# HOPF-TYPE THEOREMS FOR $f$-NEIGHBORS

A.V. Malyutin, I.M. Shirokov  Andrei Valeryevich Malyutin
iii St. Petersburg Department
iii of Steklov Mathematical Institute,
iii Fontanka, 27,
iii 191023, St. Petersburg, Russia
iii St. Petersburg State University,
iii Universitetskaya Emb., 13B,
iii 199034, St. Petersburg, Russia<EMAIL_ADDRESS>Ilya Mikhailovich
Shirokov
iii St. Petersburg Department
iii of Steklov Mathematical Institute,
iii Fontanka, 27,
iii 191023, St. Petersburg, Russia<EMAIL_ADDRESS>
Malyutin, A.V., Shirokov, I.M., Hopf-type theorems for $f$-neighbors

© 2022 Malyutin A.V., Shirokov I.M

This research project was started during the Summer Research Program for
Undergraduates 2021 organized by the Laboratory of Combinatorial and Geometric
Structures at MIPT. This research was supported by the Ministry of Science and
Higher Education of the Russian Federation, agreement 075-15-2019-1620 date
08/11/2019 and 075-15-2022-289 date 06/04/2022.

> Abstract.  We work within the framework of a program aimed at exploring
> various extended versions for theorems from a class containing Borsuk–Ulam
> type theorems, some fixed point theorems, the KKM lemma, Radon, Tverberg,
> and Helly theorems. In this paper we study variations of the Hopf theorem
> concerning continuous maps of a compact Riemannian manifold $M$ of dimension
> $n$ to $\mathbb{R}^{n}$. We investigate the case of maps $f\colon
> M\to\mathbb{R}^{m}$ with $n<m$ and introduce several notions of varied types
> of $f$-neighbors, which is a pair of distinct points in $M$ such that $f$
> takes it to a small set of some type. Next for each type, we ask what
> distances on $M$ are realized as distances between $f$-neighbors of this
> type and study various characteristics of this set of distances. One of our
> main results is as follows. Let $f\colon M\to\mathbb{R}^{m}$ be a continuous
> map. We say that two distinct points $a$ and $b$ in $M$ are visual
> $f$-neighbors if the segment in $\mathbb{R}^{m}$ with endpoints $f(a)$ and
> $f(b)$ intersects $f(M)$ only at $f(a)$ and $f(b)$. Then the set of
> distances that are realized as distances between visual $f$-neighbors is
> infinite. Besides we generalize the Hopf theorem in a quantitative sense.
>
> Keywords: Borsuk–Ulam type theorems, the Hopf theorem, winding number,
> locally injective.

##  1\. Introduction

In this paper we continue to work on the program, started in [4], of exploring
various extended versions for the so-called point-type theorems. By the point-
type theorems we mean those in which a singleton or a point plays a central
role. This class of theorems includes in particular the following subclasses.

  1. (1)

Fixed-point theorems (Brouwer, Kakutani, etc.);

  2. (2)

Theorems on the existence of common points for subsets (Radon, Tverberg,
Helly, Knaster–Kuratowski–Mazurkiewicz, etc.);

  3. (3)

Theorems on taking sets of a large diameter to a singleton (Borsuk–Ulam, Hopf,
topological versions of Radon and Tverberg theorems, etc.).

The first driving idea of the program is to cover the cases of arbitrary
dimensions and new classes of spaces by relaxing the point-type conditions.
For example, we replace conditions of the forms the map has a fixed point,
subsets have a common point, and the image of a subset that is large in some
sense is a singleton with milder ones the map has a point that moves slightly
in some sense, subsets have an equidistant point, and the image of a subset
that is large in some sense is a subset that is small in some sense
respectively.

The Borsuk–Ulam and Radon theorems, as well as the
Knaster–Kuratowski–Mazurkiewicz lemma, have counterparts obtained in this way
(see [4]). In this paper we operate in the Borsuk–Ulam subclass of theorems
(3) and concentrate on extensions related to the Hopf theorem [2]:

######  Theorem 1 (H. Hopf [2]).

Let $n$ be a positive integer, let $M$ be a compact Riemannian manifold of
dimension $n$, and let $f\colon M\to\mathbb{R}^{n}$ be a continuous map. Then
for any prescribed $\delta>0$, there exists a pair $\\{x,y\\}\in M\times M$
such that $f(x)=f(y)$ and $x$ and $y$ are joined by a geodesic of length
$\delta$.

The Hopf theorem has a rigid restriction on the codomain dimension and
involves the strict condition $f(x)=f(y)$. A question we are interested in is
whether the maps not satisfying dimension restrictions give sufficiently large
collections of pairs $\\{x,y\\}\in M\times M$ if we replace the requirement
$f(x)=f(y)$ with a requirement of the form $f(x)$ and $f(y)$ are in some sense
close to each other. We examine several versions of proximity to each other,
which are presented in the following definition of $f$-neighbors of various
types.

