Every $n-$ dimensional manifold is homemorphic to $\Bbb R^n$ This may seem like a really silly question based on the definition of a manifold (because every manifold is locally like Euclidean space), but can someone give me a counterexample where an $n-$dim manifold is not homemorphic to $\Bbb R^n$?
 A: A disjoint union of two copies of $\mathbb{R}^n$.
An $n$-dimensional sphere $S^n$.
An $n$-dimensional torus $T^n$.
A: Any n-manifold that needs more than one chart is not homeomorphic to $\mathbb R^n$. Generalizing  the answers above, by topology:
1)No compact n-manifold is homeomorphic to $\mathbb R^n$, since $\mathbb R^n$ is not compact
2) No disconnected manifold ,
3) No manifold with non-trivial n-th fundamental group or non-trivial homology;
these manifolds  are not just not homeomorphic to $\mathbb R^n$ , but they are not even homotopically equivalent (which is much weaker than homeomorphism) to $\mathbb R^n$

4) Fill in your favorite topological property of $\mathbb R^n$ and then any manifold not satisfying that property is not homeomorphic to $\mathbb R^n$.
A: The unit circle is not homeomorphic to the line.
A: As others have pointed out, there are many obstructions for an $n$-manifold to be homeomorphic to $\mathbb{R}^n$. (If there weren't, manifold topology would be really boring!) I think all of the ones that have been described are homotopy invariants coming from algebraic topology: $\mathbb{R}^n$ is contractible, so any homotopy invariant is trivial on it. Hence if you have an $n$-manifold where any such invariant is nontrivial then it cannot even be homotopy equivalent to $\mathbb{R}^n$, let alone homeomorphic. For example:


*

*the set $\pi_0$ of connected components distinguishes $\mathbb{R}^n$ from two disjoint copies $\mathbb{R}^n \sqcup \mathbb{R}^n$ of $\mathbb{R}^n$ (arguably the simplest counterexample);

*the fundamental group $\pi_1$ distinguishes $\mathbb{R}^n$ from the $n$-torus $\mathbb{R}^n/\mathbb{Z}^n$ (for example). 

*the higher homotopy groups $\pi_k$ distinguish $\mathbb{R}^n$ from the $n$-sphere $S^n$ (for example).

*the homology groups $H_k$ or cohomology groups $H^k$ also distinguish $\mathbb{R}^n$ from the $n$-sphere $S^n$, and more generally from any compact $n$-manifold.


(Compactness, which is a priori a topological condition, is actually an algebro-topological condition on $n$-manifolds as well, provided that you know $n$: for a connected $n$-manifold $X$, it's equivalent to $H^n(X, \mathbb{F}_2)$ not vanishing.)
All of these invariants vanish when an $n$-manifold $X$ is contractible, or equivalently when it is homotopy equivalent to $\mathbb{R}^n$. However, it is still not true that a contractible $n$-manifold is homeomorphic to $\mathbb{R}^n$! The Whitehead manifold is a famous counterexample here for $n = 3$. 
For positive results along these lines, see the Poincaré conjecture, Mostow rigidity, and the Borel conjecture. 
