Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?

Question

Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected.

Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?

Full disclosure: I've now crossposted the question on MO.

Various remarks

If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = \sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.
With the bijection, it holds true in $n = 1$. But this is using the order structure of $\mathbb{R}$: a bijection that preserves connectedness on $\mathbb{R}$ must be monotone.
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)

Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:

Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?
Some properties of $\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\mathbb{R}^n$ with the standard topology to itself, by self-maps of some arbitrary topological space, it is easy to make the answer go either way.
- For example, if the topological space $(X,\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.
- On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \mathbb{Z}$ equipped with the topology generated by $$ \{\mathbb{N}\} \cup \{ \{k\} \mid k \in \mathbb{Z} \setminus \mathbb{N} \} $$ then the map $k \mapsto k+\ell$ for any $\ell > 0$ maps connected sets to connected sets, but its inverse $k\mapsto k-\ell$ can map connected sets to disconnected ones.
  
  Working a bit harder one can construct in similar vein examples which are Hausdorff.

Funny how just before you posted the answer there, I was talking to one of the postdocs here, and he complained that this question got buried in heaps of triviality and no one will answer it now; but then you posted the answer and this question (which is the crux of the problem, it seems). Gotta love good timing! :-) — Asaf Karagila, Sep 30 '14 at 12:27
Just to confirm: when you say the induced power set map, you mean the mapping $S \mapsto f(S)$? — theHigherGeometer, Oct 01 '14 at 01:01
@DavidRoberts: for $S\subset X$, yes. I wrote in that strange language mostly because of the first "Note": when $f$ is not bijective $f^{-1}$ is not defined as a function from $Y\to X$ and it is more convenient to think of the map on the power sets. — Willie Wong, Oct 01 '14 at 08:45
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance. — Jyrki Lahtonen, Jun 20 '15 at 18:00
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.) — Willie Wong, Jun 22 '15 at 08:03
Can you give an example of a bijection on $\Bbb R^n$ that is connected but not continuous? — Greg Martin, Aug 18 '15 at 22:33
@Greg Martin: Presumably not, since that would provide a negative answer to the question. See http://math.stackexchange.com/a/952416/3457 — Niels J. Diepeveen, Aug 18 '15 at 23:32
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $\implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.) — Willie Wong, Aug 21 '15 at 02:24
The title and the question in the text seem to be negatives of each other. I was confused when I read the definition and then spotted the 1st remark. — JiK, Aug 25 '15 at 10:37
@JiK: you are right. But outside of semantics the two are asking about the same thing, so I ask you to forgive me for being lazy and not editing the two to perfectly align. — Willie Wong, Aug 26 '15 at 13:56
@Greg Martin: Consider $f:\mathbb{R}^2\to\mathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+\sin(\pi/s)$ for $s\neq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected. — Alan U. Kennington, Aug 29 '15 at 21:03
In my previous comment, I meant to say that $f(s,t)=(s,t+\sin(\pi/s))$ for $s\neq0$. — Alan U. Kennington, Aug 30 '15 at 00:51
My first thought was some sort of Peano curve, but those are not bijections. A related question that I once posted to mathoverflow is whether there could be a space-filling curve that preserves convexity. $\qquad$ — Michael Hardy, Apr 10 '16 at 19:25
@AlanU.Kennington: The mapping you mentioned is bijective, but are you sure it is connected? Consider a small square centered at the origin. Is the mapping of this square connected? Near $s=0,$ each vertical line of the square will be shifted wildly up and down by the mapping. — J. Heller, Dec 05 '16 at 02:55
@J.Heller I believe it is, analogously to how the [topologist's sine curve](https://en.wikipedia.org/wiki/Topologist's_sine_curve) is connected (although not locally or path connected). — Noah Schweber, Dec 07 '16 at 01:47
The image of that particular small square is indeed connected, yet $f$ is not a connected map. If $J$ is the involution $(s,t)\mapsto (-s,t)$, then $f^{-1}=J\circ f\circ J$. So, if $f$ were a connected map, so would be $f^{-1}$, and then $f$ would be continuous by the Tanaka's theorem quoted by Willie Wong. — Pietro Majer, Jan 26 '17 at 17:29
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= \{(0,1)\}\cup (0,1]\times\{0\}$ and $S:=f^{-1}(R )$. — Pietro Majer, Jan 26 '17 at 17:38
