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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.

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    For $n=1$, you might have a look at [this](https://math.stackexchange.com/questions/145639).2017-01-11

3 Answers 3

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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.

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    I would also like to point out that invariance of domain and the statement that every continuous bijection from $\mathbb{R}^n$ to $\mathbb{R}^n$ is a homeomorphism are equivalent. Therefore I would not expect there to be a proof that avoids this machinery.2011-08-25
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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.)

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    @PeteL.Clark You wrote: *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.* Would [my answer](http://math.stackexchange.com/a/155106/) to a related question be a counterexample to this claim? The map $f$ in that answer is a continuous bijection, but not a homeomorphism. (Needless to say, even if it is a counterexample; I consider your approach to this question interesting; when I thought about this problem, I was thinking about using $\sigma$-compactness and properties of continuous bijections on compacta, too.)2012-06-07
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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.

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    @gary: sorry, I wrote that comment before seeing yours addressing the same problem. That was meant to be a comment directed at Daniel.2011-08-30