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

  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 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?

  4. 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?

  5. 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.

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    $\begingroup$ I noticed some subtleties that might help prove/disprove the conjecture. If you can prove that the image of a disconnected set in $X$ must be disconnected in $Y$, then you would be done. But attempting to do this, I run into a problem: I can get separate connected sets in $Y$, but I do not actually know if their union is disconnected. I have attempted to get around this by considering their closures and using the fact that $\mathbb{R}^n$ is normal to get the separation, but then I can't ensure that the intersection of their closures is empty (and if it isn't empty, what goes wrong?) $\endgroup$ Apr 21, 2016 at 2:39
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    $\begingroup$ @Farewell: the question asks for $f$ defined on exactly $\mathbb{R}^n$, not some fancy subset of it. $\endgroup$ Jun 15, 2016 at 4:09
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    $\begingroup$ You could ask the same question more generally for a bijection $\mathbb{R}^m\to\mathbb{R}^n$: are there some pairs $(m,n)$ other than $(1,1)$ for which you know the answer? (I expect that such a bijection can be constructed by transfinite induction for $m=1$ and $n>1$, but I didn't give it too much thought.) $\endgroup$
    – Gro-Tsen
    Jan 26, 2017 at 10:11
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    $\begingroup$ By the results of this paper, the "equivalent question" is itself equivalent to asking for existence of a bijection which preserves connected subsets but not compact subsets. $\endgroup$
    – Arrow
    Jan 15, 2021 at 19:35
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    $\begingroup$ Indeed someone has asked the path-connected version of this question on MSE already math.stackexchange.com/questions/1571091/… $\endgroup$
    – D.R.
    Aug 22, 2023 at 4:31

4 Answers 4


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

Question: Does there exists $2^{\aleph_0}$ subsets $(H_\gamma)$ of the plane $\mathbb{R}^2$, each of cardinality $2^{\aleph_0}$, such that every connected $A \subseteq \mathbb{R}^2$ that is not a singleton contains one of the $H_\gamma$?

(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]

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    $\begingroup$ In fact, Tanaka's theorem which I alluded to in my question does not require that the dimensions be equal: in fact he worked at the level of biconnected mappings between topological spaces (with some mild local connectivity assumption). So a corollary of Tanaka's theorem + Invariance of domain is that there does not exist biconnected bijections from $\mathbb{R}^n \to \mathbb{R}^m$ when $n\neq m$. So in this case the demonstration boils down to the existence of a forward-connected bijection, which you showed. $\endgroup$ Jan 26, 2017 at 14:28
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    $\begingroup$ Very interesting! Let me point out that the answer to your highlighted question at the end is no. To see this, suppose otherwise. Using transfinite recursion, you can construct a Bernstein set whose complement contains a point of every $H_\gamma$. (And, as you point out, Bernstein sets are connected.) As for your second question (Do nontrivial connected subsets of the plane have cardinality c?) the answer is yes. If $X$ is a set of smaller cardinality, pick two distinct points in $X$ and observe that one of some family of (continuum many) disjoint parallel lines between them separates $X$. $\endgroup$
    – Will Brian
    Jan 26, 2017 at 14:29
  • $\begingroup$ @WillBrian: Ah, thanks for the term "Bernstein set", I probably read this somewhere already, but I couldn't remember it. I had also concluded that my question must have a negative answer, but your explanation is clearer — I guess this means this approach cannot succeed for the original question. $\endgroup$
    – Gro-Tsen
    Jan 26, 2017 at 14:39

This is not an answer to the original question of Willie Wong but a partial answer to the following related

Problem. Recognize pairs of topological spaces $X,Y$ for which every Darboux injection $f\colon X\to Y$ is continuous.

A function $f\colon X\to Y$ between topological spaces is called Darboux if for any connected subspace $C\subset X$ its image $f(C)$ is connected. An injection is any injective function.

