Given a bijection $f\colon \mathbb{Z}^2 \to \mathbb{Z}^2$, does there always exist a homeomorphism $h\colon\mathbb{R}^2\to\mathbb{R}^2$ that agrees with $f$ on $\mathbb{Z}^2$?

I don't see any immediate obstruction, but there are certain bijections that seem difficult to extend, such as $$ f(a,b) \;=\; \begin{cases} (a,b) & \text{if }b\geq 0, \\[3pt] (-a,b) & \text{if }b<0.\end{cases} $$ Note that it's possible to map any finite set of points $p_1,\ldots,p_n$ to any other finite set $q_1,\ldots,q_n$ by a homeomorphism of $\mathbb{R}^2$, so it's quite important here that $\mathbb{Z}^2$ is infinite.

Of course, one can more generally ask whether any bijection between discrete subsets of $\mathbb{R}^2$ extends to a homeomorphism.

Jim Belk
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    Interesting question. But for discrete subsets isn't the answer "no" because we can take $\{\pm 1/n : n \in \mathbb{Z}^+\} \times \{0,1\}$ and consider the bijection that is the identity on the left half-plane and switches each point $(1/n,0)$ with $(1/n,1)$ in the right half-plane? – Trevor Wilson Jun 10 '15 at 21:58
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    What about somehow pasting those locally defined homeomorphisms on the finite sets of points? – Berci Jun 10 '15 at 22:00
  • @TrevorWilson By a "discrete subset", I mean a subset that has no limit points in $\mathbb{R}^2$, as opposed to one that has no limit points in the subset itself. (I realize this terminology is ambiguous.) As you point out, the answer is "no" using the latter definition. – Jim Belk Jun 10 '15 at 22:03
  • I see. Then it seems reasonable. – Trevor Wilson Jun 10 '15 at 22:05
  • weakly related to: http://math.stackexchange.com/q/1168351/442 – GEdgar Jun 10 '15 at 22:57
  • The bijection you present seems unrealizable, I would be shocked if the answer was yes in this case. I vote no. – Olivier Bégassat Jun 10 '15 at 23:39
  • Actually, I was probably wrong, I think I can see a construction. – Olivier Bégassat Jun 10 '15 at 23:51
  • This is an example of an interpolation/extrapolation problem. One of my favorites is the related Pick-Nevanlinna interpolation problem... – DVD Jun 16 '15 at 22:48

3 Answers3



We can consider a sequence of disks $D_1,D_2,\ldots$ around $(\sqrt 2,\pi)$ such that the $n$th disk has precisely $n$ lattice points in its interior (and none on its boundary). Let $z_n$ be the lattice point in $D_n\setminus D_{n-1}$.

Construct accordingly a sequence $C_1, C_2, \ldots$ of simply connected compact sets with smooth boundary such that $C_{n+1}\setminus C_n$ is an annulus: We let $C_1$ a a small closed disk around $f(z_1)$. Given $C_n$, we can connect $f(z_{n+1})$ with $C_n$ via a straight line segment $\ell_n$ that does not pass through any other lattice point (and ends at its first intersection with $\partial C_n$. Then $C_n\cup \ell_n$ is simply connected, hence the complement is (via some map $\phi_n$) biholomorphic to $\{\,z\in\mathbb C:|z|>1\,\}$. Since almost all lattice points are "close" to $f(\infty)=\infty$, the distance $r_n$ between $f(\mathbb Z^2\setminus C_n)$ and $S^1$ is strictly positive. Let $C_{n+1}=C_n\cup \phi_n^{-1}(\{\,z\in\mathbb C:|z|\le 1+\tfrac {r_n}2\,\})$.

Now to construct the desired homeomorphism $h$:

  • Since $C_1$ and $D_1$ are disks, we can readily define $h\colon D_1\stackrel \approx \to C_1$
  • Assume we have defined $h\colon D_n\stackrel\approx \to C_n$. To extend this, we need only find a homeomorphism between the closed annuli $D_{n+1}\setminus D^\circ_n$ and $\phi_n(C_{n+1}\setminus C^\circ_n)=\{\,z\in\mathbb C:1\le |z|\le 1+\frac{r_n}2\,\}$ that agrees with what we already have as homeomorphis between the inner boundaries. Viewing both sets as $[0,1]\times S^1$, we use the identity on the first factor and the given homeomorphism between the inner boundaries on the second factor.

