bijective continuous function on $\mathbb R^n$ not homeomorphism? Suppose we have a bijective continuous map $\mathbb{R}^n\to\mathbb{R}^n$ (relative to the standard topology). Must this map be a homeomorphism?
I have little doubt about this. I think that if it happens, I guess it's true, I've heard it is true, but I can not prove it. 
 A: Every such map is open according to invariance of domain and is therefore a homeomorphism. However, invariance of domain is highly nontrivial to prove with elementary topological methods. The slickest way would be via algebraic topology.
A: I think this should also work:
We have the result that a bijective local homeomorphism between two spaces $X,Y$ is a global homeomorphism. We then show that a map $f:X\rightarrow Y$ with the given conditions is a bijective (given in the problem) , local homeomorphism.
Let's then show we get a local homeomorphism. 
We  select, for any x in $\mathbb R^n$, a closed ball $B(x,r)$;$r>0$ then
$f|_{B(x,r)}$  is a continuous bijection between the compact subset $B(x,r)$ and $f(B(x,r))\mathbb R^n$ Hausdorff (the restriction to $f(B(x,r))$ is Haudorff), is a homeomorphism.
Select , then, an open neighborhood $B^0(x,r)$ of $B(x,r)$. Then $f|_{B^0}$ is also
a homeomorphism, so you get an injective local homeomorphism $f:B^0\rightarrow f(B^o)$  between spaces X,Y, which
is a global homeomorphism. As Pete Clark said (If I understood well), you can repeat this argument when a closed, bounded subset is compact.  
A: Yes, a continuous bijection $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ must be a homeomorphism.
It suffices to check that such a map $f$ is closed.  Here is are two hints to help show this:
1) Show that a subset $K$ of $\mathbb{R}^n$ is compact iff $f(K)$ is compact.
2) Show that a subset $Y$ of $\mathbb{R}^n$ is closed iff for every compact subset $K$ of $\mathbb{R}^n$, $Y \cap K$ is closed.  
(This argument should work with $\mathbb{R}^n$ replaced by any metric space in which a subset is compact iff it is closed and bounded.)  