######  Definition 1.

Let $(\mathcal{M},d)$ be a metric space with distance function $d$, and let
$f\colon\mathcal{M}\to\mathbb{R}^{m}$ be a continuous map. Distinct points $a$
and $b$ in $\mathcal{M}$ are called

  1. (1)

_$f$ -neighbors_ if $f(a)=f(b)$;

  2. (2)

_spherical $f$-neighbors_ if either $f(a)=f(b)$ or there exists a Euclidean
ball $B^{m}\subset\mathbb{R}^{m}$ such that111We use $\partial X$ and
$\operatorname{Int}X$ to denote the boundary and interior, respectively, of a
subset $X$ in a topological space. $\\{f(a),f(b)\\}\subset\partial B^{m}$ and
$f(\mathcal{M})\cap\operatorname{Int}B^{m}=\emptyset$;

  3. (3)

_visual $f$-neighbors_ if either $f(a)=f(b)$ or the segment with endpoints
$f(a)$ and $f(b)$ intersects $f(\mathcal{M})$ only at these two points;

  4. (4)

_topological $f$-neighbors_ if either $f(a)=f(b)$ or some path in
$\mathbb{R}^{m}$ with endpoints at $f(a)$ and $f(b)$ intersects
$f(\mathcal{M})$ only at these two points.

To measure collections of $f$-neighbors of various types we use the sets of
distances that are realized as distances between $f$-neighbors.

######  Definition 2.

We introduce the following notation:

  1. (1)

$\Omega_{f}=\\{d(a,b)\in\mathbb{R}\>|$ $a$ and $b$ in $\mathcal{M}$ are
$f$-neighbors$\\}$;

  2. (2)

$\Omega_{f}^{sph}=\\{d(a,b)\in\mathbb{R}\>|$ $a$ and $b$ in $\mathcal{M}$ are
spherical $f$-neighbors$\\}$;

  3. (3)

$\Omega_{f}^{vis}=\\{d(a,b)\in\mathbb{R}\>|$ $a$ and $b$ in $\mathcal{M}$ are
visual $f$-neighbors$\\}$;

  4. (4)

$\Omega_{f}^{top}=\\{d(a,b)\in\mathbb{R}\>|$ $a$ and $b$ in $\mathcal{M}$ are
topological $f$-neighbors$\\}$.

Then
$\Omega_{f}\subseteq\Omega_{f}^{sph}\subseteq\Omega_{f}^{vis}\subseteq\Omega_{f}^{top}$.

The Hopf theorem implies that, in its settings and with respect to the natural
path-metric on $M$, each of the sets $\Omega_{f}$, $\Omega_{f}^{sph}$,
$\Omega_{f}^{vis}$, and $\Omega_{f}^{top}$ is continual and contains an
interval adjacent to $0$. Clearly, any embedding $f\colon M\to\mathbb{R}^{m}$
yields empty $\Omega_{f}$. The following theorem (see [4]) shows that
$\Omega_{f}^{sph}$ is nonempty for the case of continuous maps
$f\colon\mathbb{S}^{n}\to\mathbb{R}^{m}$ with $n<m$.

######  Theorem 2 (A. V. Malyutin, O. R. Musin [4]).

Let $m$ and $n$ be positive integers such that $n<m$, let $\mathbb{S}^{n}$ be
a Euclidean unit $n$-sphere in the Euclidean $(n+1)$-space $\mathbb{R}^{n+1}$,
and let $\mathbb{S}^{n}\to\mathbb{R}^{m}$ be a continuous map. Then there are
spherical $f$-neighbors $x$ and $y$ in $\mathbb{S}^{n}$ such that the
Euclidean distance $|x-y|$ is at least $\sqrt{(n+2)/n}$.

The main point of the present paper is further study of properties of sets
$\Omega_{f}^{sph},\Omega_{f}^{vis}$, and $\Omega_{f}^{top}$ for the case of
continuous maps $f\colon M\to\mathbb{R}^{m}$, where $M$ is a compact
Riemannian manifold of dimension $\dim M<m$.

The paper is organised as follows. First in section Negative results we give
some examples showing that in general even $\Omega_{f}^{top}$ can contain
large holes and be arbitrarily small in some sense. Next, in section Positive
results, we prove that $\Omega_{f}^{vis}$ is infinite for continuous maps
$f\colon M\to\mathbb{R}^{m}$ of compact Riemannian manifolds (with any
intrinsic metrics compatible with the topology of $M$), and obtain the same
result for $\Omega_{f}^{sph}$ in the case of continuous maps
$f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$. In the last section we generalize
the Hopf theorem in a quantitative sense.

##  2\. Negative results

First we give an example of a continuous map
$f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ where, unlike in the Hopf theorem,
the set $\Omega_{f}^{vis}$ does not cover the interval $(0,\pi]$.