The resulting map on $\mathbb R^2=\bigcup D_n$ is the desired homeomorphism if we can ensure that $\mathbb R^2=\bigcup C_n$. Right now I am not sure if the construction above warrants this auttomatically. I suppose that one should be more "greedy", that is not only add $\ell n$ to $C_n$ (and then dilate it) but add something more while still maintaining simple connectivity at that step ...

Based on Henning Makholm's and Jim Belk's excellent comments, here's a complete rewrite of the above:

Lemma. Let $Q\subset \mathbb R^2$ be countable, closed, and discrete. Let $q_1,q_2,\ldots$ be an enumeration of $Q$. Then there exists a homeomorphism $h\colon\mathbb R^2\to \mathbb R^2$ with $f(q_n)=(n,0)$ for all $n\in\mathbb N$.

Proof. Pick a point $a$ that is on none of the countably many lines through two points of $Q$. Then for each $n$, the line segment $\ell_n$ from $a$ to $q_n$ is a compact set disjoint from $Q\setminus\{q_n\}$.

For $n\in\mathbb N$, $r>0$ let $C(n,r)$ be the convex hull of $\overline{B(a;r)}\cup\overline{ B(q_n;r)}$; this is the union of two disks and a rectangle, its boundary consists of two arcs and two line segments, it is star shaped around $a$, and every ray originating at $a$ intersects its boundary in exactly one point.

For $m\in\mathbb N$ let $$ r(n,m) = \min\{\,|q_k-y|:k\ge m,k\ne n, y\in \ell_n\,\},$$ which exists and is positive because $\ell_n$ is compact and $Q$ (minus a finte set) is closed. For $0<r<r(n,m)$ we have $C(n,r)\cap Q\subseteq\{q_1,\ldots,q_{m-1}\}\cup\{q_n\}$. Clearly $m'>m$ implies $r(n,m')\ge r(n,m)$. Also $r(n,m)\to\infty$ as $m\to\infty$ because $r(n,m)>r$ for all $m$ implies that $C(n,r)$contans infinitely many elements of $Q$, contradicting discreteness.

This allows us to pick for each $n$ a sequence $\{\rho_{n,m}\}_{m=n}^\infty$ that is strictly increasing and diverges $\to\infty$ as $m\to\infty$ and such that $\rho_{n,m}<r(n,m)$ holds for all $m\ge n$. Now let $$ C_n=\bigcup_{k=1}^n C(k,\rho_{k,n}).$$ Then $C_n$ is starshaped around $a$, compact, has a useer-friendly boundary consisting of finitely many arcs and line segments, and each ray originating in $a$ intersects the boundary in exactly one point. Moreover, $C_n\subset C_{n+1}^\circ$ and $\bigcup_{n=1}^\infty C_n=\mathbb R^2$. We also have $C_n\cap Q=\{q_1,\ldots, q_n\}$ and $\partial C_n\cap Q=\emptyset$. By linear interpolation between the boundaries we obtain a homeomorphism $h_0\colon \mathbb R^2\to\mathbb R^2$ that maps $a\mapsto 0$ and $\partial C_n\to nS^1$. Note that $h_0(q_n)$ is between $(n-1)S^1$ and $nS^1$. It is a simple task to deform each such annulus (and the central disk) in such a way that the boundary remains untouched and $f(q_n)$ moves to $(n-\tfrac12,0)$. With a final translation by $\frac12$ to the right, we obtain our desired homoeomorphism. $_\square$

Now the original problem is solved by applying the lemma to an enumeration $q_1,q_2,\ldots $ of $\mathbb Z\times Z$ and also to the enumeration $f(q_1),f(q_2),\ldots$.

Hagen von Eitzen
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    Interesting! I'm not sure how easy it is to fix up the end of the argument, but at the very least your construction of the disks $D_n$ is sufficient to prove that there's a homeomorphism of $\mathbb{R}^2$ taking $\mathbb{Z}^2$ to $\mathbb{N}\times\{0\}$, which simplifies the problem considerably. – Jim Belk Jun 10 '15 at 23:28
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    I think you can maintain the invariant that $D_n$ must always be a star-domain centered on $f(z_1)$, _and_ that as soon as all points within a certain distance from $f(z_1)$ have been mapped, the next $D_n$ must contain a ball around $f(z_1)$ containing all those points, simply by "filling in the valleys". – hmakholm left over Monica Jun 10 '15 at 23:51
  • @HenningMakholm That sounds like it ought to work. – Jim Belk Jun 11 '15 at 03:54
  • Hmm, actually the star-domain should be centered on something like $(\sqrt 2,\pi)$ rather than $f(z_1)$, such that the line of sight to every point we might need to reach is clear. – hmakholm left over Monica Jun 11 '15 at 11:08
  • Wow, this is very neat! +1 – Olivier Bégassat Jun 11 '15 at 18:07

Here's an idea on how to construct a solution for the bijection you describe. It appears feasible by a diffeomorphism of the plane.