The following theorem proved in this paper gives a partial answer to the above general problem:

Theorem. A Darboux injective function $f:X\to Y$ between connected metrizable spaces is continuous if one of the following conditions is satisfied:

  1. $Y$ is a 1-manifold and $X$ is compact;

  2. $Y$ is a 2-manifold and $X$ is a closed $n$-manifold for some $n\ge 2$;

  3. $Y$ is a 3-manifold and $X$ is a closed $n$-manifold of dimension $n\ge 3$ with finite first homology group $H_1(X)$.

Corollary. For $n\le 3$ any Darboux bijection $f:S^n\to S^n$ of the $n$-dimensional sphere is a homeomorphism.

The proof of the first statement in Theorem 1 is more-or-less elementary, but 2 and 3 use heavy machinery of Algebraic Topology (Alexander Duality between homologies and Cech cohomologies, long exact sequences of Cech cohomology groups and Mayer-Vietoris exact sequence for singular homologies). I do not know if the Theorem can be generalized to higher dimensions $n>3$.

So, for example the following problem seem to be open.

Problem 1. Is any Darboux bijection of the 4-dimensional sphere continuous?

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    $\begingroup$ I look forward to seeing the proof! $\endgroup$ Sep 4, 2018 at 14:44
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    $\begingroup$ @WillieWong The update of the preprint with the proof has already appeared on arxiv: arxiv.org/pdf/1809.00401.pdf $\endgroup$ Sep 6, 2018 at 4:57
  • $\begingroup$ Thanks for letting me know, I didn't see that you updated your answer. My topology is weak enough that your proof will take me a long time to digest. $\endgroup$ Sep 6, 2018 at 16:03
  • $\begingroup$ @WillieWong For me this was also not easy -- Algebraic Topology is not my area of expertise (but nonetheless). $\endgroup$ Sep 6, 2018 at 16:38

This is a partial answer, that significantly narrows possible counterexamples.

Want to show: If $D\subset X$ is a disconnected set, then it its image, $f(D)\subset Y$ is a disconnected set.

Attempted proof: Let $\mathfrak{U}$ denote the set of connected components of $D$, then for each $U\in\mathfrak{U}$, we have, by $f$ connected, that $f(U)$ is connected. Let $f(U),f(V)$ be images of distinct connected components. Then by $f$ bijective, we have that $f(U)\cap f(V)=\emptyset$. If $\overline{f(U)}\cap\overline{f(V)}=\emptyset$, then by $\mathbb{R}^n$ normal, we have a separation of $\overline{f(U)}$ and $\overline{f(V)}$ by disjoint open sets. Those same sets separate $f(U)$ from $f(V)$, hence $f(U)\cup f(V)$ is disconnected.

The problem: What happens if $\overline{f(U)}\cap\overline{f(V)}\neq\emptyset$?

Thoughts: My suspicion is that in order for $f$ to be a connected map, it would have to be the case that the pre-image of that intersection must lie in the boundary of both $U$ and $V$, but I haven't quite been able to show that. I thought about picking an arbitrary point in the intersection and finding sequences in $f(U)$ and $f(V)$ that converge to it and then using the pre-images of those sequences, which live in $U$ and $V$ respectively to show that the pre-image of that point was indeed in the boundary of both sets, but we don't know that $f^{-1}$ preserves convergence. What happens in $Y$ if it doesn't? If I knew that the pre-image of the sequences were bounded I could perhaps use compactness to get a separation of the tail from the point, and image those to get a disconnection between the tail and the point it converges to (a contradiction), but I don't know that the pre-images of the sequences are even bounded.

  • $\begingroup$ Remark: in the case of $\mathbb{R}$, your "thoughts" is basically solved by appealing to the order structure; there are some structural similarities of your proof to the statement that connected bijections are monotone on $\mathbb{R}$. $\endgroup$ Apr 22, 2016 at 14:10
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    $\begingroup$ Remark 2: I fixed a small typo in your thoughts. And yes, your thoughts is also tied to my remarks 3 and 4 in the question. The preservation of limits, for example, would mean that $f^{-1}$ is continuous. And you can see from Tanaka's proof that your main difficulty is basically equivalent to the original question. // Still, thanks for thinking about it! $\endgroup$ Apr 22, 2016 at 14:17

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!


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