######  Proposition 1 (Example 1).

Let $\mathbb{S}^{1}$ be a Euclidean circle in $\mathbb{R}^{2}$, and let $d$ be
the angular distance on $\mathbb{S}^{1}$. Then for any prescribed
$\varepsilon\in(0,\frac{\pi}{3})$, there exists a continuous map
$f_{\varepsilon}\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ such that
$\Omega_{f_{\varepsilon}}^{vis}=(0,\varepsilon]\cup[\frac{2\pi}{3}-\varepsilon,\pi]$.

$\alpha$$\alpha$$\alpha$$\varepsilon$$\varepsilon$$\varepsilon$$A$$B$$C$$D$$F$$G$$A,D$$B,F$$C,G$$\varepsilon$$\varepsilon$$\varepsilon$$\alpha$$\alpha$$\alpha$
Fig. 1.  $\mathbb{S}^{1}$ left and $f_{\varepsilon}(\mathbb{S}^{1})$ right

###### Proof..

A construction of such $f_{\varepsilon}$ is shown in Fig. 1. The
parametrization of $f_{\varepsilon}$ on each of the red and blue arcs is
uniform. To find $\Omega_{f_{\varepsilon}}^{vis}$ we study which boundary
points of each connected component of $\mathbb{R}^{2}\setminus
f(\mathbb{S}^{1})$ can see each other. It is enough to study a connected
component whose boundary is the union of a red $\varepsilon$-arc and a blue
$\alpha$-arc, and the component whose boundary is the union of red
$\varepsilon$-arcs. If red $\varepsilon$-arcs are sufficently close to the
corresponding blue $\alpha$-arcs, then we get

$\Omega_{f_{\varepsilon}}^{vis}=(0,\varepsilon]\cup\left[\frac{2\pi}{3}-\varepsilon,\pi\right].\qed$

Next we are going to show that there exist continuous maps
$f\colon\mathbb{S}^{n}\to\mathbb{R}^{n+1}$ with $\Omega_{f}^{top}$ of
arbitrarily small Lebesgue measure.

######  Proposition 2 (Example 2).

Let $n$ be a positive integer, let $\mathbb{S}^{n}$ be a Euclidean $n$-sphere
in the Euclidean $(n+1)$-space $\mathbb{R}^{n+1}$, and let $d$ be the angular
distance on $\mathbb{S}^{n}$. Then there exists a sequence of continuous maps
$(f_{k}\colon\mathbb{S}^{n}\to\mathbb{R}^{n+1})_{k=1}^{\infty}$ such that
$\lim\limits_{k\to\infty}\mu(\Omega_{f_{k}}^{top})=0$, where $\mu$ is the
Lebesgue measure on $\mathbb{R}$.

###### Proof..

Let $k\geq 2$ be a positive integer. Then there exists a collection
$P_{k}=\\{S_{l}\\}_{l=1}^{k}$ of subsets of $\mathbb{S}^{n}$ such that each
$S_{l}$ is a closed topological $n$-ball,
$\mathbb{S}^{n}=\bigcup_{l=1}^{k}S_{l}$, and $S_{l}\cap
S_{m}=\partial{S_{l}}\cap\partial{S_{m}}$ if $l\neq m$. (Say, $P_{2}$ is a
presentation of $\mathbb{S}^{n}$ as the union of two hemispheres, and for
$k\geq 3$ one can obtain $P_{k}$ by replacing an $n$-ball in $P_{k-1}$ with
two topological $n$-balls.) For each $l\in\\{1,\dots,k\\}$, choose a point
$\xi_{l}\in$ $\operatorname{Int}S_{l}$. Then choose
$\varepsilon\in(0,1/k^{3})$ such that for each $l\in\\{1,\dots,k\\}$, $S_{l}$
contains the $\varepsilon$-neighborhood $U_{\varepsilon}(\xi_{l})$ of
$\xi_{l}$ in $\mathbb{S}^{n}$.

We construct our $f_{k}$ as follows. Set $f_{k}$ to be the identity map on
$\mathbb{S}^{n}\setminus U$, where
$U=\bigcup_{l=1}^{k}U_{\varepsilon}(\xi_{l})$. Let $T_{l}$ be a two-side
thickening of $S_{l}$ along the radius of $\mathbb{S}^{n}$. Define $f_{k}$ on
$U_{\varepsilon}(\xi_{l})$ such that $f_{k}(U_{\varepsilon}(\xi_{l}))=T_{l}$.

Then $\mathbb{S}^{n}$ is surrounded by the sets
$f_{k}(U_{\varepsilon}(\xi_{l}))$ and $f_{k}(U_{\varepsilon}(\xi_{l}))$ hides
$S_{l}\setminus U_{\varepsilon}(\xi_{l})$ from other $S_{m}\setminus
U_{\varepsilon}(\xi_{m})$.

