Is there an injective function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ that maps circles to $n$-gons?

Question

Fix $n\geq 3$. Does there exist an injective function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that the image of any circle in $\mathbb{R}^2$ under the map $f$ is a (simple) $n$-gon in $\mathbb{R}^2$?

A friend of mine showed me this problem a few days ago without an answer, and we have been trying to answer it for the better part of 3 days with no luck. We don't really have an intuition about whether or not a function like this exists, but we have managed to prove some partial results.

If we do not require that $f$ is an injection then the answer is positive.

Let $g:\mathbb{R}^2\rightarrow\mathbb{R}$ be the projection map $g(x,y)=x$ and let $P$ be any $n$-gon in $\mathbb{R}^2$. Since $|P|=|\mathbb{R}|=|\mathbb{R}/\mathbb{Q}|$, we can choose a bijection $\varphi:\mathbb{R}/\mathbb{Q}\rightarrow P$, and define $h:\mathbb{R}\rightarrow P$ by $h(x)=\varphi(x+\mathbb{Q})$. If we define $f=h\circ g$, then $f(C)=P$ for every circle $C\subseteq\mathbb{R}^2$.

If we require that $f$ is continuous and injective then the answer is negative.

Assume $f$ exists for a contradiction. Let $C$ be a circle in $\mathbb{R}^2$, and let $p\in\mathbb{R}^2$ such that $f(p)$ lies on an edge (not a vertex) of $f(C)$. Since $f$ is continuous, then its restriction $f|_C$ to the closed interior of $C$ is uniformly continuous. Take a sequence $D_m$ of circles in the tangent to $C$ at $p$ in the interior of $C$ so that $D_m$ converges to $C$ (in the Hausdorff metric). Since $f|_C$ is uniformly continuous, then $f(D_m)$ converges to $f(C)$. Since $f(D_m)$ and $f(C)$ are $n$-gons, it's clear that $\text{vertex}(f(D_m))$ converges to $\text{vertex}(f(C))$. Since each $D_m$ is tangent to $C$ at $p$, then $f(D_m)\cap f(C)=f(D_m\cap C)=\{f(p)\}$, so since $f(p)$ is not a vertex of $f(C)$ then $f(p)$ must be a vertex of $f(D_m)$, since otherwise $f(D_m)\cap f(C)$ would be infinite. The fact that $f(p)\in \text{vertex}(f(D_m))$ for all $m$ contradicts the fact that $\text{vertex}(f(D_m))$ converges to $\text{vertex}(f(C))$.

We also had a proof that there is no bijective $f$, but I may have found an error as I was typing this. Anyways, this brings us no closer to figuring out our actual problem. If anyone has any ideas, we would be happy to hear them.

Answer 1

There is in fact no such injection $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$, which (interestingly enough) we may prove using graph theory. Due to the complexity of the construction here, to hopefully provide a more clean exposition, I have made the choice to phrase things more geometrically, and leave out a few small details in the proof that should hopefully be easy to fill in for anyone interested in doing so.

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Choose an integer $m\geq 3$, and assume for contradiction's sake that there exists an injective map $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ mapping circles to $m$-gons.

It is easy to see that if two circles $C,D$ are tangent at a point $x$, then at least one of $f(C),f(D)$ must have $f(x)$ as a vertex; if $f(C)$ has $f(x)$ as a vertex, we say $C$ stabs $D$ at $x$. It is also easy to see that a circle may stab at most $m$ circles at distinct points.

Let $n\geq 3$. Now we construct a useful two index family $A_{n,k}$ of collections of colored circles:

Let $A_{n,1}$ be the collection consisting of the blue unit circle $U$ at $0$, a ring of $n$ identical red circles (each tangent to their two red neighbours) inscribed in $U$, and a smaller blue circle tangent to each of the red circles. Now, we recurse.

To create $A_{n,k}$ we replace each of the $n$ red circles $C_i$ in $A_{n,1}$ with smaller copies of $A_{n,k-1}$ each rotated appropriately by an angle $\theta_i$ so that this collection has has $n$ rotational symmetries, and for each of the $n^{k+1}$ red circles in this new collection, adding the two unique blue circles with center $0$ tangent to it (ignoring repeats); here we choose the angles $\theta_i$ so that $A_{n,k}$ does not contain three circles that are pairwise tangent at the same point (the fact that we can do this can be proven relatively easily via a cardinality argument). Depicted below are $A_{n,k}$ for a few small values of $n,k$:

We can imbue our collection of circles $A_{n,k}$ with a simple directed graph structure by defining a directed graph $G_{n,k}=(A_{n,k},E_{n,k})$ where $E_{n,k}$ is a set contianing every edge $CD$ such that $C$ stabs $D$ but $D$ does not stab $C$, and one edge out of every pair $CD,DC$ where $C$ and $D$ mutually stab eachother. Note that all information about the graph $G_{n,k}$ except for edge direction is encoded by tangency relations between the circles/vertices in $A_{n,k}$. Now, let $e_{n,k}$ and $v_{n,k}$ be the number of edges and vertices of $G_{n,k}$ respectively. For every vertex $C$ in $A_{n,k}$, we know that $C$ may stab at most $m$ other vertices in $A_{n,k}$ (since no three circles in $A_{n,k}$ are pairwise tangent at the same point), so $\text{out}(C)\leq m$ (this is the outdegree of $C$ in $G_{n,k}$). The degree sum formula then gives us that: $$e_{n,k} = \sum_{C\in A_{n,k}} \text{out}(C) \leq mv_{n,k}$$

Given how $A_{n,k}$ was constructed recursively, it is not too hard to see that $e,v$ satisfy the following recurrence inequalities: \begin{equation}\begin{split} v_{n,k+1} &\leq 2+nv_{n,k}+2n^k\\ e_{n,k+1} &\geq 3n+ne_{n,k}+2n^{k+1}\\ \end{split}\qquad\begin{split} v_{n,1}&=2+n\\ e_{n,1}&=3n\\ \end{split}\end{equation}

Applying induction on $k$, we may deduce the following asymptotic inequalities for $e,v$ with respect to $n$ (fixed $k$): \begin{equation}\begin{split} v_{n,k} &\leq n^k[1+o(1)]\\ e_{n,k} &\geq (2k+1)n^k[1+o(1)]\\ \end{split}\end{equation} Therefore, letting $2k+1>m$, for some large enough value of $n$, we will have that $e_{n,k}>mv_{n,k}$, which is a contradiction. Therefore our injective map $f$ cannot exist.