# Does there exist 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 $$f^{-1}$$?

Various remarks

1. if we remove the bijection requirement 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 is 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, 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 $$\{\mathbb{N}\} \cup \{ \{k\} | k \in \mathbb{Z} \setminus \mathbb{N} \}$$ 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.