While thinking about this question I was asking myself if a path-connected bijection $f\colon \Bbb{R}^n \to \Bbb{R}^n$ has to be continuous for $n>1$?

If we drop the requirement that $f$ is bijective, then it is not true as in the connected case.

I was wondering if this question is maybe easier? I have no intuition if it is true or not. On one hand I think there are too many connected sets, there will be some ugly counterexample, on the other hand the reals are "nice".

By a path-connected function I mean a function between topological spaces whose image of a path-connected set is path-connected.


Here's how you can get a counterexample for $n>2$ assuming the continuum hypothesis. Note that if $A\subseteq\mathbb{R}^n$ is path-connected and $x,y\in A$, there is a path-connected closed subset $B\subseteq A$ containing $x$ and $y$ (namely, the image of a path between them). So for $f$ to be path-connected, it suffices for $f(A)$ to be path-connected whenever $A$ is a closed path-connected subset of $\mathbb{R}^n$.

This is very useful, because there are only $\mathfrak{c}$ different closed subsets of $\mathbb{R}^n$. First, partition $\mathbb{R}^n$ into $\mathfrak{c}$ sets $(S_\alpha)_{\alpha<\mathfrak{c}}$, each of which intersects every closed path-connected subset with more than one point at $\mathfrak{c}$ points (you can do this by an induction of length $\mathfrak{c}$, where at the $\alpha$th stage you add a point of the $\gamma$th closed path-connected set to $S_\beta$ for all $\beta,\gamma<\alpha$). Also, enumerate all quadruples $(A,x,y,u)$ where $A$ is a path-connected closed subset of $\mathbb{R}^n$, $x,y\in A$, and $u\in\mathbb{R}^n$ with order-type $\mathfrak{c}$.

Now define $f$ by an induction of length $\mathfrak{c}$. At the $\alpha$th stage of the induction, we want to define $f$ on $\{x,y\}\cup (A\cap S_\alpha)$ such that $f(A\cap S_\alpha)$ contains a smooth path from $f(x)$ to $f(y)$, where $(A,x,y,u)$ is the $\alpha$th quadruple in our enumeration. To do this, we need the following lemma (here is where we use CH and the fact that $n>2$):

Lemma: Let $n>2$, let $a,b\in \mathbb{R}^n$ and let $g_1,g_2,\ldots:[0,1]\to\mathbb{R}^n$ be a countable collection of smooth paths and $c_1,c_2,\ldots\in\mathbb{R}^n$ be a countable collection of points distinct from $a$ and $b$. Then there is a smooth path from $a$ to $b$ that does not intersect any of the $g_i$ except possibly at $a$ or $b$ and does not pass through any of the $c_i$.

To apply this lemma, let $a=f(x)$ and $b=f(y)$ (we may have already defined these values of $f$; if not, define them arbitrarily to be some points not in the image of $f$), let the $g_i$ be the smooth paths which we have already defined to be in the image of $f$ (there is one such path from each previous stage, so by CH, there are only countably many such paths), and let the $c_i$ be the other various points we have already defined to be in the image of $f$ (there are finitely many such points from each previous stage, so countably many in total by CH). This gives us a smooth path from $a$ to $b$ which we can make the image of $f(A\cap S_\alpha)$ (except for the countably many points of $A\cap S_\alpha$ where we may have already defined $f$). In addition, let us define $f$ at a couple more points to make sure $u$ is in both the domain and image of $f$.

At the end of this induction, we will have a path-connected bijection $f:\mathbb{R}^n\to\mathbb{R}^n$. It is easy to see that we can arrange for $f$ to be discontinuous (at the $\alpha$th stage, we are free to choose any bijection at all between $A\cap S_\alpha$ and our path).

It remains to prove the Lemma. To do so, note that the space $P$ of all smooth paths from $a$ to $b$ is a complete metric space in the natural $C^\infty$ topology. For each $i$, let $U_i$ be the set of smooth paths that do not pass through $c_i$. For each $i$, let $V_i$ be the set of smooth paths that do not intersect $g_i$ unless $a$ and/or $b$ is in the image of $g_i$, in which case they intersect only at $a$ and/or $b$ and have different derivatives there. Then by standard transversality theory (using the fact that $n>2$), each $U_i$ and $V_i$ is an open dense subset of $P$. By the Baire category theorem, there is an element of $P$ that is every $C_i$ and $D_i$, and this is our desired path.

As a final note, I expect that CH is not actually necessary here. However, I don't quite see how to prove a version of the Lemma that works for a collection of $<\mathfrak{c}$ paths, rather than just for a countable collection of paths. This question on MO seems related to this issue.

  • $\begingroup$ Sorry, I am a bit lost at the moment. What does intersects at … points at $\mathfrak{c}$ points mean? For any closed path-connected $A$ there is a $\alpha$ such that $\vert A\cap S_{\alpha} \vert=\mathfrak{c}$? Could you explain in more detail why this exists? I also do not see what the $u$ is used for and how the quadruples are related to the indices $\alpha$. $\endgroup$ – user60589 Jan 14 '16 at 15:09
  • $\begingroup$ I mean that for all $\alpha$ and all $A$, $|A\cap S_\alpha|=\mathfrak{c}$. To do this, enumerate the $A$s as $(A_\alpha)_{\alpha<\mathfrak{c}}$, and run an induction of length $\mathfrak{c}$, where at the $\alpha$th stage, for each pair of ordinals $\beta,\gamma<\alpha$, you choose a point of $A_\beta$ and declare it to be an element of $S_\gamma$. You can do this because at each stage, you have only picked $<\mathfrak{c}$ points from each $A_\beta$, and $|A_\beta|=\mathfrak{c}$. At the end, you get that $|S_\gamma\cap A_\beta|=|\mathfrak{c}\setminus\max(\gamma,\beta)|=|\mathfrak{c}|$. $\endgroup$ – Eric Wofsey Jan 14 '16 at 19:04
  • $\begingroup$ The point of $u$ is just to ensure that $f$ will end up being a bijection. If we didn't include it, we might end up having some points that we never defined to be in the domain of $f$, or never made be in the image of $f$. At the beginning of the construction, we choose an enumeration $(A_\alpha,x_\alpha,y_\alpha,u_\alpha)_{\alpha<\mathfrak{c}}$ of all such quadruples, and in the $\alpha$th stage of the induction we are using the quadruple $(A,x,y,u)=(A_\alpha, x_\alpha,y_\alpha,u_\alpha)$. $\endgroup$ – Eric Wofsey Jan 14 '16 at 19:06
  • $\begingroup$ Sorry, needed some time to get comfortable with the solution. But I like the answer. So you want that $u_{\alpha}$ is in the image of $f$ at the $\alpha$th stage, right? I am not sure why we do not need CH for the induction. Don't we need the fact that $\alpha < \mathfrak{c}$ is countable? $\endgroup$ – user60589 Apr 11 '16 at 9:56
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    $\begingroup$ I'll never understand why answers like these don't get more up votes. Excellent answer, very impressive. $\endgroup$ – M10687 Aug 12 '16 at 7:48

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