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
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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.
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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.
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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?
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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?
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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.
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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.
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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.
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Best Answer
This is only a partial answer; but it's too long for a comment, and I believe the following facts, which come close to the question being asked and might provide a clue to answering it, are worth pointing out.
Let me show that there exists a bijection $\mathbb{R} \to \mathbb{R}^2$ such that the forward map is connected but the inverse is not.
Let me call a subset $E$ of the plane "hyperdense" when it meets every perfect set of the plane (perfect = nonempty, closed, with no isolated point). In particular, it meets every nonempty open set and every uncountable closed set (by Cantor-Bendixson). Let me reproduce the argument given here to show that every hyperdense set of the plane is connected: if not, there would be $U,V\subseteq\mathbb{R}^2$ open such that $E \subseteq U\cup V$, $E$ meets both $U$ and $V$ and does not meet $U\cap V$; but since $E$ is hyperdense, not meeting the open set $U\cap V$ is possible only if $U\cap V = \varnothing$, so $U\cup V$ is disconnected, so the closed set $\mathbb{R}^2 \setminus (U\cup V)$ must be uncountable (because the complement of a countable set in the plane is connected), so $E$ must meet it, a contradiction.
Now I construct a bijection $\mathbb{R} \to \mathbb{R}^2$ such that $f(I)$ is hyperdense for every nonempty open interval $I$. Let $\mathfrak{c} = 2^{\aleph_0}$ (seen as the smallest ordinal of that cardinality); let $(I_\beta,P_\beta)_{\beta<\mathfrak{c}}$ be an enumeration of pairs consisting of a nonempty open interval $I \subseteq \mathbb{R}$ and a perfect set $P \subseteq \mathbb{R}^2$ (recall that there are precisely continuum-many perfect sets), and let $(z_\beta)_{\beta<\mathfrak{c}}$ and $(t_\beta)_{\beta<\mathfrak{c}}$ be an enumeration of $\mathbb{R}$ and $\mathbb{R}^2$ respectively. Construct $(x_\alpha,y_\alpha) \in \mathbb{R} \times \mathbb{R}^2$ by induction on $\alpha<\mathfrak{c}$ such that $x_\alpha$ is different from all the (previously constructed) $x_{\alpha'}$ for $\alpha'<\alpha$ and $y_\alpha$ is different from all $y_{\alpha'}$ for $\alpha'<\alpha$, and additionally: (A) if $\alpha=2\beta$ is even, then choose $x_\alpha \in I_\beta$ and $y_\alpha \in P_\beta$ (these conditions can be met because $I_\beta$ and $P_\beta$ have cardinality $2^{\aleph_0}$ and there are $<2^{\aleph_0}$ previously constructed $x_{\alpha'}$ and $y_{\alpha'}$ to be avoided); (B₁) if $\alpha=4\beta+1$ then take $x_\alpha=z_\beta$ unless $z_\beta$ is already among the previous $x_{\alpha'}$ (otherwise, the choice is unconstrained), and (B₂) if $\alpha=4\beta+3$ then take $y_\alpha=t_\beta$ unless $t_\beta$ is already among the previous $y_{\alpha'}$ (otherwise, the choice is unconstrained; conditions (B₁) and (B₂) are only there to guarantee that $f$ will be defined everywhere and surjective). Clearly the choice can be made at every stage, so the sequence $(x_\alpha,y_\alpha)_{\alpha<\mathfrak{c}}$ exists. By construction, the $x_\alpha$ are distinct and by (B₁) every real $z_\beta$ is equal to some $x_\alpha$, so we can define $f(x_\alpha) = y_\alpha$, giving a function $\mathbb{R} \to \mathbb{R}^2$, which is injective because the $y_\alpha$ are distinct and surjective by (B₂). Now if $I$ is any nonempty open interval and $P$ is some perfect set, we can write $(I,P) = (I_\beta,P_\beta)$ and if $\alpha = 2\beta$ we have $f(x_\alpha) = y_\alpha \in P$ for $x_\alpha \in I$, which shows $f(I) \cap P \neq \varnothing$. So $f(I)$ is always hyperdense.
Now if $I \subseteq \mathbb{R}$ is connected, either it is empty or a singleton, in which case $f(I)$ is trivially connected, or it is an interval with nonzero length, but then it contains a nonempty open interval, so $f(I)$ is hyperdense, so it is connected as explained above. This shows that the forward map of $f$ is connected.