Clearly, $\Omega_{f_{k}}^{top}$ is contained in the union of the interval
$(0,D_{k}^{max})$, where $D_{k}^{max}$ is the maximum of diameters of $S_{l}$,
$l\in\\{1,\dots,k\\}$, and the intervals of the form

$(d(\xi_{i},\xi_{j})-2/k^{3},d(\xi_{i},\xi_{j})+2/k^{3}),\quad
i,j\in\\{1,\dots,k\\}.$

Thus we have

$\mu(\Omega_{f_{k}}^{top})\leq D_{k}^{max}+4k(k-1)/k^{3}.$

Taking $P_{k}$ with increasing $k$ and with $D_{k}^{max}$ tending to $0$ as
$k$ tends to infinity, we obtain a sequence of maps with the desired property.
∎

The construction of the above proof of Proposition 2 implies the following
proposition.

######  Proposition 3.

Under assumptions of Proposition 2, for any finite set $F\subset\mathbb{R}$
there exists a continuous map $f_{F}\colon\mathbb{S}^{n}\to\mathbb{R}^{n+1}$
such that $F\cap\Omega_{f_{F}}^{top}=\varnothing$.

We describe an example of another kind with the same result as above for the
case of continuous maps $f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$. Namely, for
each pair of coprime integers $(3,q)$ with $q\geq 4$, we define a continuous
map $f_{q}\colon\mathbb{S}^{1}\to T_{3,q}$, where
$T_{3,q}\subset\mathbb{R}^{2}$ is a standard diagram of torus knot of type
$(3,q)$ (see Fig. 2).

$B$$D$$F$$H$$A$$C$$E$$G$$A,D$$C,F$$B,G$$H,E$ Fig. 2. Example of $f_{4}$:
$\mathbb{S}^{1}$ left and $f_{\alpha}(\mathbb{S}^{1})$ right

We take red arcs with centers at vertices of a regular $q$-polygon inscribed
in $\mathbb{S}^{1}$ and construct our $f_{q}$ with uniform parametrization on
each of red and blue arcs as shown in Fig. 2. The image of the union of red
arcs forms a regular $q$-polygon, and the endpoints of the image of each blue
arc are vertices of this $q$-polygon. We construct a sequence
$(f_{q})_{q=4}^{\infty}$, with $q$ coprime to $3$, such that the total length
of red arcs decreases as $O(1/q^{3})$ as $q$ tends to infinity.

One can find $\Omega_{f_{q}}^{sph}$, $\Omega_{f_{q}}^{vis}$, and
$\Omega_{f_{q}}^{top}$ by studying colorings of the set of pairs of points
$T^{2}=\mathbb{S}^{1}\times\mathbb{S}^{1}$. Namely, we color a pair $(a,b)$
red if $a$ and $b$ are $f$-neighbors of corresponding type and green
otherwise. Fig. 3 with an unfolding of colored torus provides an example for
$\Omega_{f_{4}}^{vis}$, where we color a pair $(a,b)$ red if $f(a)$ and $f(b)$
see each other.

Fig. 3. Coloring of $\mathbb{S}^{1}\times\mathbb{S}^{1}$

In Fig. 3, for all points on a line that is parallel to a diagonal, angular
distances between corresponding points on $\mathbb{S}^{1}$ are the same. When
we increase $q$, the number of red squares increases and we get more red
stripes, but they become narrower and the total area of red squares decreases.
As a result we have ${\mu(\Omega_{f_{q}}^{vis})\to 0}$.

A variation of this construction and similar arguments work for
$\Omega_{f_{q}}^{top}$.

It would be interesting to apply the colored torus method to study the sets
$\Omega_{f}^{sph}$, $\Omega_{f}^{vis}$, and $\Omega_{f}^{top}$ for the case of
more complex closed braid diagrams.

##  3\. Positive results

In the previous section we saw that even $\Omega_{f}^{top}$ can have
arbitrarily small Lebesgue measure. It is natural to ask, Can
$\Omega_{f}^{top}$, $\Omega_{f}^{vis}$, or $\Omega_{f}^{sph}$ be finite? In
this section we first prove that $\Omega_{f}^{vis}$ is infinite for the case
of continuous maps $f\colon M\to\mathbb{R}^{m}$, where $M$ is a compact
Riemannian manifold (with an intrinsic metric compatible with the topology on
$M$), and then we show that $\Omega_{f}^{sph}$ is infinite for the case of
continuous maps $f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$. We need several
auxiliary lemmas and definitions.

######  Definition 3.

Let $A$ and $B$ be topological spaces. A continuous map $f\colon A\to B$ is
called _locally injective_ if any point in $A$ has a neighborhood such that
the restriction of $f$ to this neighborhood is injective.

######  Definition 4 (cf. Definition 1).

Let $Y$ be a subset of $\mathbb{R}^{m}$. We say that two points $p$ and $q$ in
$Y$ are

  1. (1)

_visual neighbors_ (with respect to $Y$) if $p\neq q$ and the segment with
endpoints $p$ and $q$ intersects $Y$ only at these two points;

  2. (2)

_spherical neighbors_ (with respect to $Y$) if $p\neq q$ and there exists a
Euclidean ball $B^{m}\subset\mathbb{R}^{m}$ such that
$\\{p,q\\}\subset\partial B^{m}$ and $Y\cap\operatorname{Int}B^{m}=\emptyset$.