First we solve the following problem : extend the bijection \begin{array}{RCL} \phi:\Bbb Z\times\lbrace-1,0\rbrace & \longrightarrow & \Bbb Z\times\lbrace-1,0\rbrace\\ (n,\epsilon) & \longmapsto & \begin{cases} (n,\epsilon) & \text{if }\epsilon=0\\ (-n,\epsilon) & \text{if }\epsilon=-1\end{cases} \end{array} to a diffeomorphism of the plane using a vector field $X$ defined as follows: it is supported in the open set $$\bigcup_{n\geq 1}~H_n+D_\frac13$$ where $H_n$ is the bottom half circle with diameter the segment $[(-n,-1),(+n,-1)]$, and $D_\frac13$ is the open disk centered at the origin with radius $1/3$. For every $n\geq 1$, the vector field inside $H_n+D_\frac13$ is chosen so that it switches $(-n,-1)$ with $(+n,-1)$ and vice versa after flowing for one second, and, importantly, is chosen so that it is zero on a neighborhood of every point $(n,k)$ with $k\leq -2$. Then $\Phi=\mathrm{Fl}^X_1$ extends $\phi$ to the whole plane.

Now consider the map $\ell:\Bbb R \to\Bbb R^2,(x,y)\mapsto(x,y+1)$. Then the infinite composition

$$\cdots\circ(\ell^{-n} \circ\Phi\circ\ell^n)\circ\cdots\circ(\ell^{-1}\circ\Phi\circ\ell)\circ\Phi$$

EDIT. To make this work one should instead consider the infinite composite in the opposite direction $$\Phi\circ(\ell^{-1}\circ\Phi\circ\ell)\circ\cdots\circ(\ell^{-n} \circ\Phi\circ\ell^n)\circ\cdots$$ This will work and provide a (well-defined this time?!) continuous bijection of $\Bbb R^2$ onto itself, hence a homeomorphism by invariance of domain.

makes sense, and defines a smooth diffeomorphism of the plane that realizes the bijection you describe. The reason this makes sense is that for every $n\geq 1$ the region $\lbrace(x,y)\in\Bbb R^2\mid y\geq-\frac12\rbrace\cup D((0,-1),n+\frac12)$ it stable, and there the infinite composition is actually a finite composite of diffeomorphisms, and is smooth.

I think something similar is possible for any bijection of $\Bbb Z^2$, to get a diffeomorphism that extends it defined as an infinite composition of time one flows of vector fields whose supports tend to infinity, i.e. lie outside of bigger and bigger balls, and hence, after a while, will not do anything on a fixed ball, I might come back to this tomorrow.

I think you could do it like so : first, write $V_\epsilon(A)$ for the $\epsilon$ neighborhood of a set $A\subset\Bbb R^2$. Now enumerate the elements of $\Bbb Z^2$ in a spiral fashion by a bijection $\varphi:\Bbb N\to\Bbb Z\times\Bbb Z$. Set $m_n=\min\lbrace|\phi(n)|,|f(\phi(n))|\rbrace$ and $M_n=\max\lbrace|\phi(n)|,|f(\phi(n))|\rbrace$. For every $n$, construct a compactly supported vector field $X_n$ so that

  1. $X_n$ is compactly supported with $$\mathrm{supp}(X_n)\subset\lbrace x\in\Bbb R^2\mid m_n-1\leq|x|\leq M_n+1\rbrace\setminus V_\frac14\left(\Bbb Z^2\setminus\lbrace\phi(n),f(\phi(n))\rbrace\right)$$
  2. The time one flow $\Phi_n=\mathrm{Fl}^{X_n}_1$ of $X_n$ sends $\phi(n)$ to $f(\phi(n))$

Then the infinite composition $$\cdots\circ\Phi_n\circ\cdots\circ\Phi_{1}\circ\Phi_{0}$$ makes sense and is smooth because on every compact disk it actually is a finite composite.