On the other hand, the inverse map of $f$ cannot be connected, because if it were, the inverse image of an open disk in $\mathbb{R}^2$ would be connected, hence an interval, which cannot be trivial since $f$ is bijective, so it must contain an open interval, so $f$ is continuous; but $f$ is certainly not continuous since it is bounded on no interval.
In a related vein, let me show that there exists a bijection $f\colon \mathbb{R}^2 \to \mathbb{R}^2$ such that the forward image of every PATH-connected set is connected and such that the inverse map is not connected.
The construction above easily gives a bijection $f\colon \mathbb{R}^2 \setminus\Delta \to \mathbb{R}^2 \setminus Z$, where $\Delta$ is the diagonal and $Z = (0,1)\times\{0\}$, such that $f(Q) \cap P \neq\varnothing$ for every perfect set $Q$ in $\mathbb{R}^2 \setminus\Delta$ and every perfect set $P$ in $\mathbb{R}^2 \setminus Z$, i.e., $f(Q)$ is hyperdense in $\mathbb{R}^2 \setminus Z$ for every perfect $Q$ in $\mathbb{R}^2 \setminus\Delta$.
Extend $f$ to a bijection $\mathbb{R}^2 \to \mathbb{R}^2$ by using a homeomorphism between $\Delta$ and $Z$.
Clearly, the inverse map of $f$ is not connected, since $f^{-1}(\mathbb{R}^2 \setminus Z) = \mathbb{R}^2 \setminus\Delta$ and $\mathbb{R}^2 \setminus\Delta$ is not connected whereas $\mathbb{R}^2 \setminus Z$ is.
Now consider the forward image of a path-connected set $A \subseteq \mathbb{R}^2$. If $A \subseteq \Delta$ then $f(A)$ is connected because $f$ restricts to a homeomorphism between $\Delta$ and $Z$. On the other hand, if $A \not\subseteq \Delta$, clearly $A\setminus\Delta$ is not a single point, so $A$ must contain a path between two points not in $\Delta$, and at least part of this path is not in $\Delta$, so it contains a perfect set $Q \subseteq \mathbb{R}^2 \setminus\Delta$. So $f(Q)$ is hyperdense in $\mathbb{R}^2 \setminus Z$; in particular, it is dense in $\mathbb{R}^2$, and connected (the same argument showing that hyperdense sets in $\mathbb{R}^2$ are connected still works for $\mathbb{R}^2\setminus Z$ since the complement of a countable set in the latter is connected). Now $f(A)$ contains $f(Q)$ which is dense in $\mathbb{R}^2$ connected, and anything containing a dense connected set is still connected (because if $D$ is dense connected and $D \subseteq B$, then any continuous $B\to\{0,1\}$ has a constant restriction to the dense subset $D$ so it is constant). So $f(A)$ is connected.
It is conceivable that the same kind of arguments give a positive answer to the original question, provided we can find something to play the role of the intervals $I$ in the first part and the perfect sets $Q$ in the second. But the naïve approach fails: indeed, one might ask the following question
(Then we could define a bijection $f\colon\mathbb{R}^2\to\mathbb{R}^2$ by transfinite induction as above so that $f(H_\gamma)$ is hyperdense for each $H_\gamma$. Its forward map would be connected, but it wouldn't be continuous so this would answer the "equivalent question" cited in the original post.)
However, the answer to this question must be "no", because if such $(H_\gamma)$ existed, we could also use them to construct a bijection $f\colon\mathbb{R}^2\to\mathbb{R}^2$ such that $f(H_\gamma)$ and $f^{-1}(H_\gamma)$ are both hyperdense for each $H_\gamma$, which would make both the forward and inverse maps of $f$ connected, but $f$ would not be continuous (say we include all perfect sets among the $H_\gamma$ to be sure), contradicting the result of Tanaka cited by Willie Wong's question. (Probably there exists a more direct proof that the question has a negative answer. [EDIT: Will Brian gives such a proof in the comments]) So either this approach is doomed or we must be smarter in how to use it.
Incidentally, does ZFC prove that connected subsets of the plane satisfy the continuum hypothesis? [EDIT: again, Will Brian gave a (positive) answer in the comments]