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

connectedif 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.

**Answer**

**Attribution***Source : Link , Question Author : Willie Wong , Answer Author : Community*