TITLE: Does there exist a bijection of $\mathbb{R}^n$ to itself such that the forward map is connected but the inverse is not?
QUESTION [146 upvotes]: 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 originally asked this on Math.SE a year and a half ago; there were some discussion in the comments as well as a few answer attempts that were unfortunately flawed. Remarks 3 and 5 below capture the essences of many of those attempts. 
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 https://math.stackexchange.com/q/949168/1543 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? 

In fact, it is a Theorem of Tanaka's (see my answer here) that if $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection such that both $f$ and $f^{-1}$ are connected, then $f$ is a homeomorphism. So an equivalent formulation of the question is
Equivalent Question: Does there exist a bijection $f:\mathbb{R}^n\to\mathbb{R}^n$ such that $f$ is connected but discontinuous?
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.

REPLY [4 votes]: I have a partial answer which should at least help to restrict our attention to a certain case. 
As noted, if either $f$ of $f^{-1}$ is continuous, then we are done by Invariance of Domain. So suppose $f$ is discontinuous at $x\in \mathbb{R}^n$. Then there exists $\epsilon_0>0$ such that for all $\delta>0$, there exists $y_\delta\in\mathbb{R}^n$ with $|x-y_\delta|<\delta$ and $|f(x)-f(y_\delta)|\geq \epsilon_0$. Note that
$$f(B(x,\delta))=\Big[f(B(x,\delta))\cap (B(f(x),\epsilon_0))\Big]\coprod\Big[f(B(x,\delta))\cap (B(f(x),\epsilon_0))^c\Big]=:U_\delta\coprod V_\delta$$
We know that each of these disjoint sets is non-empty since $f(x)\in U_\delta$ and $y_\delta\in V_\delta$. Furthermore, $U_\delta\subseteq f(B(x,\delta))$ is clearly open in the subspace topology. If only $V_\delta$ were open in $f(B(x,\delta))$ then $(U_\delta,V_\delta)$ would be a separation of $f(B(x,\delta))$, hence $f$ would not be connected!
But the problem happens at the boundary where $|x-y|<\delta$ and $|f(x)-f(y)|$ is precisely equal to $\epsilon_0$. If there are no such $y$, then $V_\delta=f(B(x,\delta))\cap \mathrm{Int}\big(B(f(x),\epsilon_0)\big)$ which is open in $f(B(x,\delta))$ and we are done. 
Otherwise, $\forall \delta>0, \exists y_\delta\in \mathbb{R}^n$ such that $|x-y_\delta|<\delta$ and $|f(x)-f(y_\delta)|=\epsilon_0$. Now, define 
$$E:=\{\epsilon>0:\hspace{.3cm}\forall\delta>0, \exists y_\delta\in\mathbb{R}^n\text{ such that }|x-y_\delta|<\delta\text{ and }|f(x)-f(y_\delta)|=\epsilon \}$$
Note that $E\neq\varnothing$ since $\epsilon_0\in E$. Moreover, $E$ is bounded below by zero so (by completeness of $\mathbb{R}^n$),  $\hat{\epsilon}:=\inf(E)$ exists and $0\leq \hat{\epsilon}<\infty$. 
Case 1: Suppose $\hat{\epsilon}>0$. 
Then $\frac{\hat{\epsilon}}{2}<\hat{\epsilon}$, so $\frac{\hat{\epsilon}}{2}\notin E$. Thus, $\exists \hat{\delta}>0$ such that  $|x-y|<\hat{\delta}\Longrightarrow |f(x)-f(y)|\neq \frac{\hat{\epsilon}}{2}$. 
But by the discontinuity of $f$, we know that $\exists y_{\hat{\delta}}\in\mathbb{R}^n$ such that $|x-y_{\hat{\delta}}|<\hat{\delta}$ and $|f(x)-f(y_{\hat{\delta}})|\geq  \epsilon_0\geq \hat{\epsilon}>\frac{\hat{\epsilon}}{2}$. 
Then we have
\begin{align*}f(B(x,\hat{\delta}))&=\Big[f(B(x,\hat{\delta}))\cap B(f(x),\frac{\hat{\epsilon}}{2})\Big]\coprod \Big[f(B(x,\hat{\delta}))\cap B(f(x),\frac{\hat{\epsilon}}{2})^c\Big]\\&=\Big[f(B(x,\hat{\delta}))\cap B(f(x),\frac{\hat{\epsilon}}{2})\Big]\coprod\Big[f(B(x,\hat{\delta}))\cap \big\{y\in \mathbb{R}^n: |f(x)-y|>\frac{\hat{\epsilon}}{2}\big\}\Big]\end{align*}
Thus, $f(B(x,\hat{\delta})$ is disconnected. 
Case 2: Suppose $\hat{\epsilon}=0$. 
This is where the trouble lies. If $f$ is not assumed to be bijective, then this can actually happen. The classic example of  $f(x)=\begin{cases}\sin(\frac{1}{x}),&x\neq 0\\0,&x=0\end{cases}\hspace{.3cm}$ satisfies $\hat{\epsilon}=0$. For any $0<\epsilon< 1$, any neighborhood of the origin contains a point $x$ for which $|f(x)|=\epsilon$ (in fact infinitely many). 
Therefore, if there exists a bijective discontinuous connected function, it must be of this form. That is, the $\epsilon$ which witnesses the discontinuity of $f$ can be taken arbitrarily close to zero. 
I hope this helps the search for a proof/counter-example!