######  Proposition 4.

Let $f\colon X\to Z$ be a continuous map of metric spaces. If $X$ is compact
and $f$ is locally injective, then

  1. (1)

For any $z\in f(X)$, the inverse image $f^{-1}(z)$ is finite;

  2. (2)

Each point $z\in f(X)$ has a neighborhood $U=U(z)$ such that the restriction
of $f$ to each connected component of $f^{-1}(U)$ is injective.

###### Proof..

  1. (1)

Assume that $f^{-1}(z)$ is infinite for some $z\in f(X)$. Then $f^{-1}(z)$ has
an accumulation point since $X$ is compact. Clearly, no accumulation point of
$f^{-1}(z)$ has a neighborhood such that the restriction of $f$ to this
neighborhood is injective.

  2. (2)

By assertion (1) of the proposition, $f^{-1}(z)$ is finite. Suppose
$f^{-1}(z)$ consists of $k$ points and denote them by $x_{1},\dots,x_{k}$. Due
to the assumed local injectivity of $f$, each of $x_{i}$’s has an open
neighborhood $U_{i}$ such that the restriction of $f$ to $U_{i}$ is injective.
Taking (if necessary) smaller neighborhoods of $x_{1}$, $\dots$, $x_{k}$, we
can assume that $U_{1}$, $\dots$, $U_{k}$ are pairwise disjoint. Observe that
the complement $C:=X\setminus(U_{1}\cup\dots\cup U_{k})$ is compact and
$z\notin f(C)$. Hence, since a continuous image of a compact set is compact,
it follows that there is a neighborhood $W=W(z)$ of $z$ such that $f(C)\cap
W(z)=\emptyset$. Therefore, $f^{-1}(W)$ is contained in $U_{1}\cup\dots\cup
U_{k}$. Since $U_{1}$, $\dots$, $U_{k}$ are all open and pairwise disjoint, it
follows that each connected component of $f^{-1}(W)$ is contained in $U_{i}$
for some $i$. The statement follows. ∎

######  Proposition 5.

Let $f\colon X\to Z$ be a continuous map of metric spaces. If $X$ is compact
then the following conditions are equivalent:

  1. (1)

$f$ is not locally injective;

  2. (2)

There exist arbitrarily close $f$-neighbors.

###### Proof..

If $f$ is not locally injective then there exists $x\in X$ such that for each
positive integer $i$ the open metric ball $B_{1/i}(x)$ of radius $1/i$
centered at $x$ contains distinct points $a_{i}$ and $b_{i}$ with
$f(a_{i})=f(b_{i})$. These $a_{i}$ and $b_{i}$ are $f$-neighbors within
distance at most $2/i$ from each other.

Conversely, suppose there exist arbitrarily close $f$-neighbors. This means
that for each positive integer $i$, there exist $f$-neighbors $a_{i}$ and
$b_{i}$ with $d(a_{i},b_{i})<1/i$. Since $X$ is compact, the sequence
$(a_{i})_{i=1}^{\infty}$ has converging subsequences. Clearly, no limit point
of a converging subsequence of $(a_{i})_{i=1}^{\infty}$ has a neighborhood
such that the restriction of $f$ to this neighborhood is injective. ∎

######  Lemma 1.

Let $X$ be a topological space, let $k$ be a positive integer, and let
$(A_{1},\dots,A_{k})$ be a $k$-element sequence of closed sets in $X$. Suppose
that at least one of these sets is nonempty. A point $P$ in the union
$A:=A_{1}\cup\dots\cup A_{k}$ is called _properly covered_ if there exists a
neighborhood $U=U(P)$ of $P$ such that all nonempty sets in the sequence
$(U\cap A_{1},\dots,U\cap A_{k})$ coincide. Then the set of properly covered
points is open and everywhere dense in $A$.

###### Proof..

First we prove that $A$ contains at least one properly covered point. We use
induction on $k$. For $k=1$ the statement is obvious. Let $k>1$. If all
nonempty sets in $(A_{1},\dots,A_{k})$ coincide, then taking the whole space
$X$ as a suitable neighborhood, we get that all points in $A$ are properly
covered. Otherwise there are nonempty noncoinciding sets in
$(A_{1},\dots,A_{k})$, and hence there exists $j\in\\{1,\dots,k\\}$ such that
$A_{j}$ is nonempty and $A_{j}\neq A$. Then at least one element in the
$(k-1)$-element sequence

$(A_{1}\setminus A_{j},\dots,A_{j-1}\setminus A_{j},A_{j+1}\setminus
A_{j},\dots,A_{k}\setminus A_{j})$

is nonempty and all elements of this sequence are closed subsets in
$X\setminus A_{j}$.

Applying the inductive hypothesis to the space $X\setminus A_{j}$ and the
$(k-1)$-element sequence above, we get that there exists a properly covered
(in the corresponding sense) point $P$ in $A\setminus A_{j}$. Since $A_{j}$ is
closed, the sets (in particular neighborhoods of points) that are open in
$X\setminus A_{j}$, are open in $X$ as well. Hence $P$ is properly covered for
the initial sequence $(A_{1},\dots,A_{k})$.