Olivier Bégassat
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  • There is a slight problem with the final composite (it may not be finite on every compact set), but it can be fixed. You just need a collection of **disjoint**, banana-shaped open sets $B_n$ that contain $\phi(n)$ and $f(\phi(n))$ for all $n$ and no other lattice points, and have the $X_n$ have their supports in there. Actually, one could forego the infinite composition altogether. – Olivier Bégassat Jun 11 '15 at 00:56
  • Actually, one could forego the infinite composition altogether by just integrating $\sum_{n=1}^\infty X_n$. – Olivier Bégassat Jun 11 '15 at 01:01
  • This is very nice! I'm not entirely convinced by the infinite composite, though, and I don't see why it works to add the vector fields together. Each point is only in the support of finitely many $\Phi_i$, but it seems possible that, by applying the $\Phi_i$'s in sequence, a point could "wander" during the process so that it is moved by infinitely many different $\Phi_i$, and never arrive at a final destination. (Though it does work for the specific $f$ I gave, since it's possible to find disjoint banana-shaped regions.) – Jim Belk Jun 11 '15 at 03:44
  • @JimBelk You are right, and in any case the process I propose doens't do what is expected, it's a little more complicated than I originally thought: the banana shaped regions should extend along the complete positive and negative orbit of a point. I think there might be a problem if $f$ isn't nice enough, in particular, it should work if all of $f$'s orbits are finite. – Olivier Bégassat Jun 11 '15 at 03:52

Here is another solution, somewhat in the same spirit as Olivier Bégassat's answer. In fact, we will prove a stronger result : any bijection $f : \mathbb{Z}^2 \to \mathbb{Z}^2$ comes from a smooth isotopy $h : \mathbb{R}^2 \times [0,1] \to \mathbb{R}^2$ such that $h(x,0) = x$ and $h(-,1) = f : \mathbb{Z}^2 \to \mathbb{Z}^2$.

As in Hagen von Eitzen's answer, let us consider first the point $O = (\sqrt{2}, \pi)$. It is not in the lattice $\mathbb{Z}^2$ as $\sqrt{2}$ and $\pi$ are irrational numbers. What is more is that any circle centred at $O$ and any straight line through $C$ contains at most one element of the lattice, since $\sqrt{2}$ is algebraic whereas $\pi$ is transcendental.

Fix some bijection $l : \mathbb{N} \to \mathbb{Z}^2$ ; This induces a bijection $g : \mathbb{N} \to \mathbb{N} : n \mapsto (l^{-1} \circ f \circ l)(n)$. Let $C_n$ denote the circle centred at $O$ passing through $l(n)$ ; Similarly, let $L_n$ be the straight line spanned by the segment $\overline{l(n)O}$. We define a continuous path $\beta_n : [0,1] \to \mathbb{R}^2$ as follow :

1) $\beta_n : [0, 1/2] \to C_n$ is a smooth path joining $l(n)$ to the point of intersection of $C_n$ with $L_{g(n)}$.

2) $\beta_n : [1/2, 1] \to L_{g(n)}$ is a smooth path joining the point of intersection of $C_n$ with $L_{g(n)}$ to the $f(l(n)) = l(g(n))$.

A smooth reparametrization of $\beta_n$ yields a smooth path $\gamma_n : ([0,1], \{0\}, \{1\}) \to (C_n \cup L_{g(n)}, l(n), f(l(n)))$.

By construction, at any given time $t$, none of the points $\gamma_n(t)$ coincide. Moreover, since each $\mathrm{Im}\, \gamma_n$ is compact, the set $\{ \gamma_n(t) \,| \, n \in \mathbb{N} \}$ has only isolated points : since $\mathbb{R}^2$ is a normal space, we can find reccursively a sequence of positive reals $\epsilon_n$ such that at each time $t$, none of the balls $B(\gamma_n(t), \epsilon_n)$ intersect. Remark : We are implicitly using here properties of the $C_n$'s and of the $L_n$'s.

