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

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

1. 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.

2. 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.

3. 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?

1. 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, 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 the 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 $$X = \mathbb{Z}$$ be 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.

• 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. Commented Jun 20, 2015 at 18:00
• @JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.) Commented Jun 22, 2015 at 8:03
• @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.) Commented Aug 21, 2015 at 2:24
• @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. Commented Aug 29, 2015 at 21:03
• 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 $S:=f^{-1}(R )$. Commented Jan 26, 2017 at 17:38