To check that the set of properly covered points is everywhere dense in $A$,
it suffices, for any open set $O$ that intersects $A$, to apply the proven
statement to the space $O$ and the sequence $(O\cap A_{1},O\cap
A_{2},\dots,O\cap A_{k})$.

The set of properly covered points is open because some open neighborhood (in
$A$) of any properly covered point consists of properly covered points due to
the definition. ∎

######  Lemma 2.

Let $K$ be a compact subset of a Euclidean space $E$. If $K$ is not convex,
then $E$ contains a nondegenerate segment that intersects $K$ only by its
endpoints.

###### Proof..

Since $K$ is compact and not convex, it follows that the relative interior of
the convex hull of $K$ contains a point $P$ such that $P\notin K$. By
compactness argument it follows that $E$ contains a closed Euclidean ball $B$
centered at $P$ and such that $K\cap\operatorname{Int}B=\emptyset$ while
$\partial{B}$ intersects $K$. Let $Q$ be a point in $B\cap K$, and let $H$ be
the hyperplane in $E$ that touches $B$ at $Q$. Denote by $H^{+}$ the closed
half-space bounded by $H$ and containing $B$. Since $\operatorname{Int}H^{+}$
contains $P$ and $P$ lies in the convex hull of $K$, it follows that
$\operatorname{Int}H^{+}$ contains points of $K$ as well. Let $T$ be a point
in $K\cap\operatorname{Int}H^{+}$, and let $I$ be the segment with endpoints
$Q$ and $T$. Observe that some punctured neighborhood of $Q$ in $I$ lies in
$\operatorname{Int}B$ and hence does not intersect $K$. Then, since $K$ is
compact, it follows that $I$ contains a segment with desirable properties. ∎

######  Theorem 3.

Let $M$ be a compact Riemannian manifold, let $d\colon M\times M\to[0,\infty)$
be any intrinsic metric on $M$ compatible with its topology, and let $f\colon
M\to\mathbb{R}^{m}$ be a continuous map. Then $\Omega_{f}^{vis}$ for $(M,d)$
is infinite.

###### Proof..

Let $n=\dim M$ be the dimension of $M$. The case where $m\leq n$ is covered by
the Hopf theorem. We study the case where $n<m$ and prove the theorem by case
analysis.

  1. (1)

If $f$ is not locally injective, then the statement of the theorem follows by
Proposition 5 because $f$-neighbors are visual $f$-neighbors.

  2. (2)

If $f$ is locally injective, let $P$ be any point in $f(M)$. Since $f$ is
locally injective and $M$ is compact, it follows that $f^{-1}(P)$ is finite
(see Proposition 4(1)). Suppose $f^{-1}(P)$ consists of $k$ points and denote
them by $x_{1},\dots,x_{k}$. Due to the assumed local injectivity of $f$, each
$x_{i}$ has an open neighborhood $U_{i}$ such that the restriction of $f$ to
$U_{i}$ is injective (cf. the above proof of assertion (2) of Proposition 4).
Taking (if necessary) smaller neighborhoods of $x_{1}$, $\dots$, $x_{k}$, we
can assume that $U_{1}$, $\dots$, $U_{k}$ are pairwise disjoint. Let
$U^{\prime}_{i}$ be an open neighborhood of $x_{i}$ such that the closure
$\overline{U^{\prime}_{i}}$ of $U^{\prime}_{i}$ is contained in $U_{i}$.
Observe that the complement
$\mathcal{C}:=M\setminus(U^{\prime}_{1}\cup\dots\cup U^{\prime}_{k})$ is
compact and $P\notin f(\mathcal{C})$. Hence, since a continuous image of a
compact set is compact, it follows that there is a neighborhood $W=W(P)$ of
$P$ in $\mathbb{R}^{m}$ such that $f(\mathcal{C})\cap W(P)=\emptyset$. Thus
the intersection of $W(P)$ with $f(M)$ is the union of intersections of $W(P)$
with compact sets $f(\overline{U^{\prime}_{1}})$, …,
$f(\overline{U^{\prime}_{k}})$. By Lemma 1 there exists an open set $V\subset
W(P)$ such that $V\cap f(M)$ is nonempty and all nonempty sets in the sequence
$(V\cap f(\overline{U^{\prime}_{1}})$, $\dots$, $V\cap
f(\overline{U^{\prime}_{k}}))$ coincide. Let $D^{\prime}$ be a closed
Euclidean ball in $\mathbb{R}^{m}$ such that $D^{\prime}\subset V$ and
$f(M)\cap\operatorname{Int}D^{\prime}\neq\emptyset$. We have two subcases:

    1. (a)