@Teddy38: This question was posted at mathoverflow, received quite a bit of attention there, but only partial solutions were proposed. What are the chances that your bounty will succeed here? — Moishe Kohan, Nov 13 '17 at 21:23
@AndréLevy: how do you envision Banach-Tarski to be helpful here? — Willie Wong, Jun 27 '18 at 21:16
Isn't that what Banach-Tarski is precisely about; a bijection between two sets, a connected one (a single sphere) and a disjoint one (two spheres)? — André Levy, Jun 27 '18 at 22:38
@AndréLevy: not at all. (In multiple senses of the phrase: your characterization of Banach-Tarski is not what it is really about, and your characterization of Banach-Tarski says nothing about my question.) — Willie Wong, Jun 28 '18 at 13:25
Well, @WillieWong, isn't the disassembling and reassembling of the pieces in B-T a bijection, isn't this bijection connected from the 2 spheres to the 1 sphere, but not the reverse? — André Levy, Jun 30 '18 at 00:36
@AndréLevy: Can you justify that the bijection is connected from the 2 spheres to the 1 sphere? Each of the paradoxical pieces is dense in the sphere, this suggests that the corresponding bijection cannot be connected. (Let $P,Q\subset \mathbb{S}^2$ be two disjoint dense sets. And let $\rho$ be a non-trivial rotation. Then for any $\epsilon > 0$ there exists $p\in P$ and $q\in Q$ such that $d(p,q) < \epsilon d(p,\rho(q))$.) — Willie Wong, Jul 02 '18 at 14:06
@RobertFrost: you are half right. See [this previous comment](https://math.stackexchange.com/questions/952466/does-there-exist-a-bijection-of-mathbbrn-with-itself-such-that-the-forward?noredirect=1#comment2870188_952466) — Willie Wong, Aug 19 '18 at 19:54
@RobertFrost: mostly for my curiosity. But I note that the answer to a [related question](https://math.stackexchange.com/questions/949168/is-bijection-mapping-connected-sets-to-connected-homeomorphism) is known, and that question is what inspired me to ask this one. — Willie Wong, Aug 20 '18 at 12:59
@WillieWong I'm not competent with the subtleties of the concept of continuity but this reminds me of a function I'm doing some work with. The branch cuts of its inverse are all superimposed on top of each other so for any $x+\epsilon$ there are $f^{-1}(x+\epsilon)$ not necessarily nearby to $f^{-1}(x)$ but I'm not totally clear how the concept of continuity treats this and whether it would fit your conditions. — samerivertwice, Aug 22 '18 at 12:45
@RobertFrost: this particular question I am asking about is turning out to be very subtle. [This particular answer of the cross post to MathOverflow](https://mathoverflow.net/a/260589/3948) shows how to answer the question if we change some small number of conditions. But the original question is still, as far as I know, open. — Willie Wong, Aug 22 '18 at 14:13
@WillieWong I think I can prove this if I succeed to prove that f maps closed connected sets onto closed connected sets. — Right, Aug 27 '18 at 02:55
I suspect that there is no such map in the analogous problem for $S^2$ instead of $\mathbb R^2$; the reason being that if we have a separating set $X$ (i.e. one such that $S^2\setminus X$ is disconnected), then its preimage has to be a separating set as well. Separating sets are kinda circle-like and I suspect have properties like "there are connected subsets $A,B,C\subseteq X$, no pair disjoint, such that $A\cap B\cap C=\emptyset$". Given this lemma or a similar ones, one can show the non-existence of such a forward-but-not-backwards connected map $S^2\rightarrow S^2$. — Milo Brandt, Jan 07 '19 at 04:48
@MiloBrandt: for $S^2$ your suspicion is sound. See [this answer](https://mathoverflow.net/a/309671/3948) on the MathOverflow version of this question. Thanks! — Willie Wong, Jan 07 '19 at 19:06
For n=1 with f(x) = x, if x is rational; and x+1, if x is irrational, f is a bijection that is nowhere continuous. The usual topology on R begins its life connected. What is the definition of connected that lets us discard this f and conclude n > 1 for R^n? — mmfrgmpds, Sep 20 '19 at 18:33
Ok I think I see the issue. The piecewise assignment is a priori and causes us to voluntarily disconnect the domain. So the pre-image and the image are both disconnected. But the problem asks for a bijection with a connected pre-image but disconnected image, for example. — mmfrgmpds, Sep 22 '19 at 02:44
@WillieWong might it be possible to send borromian rings in $\Bbb R^3$ to trefoil knots? I can see the "rope" would need to be spliced into possibly three strands, with twists in such a way that it continuously morphs between knots and rings. I've no idea if this can hang together rigorously. — samerivertwice, Dec 24 '19 at 10:55

Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?

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