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.

  • $\begingroup$ From where to where? From $\mathbb R^n$ to itself, take any invertible linear operator. $\endgroup$
    – Asaf Karagila
    Commented Aug 24, 2011 at 19:48
  • $\begingroup$ R^n --> R^n both with the same topology. $\endgroup$
    – Daniel
    Commented Aug 24, 2011 at 19:50
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    $\begingroup$ "I guess that if it happens, I guess it's true." If what happens, then what is true? I've tried to guess by adding what seems to be the question, but this is still somewhat confusing. Could you please clarify? $\endgroup$ Commented Aug 24, 2011 at 19:52
  • $\begingroup$ The question is this. Prove or disprove the following: Every bijective continuous function $$ f:R^n \to R^n $$ it´s a homeomorphism. ( both with the standard topology) $\endgroup$
    – Daniel
    Commented Aug 24, 2011 at 19:55
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    $\begingroup$ @Daniel: Thank you! I've edited the question to try to clarify it, though the second paragraph is still a bit hard to parse. $\endgroup$ Commented Aug 24, 2011 at 19:56

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 25, 2011 at 17:17

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.)

  • 2
    $\begingroup$ Oh Pete your idea is so great! but something bothers me, first I'm not using the fact that the $ $ R ^ n \ to R ^ n $ $ (Same n) is not true with "n" other than that there is a homeomorphism. And I do not know how to end the case when is unbounded T_T, sorry for being so stupid $\endgroup$
    – Daniel
    Commented Aug 24, 2011 at 20:41
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    $\begingroup$ @Daniel: I am sorry to say that I am having trouble carrying the proof of 1) through myself. (The argument I originally had in mind turned out to need Invariance of Domain, which should of course be off limits in the context of this proof.) It seems like there is some chance to repair it using the Baire Category Theorem: e.g. if $C_n$ is the image under $f$ of the closed disk of radius $n$ about zero, then by BCT the union of the interiors of the $C_n$ is dense in $\mathbb{R}^n$. $\endgroup$ Commented Aug 24, 2011 at 22:45
  • $\begingroup$ What I need to prove is that any compact subset $K$ is contained in $C_n$ for some $n$, because then no unbounded subset could map under $f$ to a bounded set. If the union of the interiors of the $C_n$'s were all of $\mathbb{R}^n$, then we would be done by compactness. With the weaker conclusion of the last comment, I'm not yet sure what to do. More later when I get the time to think about it... $\endgroup$ Commented Aug 24, 2011 at 22:48
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    $\begingroup$ I would now like to classify this as a reasonable idea which doesn't seem to pan out. Sorry for giving it as "hints". But there are actually some interesting issues here which I hope to return to eventually... $\endgroup$ Commented Aug 25, 2011 at 16:18
  • $\begingroup$ @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 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.) $\endgroup$ Commented Jun 7, 2012 at 13:49

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.

  • $\begingroup$ nice gary, but I have a question, why this argument does not hold on a continuous bijective map between $$ R^n \to R^k \,\,\,\,n \ne k $$ $\endgroup$
    – Daniel
    Commented Aug 24, 2011 at 21:32
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    $\begingroup$ I think then the image f of an open set in $\mathbb R^n$ is not open in $\mathbb R^k$ , if $k\neq n$. Actually, I think there is no continuous bijection between $\mathbb r^n $ and $\mathbb R^k$. There are definitely bijections, but I do not think they can be continuous, but I can't think of a good argument for why. $\endgroup$
    – gary
    Commented Aug 24, 2011 at 21:40
  • $\begingroup$ I think the idea is this: the image of a continuous map $\mathbb R\rightarrow \mathbb R^2$ is a curve; the image of a continuous (even smooth, etc.) map from $\mathbb R^2$ into $\mathbb R^3$ is a surface, i.e., a 2-dimensional object, etc. $\endgroup$
    – gary
    Commented Aug 24, 2011 at 21:49
  • $\begingroup$ That is my claim $\endgroup$
    – Daniel
    Commented Aug 24, 2011 at 21:58
  • $\begingroup$ Well, but my point is that since the image is meager, i.e., the image contains no open sets in the target space, the map cannot be globally--I think not even locally--bijective, so that we cannot apply the result we did for maps $f:\mathbb R^n\rightarrow \mathbb R^n$ $\endgroup$
    – gary
    Commented Aug 24, 2011 at 22:04

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