If $f(M)$ has arbitrarily close visual neighbors (see Definition 4) lying in
$D^{\prime}$, then, since due to the choice of $V$ all nonempty sets in the
sequence $(D^{\prime}\cap f(\overline{U^{\prime}_{1}}),\dots,D^{\prime}\cap
f(\overline{U^{\prime}_{k}}))$ coincide with $D^{\prime}\cap f(M)$, we
conclude that in each $U^{\prime}_{i}$ with nonempty $D^{\prime}\cap
f(\overline{U^{\prime}_{i}})$ there is an infinite sequence of pairs of visual
$f$-neighbors $(a_{j},b_{j})$ such that the distance between $f(a_{j})$ and
$f(b_{j})$ tends to $0$ as $j$ tends to infinity. Since the restriction of $f$
to $\overline{U^{\prime}_{i}}$ is an embedding and $\overline{U^{\prime}_{i}}$
is compact, we conclude that $d(a_{j},b_{j})\to 0$ when $j\to\infty$ as well.
This proves the statement for this subcase.

    2. (b)

If there exists $\delta>0$ such that $D^{\prime}\cap f(M)$ contains no visual
neighbors at distance less than $\delta$ from each other, then take a closed
Euclidean ball $D^{\prime\prime}\subset D^{\prime}$ of diameter less than
$\delta$ and such that $\operatorname{Int}D^{\prime\prime}$ intersects $f(M)$.
Then $D^{\prime\prime}\cap f(M)$ is convex by Lemma 2. Let $H$ be the affine
hull of $D^{\prime\prime}\cap f(M)$ in $\mathbb{R}^{m}$, and let $\dim H$ be
its dimension (or, which is the same, the dimension of $D^{\prime\prime}\cap
f(M)$). Observe that $\dim H=n=\dim M$. Indeed notice that due to the choice
of $V$ (which contains $D^{\prime}$ and $D^{\prime\prime}$) all nonempty sets
in the sequence $(D^{\prime\prime}\cap
f(\overline{U^{\prime}_{1}}),\dots,D^{\prime\prime}\cap
f(\overline{U^{\prime}_{k}}))$ coincide with $D^{\prime\prime}\cap f(M)$ while
for each $i\in\\{1,\dots,k\\}$ the restriction
$f|_{\overline{U^{\prime}_{i}}}$ of $f$ to $\overline{U^{\prime}_{i}}$ is an
embedding of a compact set to a Hausdorff space and hence
$f|_{\overline{U^{\prime}_{i}}}$ is a homeomorphism between
$\overline{U^{\prime}_{i}}$ and $f(\overline{U^{\prime}_{i}})$. Thus for an
arbitrary point $x$ in the relative interior of $D^{\prime\prime}\cap f(M)$
the map $f$ gives a homeomorphism between some neighborhood of any point in
$f^{-1}(x)$ and a neighborhood of $x$ in $D^{\prime\prime}\cap f(M)$. Then the
desired equality of dimensions $\dim H=n$ follows by the domain invariance
theorem. Now the proof splits into two subsubcases:

      1. (i)

If $H$ contains $f(M)$ then the statement readily follows from the Hopf
theorem. (Supplemented by the fact that, due to compactness, a sequence of
pairs of points arbitrarily close to each other with respect to the standard
metric preserves this property when passing to any other metric compatible
with the topology.)

      2. (ii)

If $H$ does not contain $f(M)$, let $T$ be a point in $f(M)\setminus H$, let
$H_{+}$ be the half-space of dimension $n+1$ bounded by $H$ and containing
$T$, and let $Q$ be an arbitrary point in the relative interior of
$D^{\prime\prime}\cap f(M)$. Next let $\mathcal{B}$ be the family of closed
$m$-dimensional Euclidean balls centered at points of $H_{+}$ and touching $H$
at $Q$. It is clear that balls in $\mathcal{B}$ with sufficiently small radii
lie in $D^{\prime\prime}$ and intersect $f(M)$ just at $Q$, while balls in
$\mathcal{B}$ with sufficiently large radii contain $T$. By compactness
arguments and since a part of $f(M)$ that is close to $Q$ is contained in $H$,
we conclude that there is a ball $B$ in $\mathcal{B}$ such that
$f(M)\cap\operatorname{Int}B=\emptyset$ and $(\partial B\cap
f(M))\setminus\\{Q\\}\neq\emptyset$. Let $Q^{\prime}$ be a point in $(\partial
B\cap f(M))\setminus\\{Q\\}$. It is easily seen by construction that all
points of $f(M)$ in a small neighborhood of $Q$ are visual neighbors of
$Q^{\prime}$. Now by virtue of the above-mentioned homeomorphisms between
neighborhoods of points in $f^{-1}(Q)$ and a neighborhood of $Q$ in
$D^{\prime\prime}\cap f(M)$ it follows that all points in some sufficiently
small neighborhood of $f^{-1}(Q)$ are visual $f$-neighbors for all points in
$f^{-1}(Q^{\prime})$. Let $q$ be a point in $f^{-1}(Q)$, let $q^{\prime}$ be a
point in $f^{-1}(Q^{\prime})$, and let $\rho$ be a shortest path joining $q$
and $q^{\prime}$ (this $\rho$ exists due to compactness of $M$). Then subpaths
of $\rho$ are also shortest paths, which shows that in this subsubcase
$\Omega_{f}^{vis}$ is continual for intrinsic metrics. The theorem is proved.
∎

To prove the result about spherical $f$-neighbors for the case of continuous
maps $\mathbb{S}^{1}\to\mathbb{R}^{2}$, we need several additional
constructions and statements.

