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If $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ is a bijection, mapping connected sets to connected, is $f$ necessarily a homeomorphism?

The converse is true, a well known property of homeomorphisms.

I know the result is true for $n = 1$. In that case, it's not very hard to see, that $f$ maps open intervals to open intervals: Image is going to be interval, and if it had a boundary point, some $x$ in the open interval would have to map to the boundary point. But now images of $(x, \infty)$ and $(-\infty, x) $ couldn't both be connected.

Also, pre-image of open interval is also open: If $x < y$ are pre-images of endpoints of the interval, the original interval is subset of image of $[x,y]$, but then clearly image of $(x,y)$ is the open interval.

Those two observations prove that $f$ is homeomorphism when $n = 1$. Connectedness is however a quite different phenomenon in higher dimensions, so I don't even have a clear idea if it should be true or not.

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  • $\begingroup$ A function $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ that is both bijective and continuous must map open sets to open sets. So it suffices to show your function is continuous. $\endgroup$ – Michael Sep 28 '14 at 7:58
  • $\begingroup$ @Michael, can you give a reference? $\endgroup$ – user40167 Sep 28 '14 at 7:58
  • $\begingroup$ @krey : You can prove it the same way the asker proves his sub-result. Take an open set that maps to a set that is not open. So there must be a boundary point in the set. $\endgroup$ – Michael Sep 28 '14 at 7:59
  • $\begingroup$ Let me see...I did the proof in my head, perhaps I should do a full proof before making my claim.... $\endgroup$ – Michael Sep 28 '14 at 8:02
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    $\begingroup$ This paper comes pretty close to what you want:ams.org/journals/proc/1958-009-03/S0002-9939-1958-0095468-2/… $\endgroup$ – Henno Brandsma Sep 28 '14 at 9:24
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Using the converse you mentioned, it is clear that if $f$ or $f^{-1}$ fails to map a connected set to another connected set, $f$ cannot be a homeomorphism. Thus, it suffices to show that

Theorem: Fix $n > 1$. Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be a bijection, such that $f$ maps connected sets to connect sets, and $f^{-1}$ maps connected sets to connected sets. Then both $f$ and $f^{-1}$ are continuous.

This is a corollary of Theorem 1 in a paper of Tanaka. I reproduce the proof below.

Proof: since the hypothesis is symmetric in $f$ and $f^{-1}$, it suffices to prove that $f$ is continuous. We proceed by contradiction. Suppose $p\in \mathbb{R}^n$ is such that $f$ is discontinuous at $p$. Then there exists a sequence of points $(p_n)$ such that $p_n \to p$ and $f(p_n) \not\to f(p)$; without loss of generality we assume that $f(p) = 0$. Thus, up to a subsequence we can assume that there exists $\epsilon > 0$ such that $f(p_n) \in B_\epsilon^c = \mathbb{R}^n \setminus B_\epsilon$. Clearly $\{0\} \cup B_\epsilon^c$ has two connected components. On the other hand, $f^{-1}( \{0\} \cup B_\epsilon^c)$ is connected: by assumption $f^{-1}(B_\epsilon^c)$ is connected since $B_\epsilon^c$ is connected, and we have that every open neighborhood of $p = f^{-1}(0)$ intersects $f^{-1}(B_\epsilon^c)$. Thus we forced a contradiction.

Remark: In the case $n = 1$, $B_\epsilon^c$ is not connected. However, it has two connected components $B_\epsilon^{c+}$ and $B_\epsilon^{c-}$. The proof can be carried through if we replace every instance of $B_\epsilon^c$ with either $B_\epsilon^{c+}$ or $B_\epsilon^{c-}$: at least one of the two must contain infinitely many $f(p_n)$.

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  • $\begingroup$ Thank you very much! That is a nice use of the fact that for each connected set $A$, each $A \subset B \subset \overline{A}$ is also connected. You should maybe note that the proof only works for $n>1$ (which is no problem since the claim is easy for $n=1$), because for $n=1$, $B_\varepsilon^c$ is not connected. $\endgroup$ – PhoemueX Sep 30 '14 at 11:35
  • $\begingroup$ @PhoemueX: good point. Edited, and also added a remark for the case $n = 1$ where the proof goes through with a minor modification. $\endgroup$ – Willie Wong Sep 30 '14 at 11:38
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    $\begingroup$ The proof surely works if also $f^{-1}$ has given property, but I can't see why it is implied by assumptions. If not, the function couldn't be homeomorphism, but does there still exist such function ($f$ maps connected sets to connected, but $f^{-1}$ doesn't)? $\endgroup$ – gtrrebel Sep 30 '14 at 11:44
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    $\begingroup$ @gtrrebel: indeed that is the better question. And I have asked it. If I were to guess somewhat haphazardly, maybe it is true in 2D but not higher. I am imagining something that looks like the horned sphere, but can't quite get it to work. $\endgroup$ – Willie Wong Sep 30 '14 at 12:22
  • $\begingroup$ @WillieWong great! I also have similar intuition on the problem, but haven't really had any good ideas for counterexamples; they would have to be quite awful $\endgroup$ – gtrrebel Sep 30 '14 at 12:41
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I think that answer is no and introduce some example.

Assume that $h:\mathbb{R} \rightarrow \mathbb{R}$ is a function such that satisfies in intermediate value property (thus maps connected sets to connected) but is not continous. Now define $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as $f(x,y) = (x,y+h(x))$. $f$ is an example that says NO to above question.

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    $\begingroup$ I am not sure that $f$ necessarily maps connected sets to connected sets. If that is true, note that $f^{-1}(x,y) = (x, y - h(x))$, so that (essentially) the same argument should yield that also $f^{-1}$ maps connected sets to connected sets, but this is impossible by the argument given by @Willie Wong. Still, an interesting idea! $\endgroup$ – PhoemueX Oct 1 '14 at 11:34
  • $\begingroup$ You're right, thanks. I try to construct some thing that $\endgroup$ – Abdollah Oct 2 '14 at 6:57
  • $\begingroup$ (... some thing that) maps $\mathbb{R}$ to a connected set but not homeomorphic to $\mathbb{R}$. $\endgroup$ – Abdollah Oct 2 '14 at 7:04

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