimension is globally the same)
If you have not heard about homology, you can probably show $\mathbb R \not \cong \mathbb R^2$. Maybe also (still not very hard) that $\mathbb R^2 \not \cong \mathbb R^3$ (and of course $\mathbb R \not \cong \mathbb R^3$). Beyond that it gets very hard and highly non trivial.
This is usually one of the first examples in lectures about homology, to show how powerful the presented tool is. You will get a homotopy invariant (in particular a topological invariant) which yields the following groups (indexed with $n\geq 0$) applied to the desired spaces: 
$$
H_i(\mathbb R^n,\mathbb R^n -0) \cong \begin{cases} \mathbb Z, & i=0,n-1 \\ 0,& \text{else} \end{cases}
$$
Hence you get that $H_*(\mathbb R^n,\mathbb R^n-0)=H_*(\mathbb R^k,\mathbb R^k-0)$ (i.e. all groups with same index coincide)$ \implies $k=n$, in particular there is no homeomorphism for euclidean space of different dimension (it would induce an iso on homology).
Also note that there is a very important theorem called "Invariance of domain", but also that this is harder to show than the statement above.