The $\gamma_n$'s define a vector field $V$ on the set $\{ (x,t) \in \mathbb{R}^2 \times [0,1] \, | \, \exists n \in \mathbb{N} \; \mbox{ s.t. } \; x = \gamma_{n}(t) \}$. More explicitly, $V_{(\gamma_n(t),t)} = \dot{\gamma_n}(t) + \partial/\partial t$. This vector field can be smoothly extended to a vector field $X$ on $\mathbb{R}^2 \times [0,1]$ of the form $\pi^{\ast}Y + \partial/\partial t$ (where $\pi : \mathbb{R}^2 \times [0,1] \to \mathbb{R}^2$ is the canonical projection and $Y$ is a smooth non-autonomous vector field on $\mathbb{R}^2$) in such a way that it equals $\partial/\partial t$ outside the set $\{ (x,t) \in \mathbb{R}^2 \times [0,1] \, | \, \exists n \in \mathbb{N} \; \mbox{ s.t. } \; d(x,\gamma_{n}(t)) \le \epsilon_{n} \}$. We define the isotopy $h$ as the flow of the vector field $Y$.

Remark : $\mathbb{R}^2$ admits a standard symplectic form $\omega = \mathrm{d}x \wedge \mathrm{d}y$. At each point $p \in \mathbb{R}^2$, this 2-form defines an isomorphism $\Omega_p : T_p\mathbb{R}^2 \to T^{\ast}_p \mathbb{R}^2 : v \mapsto \omega_p(v, -)$. One can extend such a linear 1-form in a constant way in an $\epsilon$-neighbordhood of $p$ in order to get a closed differential 1-form in this neighbordhood. Applying Poincaré's lemma, we get in a smaller neighbordhood of $p$ a smooth function whose exterior derivative equals the given 1-form. Using an appropriate bump function, we obtain a global smooth function $f_p$ such that $(\Omega^{-1} \mathrm{d}f_p)_p = v$ with support in $B(p, \epsilon)$. In fact, $p \mapsto f_p$ can be chosen in a smooth way. As such, at any time $t$, we can define a smooth function $F_t = \sum_{n=1}^{\infty} \, f_{\gamma_{l(n)}(t)}$ smoothly varying with $t$ such that $Y_t := \Omega^{-1} \mathrm{d}F_t$ is as above. The resulting isotopy $h_t$ is a Hamiltonian one ; We conclude that $h$ can be chosen area-preserving.

Jordan Payette
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  • This is very nice, and the existence of an area-preserving homeomorphism isotopic to the identity is very interesting. But as with Olivier Bégassat's answer there seem to be a couple of issues. First, it's not clear that smooth paths $\gamma_n$ satisfying the given conditions exist, even in the case where $f$ has order two. In the case where $f$ doesn't have order two, it's clear that they don't exist, since the $\epsilon$-neighborhood of a path from $f(p)$ to $f(f(p))$ will have to intersect the $\epsilon$-neighborhood of a path from $p$ to $f(p)$. – Jim Belk Jun 15 '15 at 16:53
  • I edited my answer in order to be more careful about the neighborhoods of the paths before I read your comment. I did not think about your counter-example, so there is still some corrections to be made. However, I think what is really useful is for the neighborhoods of the paths to be disjoint at any given time, but there might be intersection at different times. – Jordan Payette Jun 15 '15 at 17:04
  • That sounds true. In fact, it should suffice to get the first half of each path to be far away from the first halves of other paths, and similarly for the second halves. You can then implement the flow in two stages, first moving each $l(n)$ to the midpoint of $\gamma_n$, and then moving that midpoint to $f(l(n))$. – Jim Belk Jun 15 '15 at 17:16
  • Indeed! I found an explicit set of paths based on one of Hagen von Eitzen's ideas. Consider the point $C = (\sqrt{2}, \pi)$ as the origin of the plane. On any straight ligne through $C$, there is at most one element of the lattice ; Similarly, on any circle centred on $C$, there is at most one element of the lattice. We can define $\gamma_n$ as a path whose first half is at constant radius and whose second half is at constant angle. We can associate to $\gamma_n$ an $\epsilon_n > 0$ such that at any given time $t$, the disks $B(\gamma_n(t) , \epsilon_n)$ are disjoints. – Jordan Payette Jun 16 '15 at 13:56
  • In any case, my original answer is incorrect at the moment. Should I modify it (and hide my 'crime'!) or would it be more appropriate to add an 'erratum' after the *remark*? – Jordan Payette Jun 16 '15 at 14:07
  • @JordanPayette You should hide your 'crime'; interested parties can always look up the history. If you view your answer as a miniature math paper, it is clear that you always want to present the most logical and correct progression - most people won't care that an earlier version had a mistake. – Mario Carneiro Jun 17 '15 at 00:55
  • @MarioCarneiro Thanks for the advice, it seems indeed the best thing to do. – Jordan Payette Jun 17 '15 at 13:59