# Bijection $f: \mathbb R^n \to \mathbb R^n$ that maps connected onto connected sets must map closed connected onto closed connected sets?

Willie Wong asked here (MO) and here (MSE) very interesting question.

As he phrased it:

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 $$f^{-1}$$?

When I did some research on that question another question became important to me, and here it is:

Assume $$f:\mathbb{R}^n\to\mathbb{R}^n$$ is a bijection, where $$\mathbb{R}^n$$ is equipped with the standard topology and for every $$S \subset \mathbb R^n$$ connected we have that $$f(S)$$ is connected. Does that imply that if $$T \subset \mathbb R^n$$ is closed and connected that then $$f(T)$$ is closed and connected?

So, basically, I believe that this is really true, that is, that in requirements that $$f$$ maps connected sets onto connected sets and that $$f$$ is a bijection there is hidden a theorem that $$f$$ maps closed connected sets onto closed connected sets.

If this were settled we would be closer to a solution of Willie´s problem, but even if this problem stands on its own it could be of interest to someone.

• @BrevanEllefsen: If $T$ is a non-connected closed set, then we are not asking for $f(T)$ to be closed. Aug 27, 2018 at 5:36
• @BrevanEllefsen: The link addresses the question in the first quote box, which as you say has been answered by Willie. It's included for context, but the point of this post is the question in the second box. Aug 27, 2018 at 5:39
• @NateEldredge right you are about the closed map remark. Skimmed through too quickly and thought too little. Regardless, the link I gave in my comment above should still be applicable. The link answers the question in the second quote box. The question in the first quote box is still an open problem Aug 27, 2018 at 5:39
• @NateEldredge The linked post addresses this question as follows: We know $f$ is a bijection that maps connected sets to connected sets. We wish to show $f$ is also a homeomorphism, as then $f$ is a closed map and this question is trivial. Due to a property of homeomorphisms, we can show $f$ cannot be a homeomorphism unless both $f$ and $f^{-1}$ map connected sets to connected sets, and under this hypothesis we can show $f$ is indeed a homeomorphism. We are thus dependent on the answer to Willie Wong's original post, which is still open. Aug 27, 2018 at 5:49
• @NateEldredge this is what I meant when I gave the closed map remark above - I had this argument in my head, and wrote too quickly without explanation. Of course, it might very well turn out that we can independently prove $f$ is a closed map or even merely that its restriction to connected sets is a closed map, and thus bypass the need for a positive answer to Willie Wong's question. I've now deleted my unclear comments above, and repeat the link for those interested math.stackexchange.com/questions/949168/… Aug 27, 2018 at 5:51

This question is equivalent to the question of Willie Wong because of the following theorem of Jones.

Theorem (Jones, 1967). Each bijective semicontinuous map from a topological space to a semilocally locally connected Hausdorff space is continuous.

A topological space $$X$$ is semilocally connected if if has a base of the topology consisting of open sets whose complements have finitely many connected components.

A function $$f:X\to Y$$ is called

$$\bullet$$ Darboux if for any connected subspace $$C\subset X$$ the image $$f(C)$$ is connected in $$Y$$;

$$\bullet$$ connected if for any connected subspace $$B\subset Y$$ the preimage $$f^{-1}(B)$$ is connected in $$Y$$;

$$\bullet$$ semiconnected if for any connected closed subset $$B\subset Y$$ the preimage $$f^{-1}(B)$$ is connected and closed in $$X$$.

In this terms the problems of @Right and Wong read as follows:

Problem (@Right). Is each connected bijection of $$\mathbb R^n$$ semiconnected?

Problem (Wong). Is each connected bijection of $$\mathbb R^n$$ a homeomorphism?

Now I explain why these two problems are equivalent:

If the answer to the problem of @Right is affirmative, then any connected bijection $$f$$ of $$\mathbb R^n$$ is semiconnected and by Jones Theorem is continuous. By the Invariance of Domain Principle, $$f$$ is open and hence a homeomorphism. So, we get an affirmative answer to the problem of Wong.

If the problem of Wong has an affirmative answer, then any connected bijection $$f$$ of $$\mathbb R^n$$ is a homeomorphism and hence it is both semiconnected and Darboux.

Concerning (partial) answers to the equivalent problems of Wong and @Right let us mention the following two results:

Theorem (Tanaka, Pervine, Levine). A connected Darboux bijection $$f:X\to Y$$ from a Hausdorff topological space $$X$$ to a semilocally connected Hausdorff space $$Y$$ is continuous.

and

Theorem (Banakhs). A Darboux injective map $$f:X\to Y$$ between connected metrizable spaces is continuous if one of the following conditions is satisfied:

1) $$Y$$ is a 1-manifold and $$X$$ is compact;

2) $$Y$$ is a 2-manifold and $$X$$ is a closed $$n$$-manifold of dimension $$n\ge 2$$;

3) $$Y$$ is a 3-manifold and $$X$$ is a closed $$n$$-manifold of dimension $$n\ge 3$$ with finite homology group $$H_1(X)$$.

• You did some intellectual work and have shown an interest and gave an answer that highlights some aspects of this problem of mine, so it is fair that you earn +100. Sep 12, 2018 at 12:17
• If you wish, we can also chat about this problem in order to try to resolve it finally. Or I can send you an e-mail. Sep 12, 2018 at 12:21
• @Right In fact, there are substantial problem that do not allow to move from dimension 3 to dimension 4 and also a bit smaller (but nonetheless) problems that do not allow to apply the methods working for closed manifolds to the case of $\mathbb R^n$. I have no idea how to struggle them (especially the transition from dimension 3 to dimension 4). Sep 12, 2018 at 17:17
• @Right I you have any ideas, please write me to the e-mail t.o.banakh@gmail.com Sep 12, 2018 at 17:18