Let $\mathbb{R}^{2}$ be the Euclidean plane. Suppose
$\mathcal{O}\subset\mathbb{R}^{2}$ is a nonempty bounded open connected set.

######  Definition 5.

A closed Euclidean disk $B(x)\subset\overline{\mathcal{O}}$ centered at
$x\in\mathcal{O}$ is called a _good disk_ if
$\operatorname{card}(B(x)\cap\partial{\mathcal{O}})\geq 2$ (see Fig. 4).
Points of $B(x)\cap\>\partial{\mathcal{O}}$ are called _good points_. The
union of good points corresponding to all good disks in
$\overline{\mathcal{O}}$ we denote by $\mathcal{G}_{\mathcal{O}}$. Points of
$\partial{\mathcal{O}}\setminus\mathcal{G}_{\mathcal{O}}$ are called _bad
points_.

Fig. 4. An example of good disks in $\overline{\mathcal{O}}$. Notice that a
good point can belong to uncountably many good disks.

For example, an ellipse in $\mathbb{R}^{2}$ has two bad points and a square
has four bad points. For a triangle whose sides are the Koch curve, bad points
are everywhere dense in $\partial{\mathcal{O}}$. The following lemma shows
that good points are always everywhere dense in $\partial{\mathcal{O}}$.

######  Lemma 3.

$\overline{{\mathcal{G}}_{\mathcal{O}}}=\partial{\mathcal{O}}$.

###### Proof..

It suffices to show that if $b$ is a bad point in $\partial{\mathcal{O}}$ then
a sequence of good points converges to $b$. Take a sequence
$(x_{k})_{k=1}^{\infty}$ in $\mathcal{O}$ converging to $b$. Note that, since
$\mathcal{O}$ is the union of interiors of all good disks in
$\overline{\mathcal{O}}$ (see [3, 5, 1]), each $x_{k}$ lies in some good disk.
Let $B_{k}$ be a good disk containing $x_{k}$. Passing to a subsequence if
necessary, we may assume that the sequence $(B_{k})_{k=1}^{\infty}$ converges
in the disk space. Let $B_{\infty}$ be the limit of $(B_{k})_{k=1}^{\infty}$.
If $B_{\infty}=\\{b\\}$, then the statement of the lemma easily follows.
Otherwise $B_{\infty}$ is nondegenerate. It is clear that
$B_{\infty}\subset\overline{\mathcal{O}}$, because $B_{k}$ is in
$\overline{\mathcal{O}}$ for each $k\in\mathbb{N}$. Next observe that
$B_{\infty}\cap\overline{\mathcal{O}}=b$, since otherwise $b$ is not a bad
point. Now for each good disk $B_{k}$, choose a good point $\xi_{k}\in B_{k}$.
Any accumulation point of $\\{\xi_{k}\\}_{k=1}^{\infty}$ is a point in
$\overline{\mathcal{O}}$ and a point in $\partial{B_{\infty}}$, i. e. it is
the point $b$. This completes the proof. ∎

######  Proposition 6.

  1. (1)

Any open connected subset of $\mathbb{R}^{2}$ is path-connected.

  2. (2)

If a subset $K$ of $\mathbb{R}^{2}$ is closed and connected then each bounded
component of the complement $\mathbb{R}^{2}\setminus K$ is simply connected.

  3. (3)

If $O$ is an open, connected, and simply connected subset of $\mathbb{R}^{2}$
and $H$ is a nondegenerate segment in $\mathbb{R}^{2}$ with relative interior
in $O$ and with endpoints in $\partial{O}$, then the set $O\setminus H$ has
precisely two connected components.

###### Proof..

  1. (1)

This readily follows from the fact that $\mathbb{R}^{m}$ is locally path-
connected (i. e., it has a basis of path-connected neighbourhoods). Each open
subset of a locally path-connected space is also locally path-connected. Each
path-connected component of a locally path-connected space is open.
Consequently, connected components of a locally path-connected space are the
same as its path-connected components.

  2. (2)

Assume to the contrary that a bounded component $O$ of
$\mathbb{R}^{2}\setminus K$ is not simply connected. Then there exists a loop
$\gamma\colon[0,1]\to O$ noncontractible in $O$. Since $\gamma([0,1])$ is
compact and $O$ is open, it follows that $\gamma$ is homotopic in $O$ to a
polygonal loop in general position. Then we see that there exists a simple
noncontractible loop $\gamma_{s}\colon[0,1]\to O$. This means that the region
$B$ in $\mathbb{R}^{2}$ bounded by $\gamma_{s}$ (here, we use the
Jordan–Schoenflies theorem) is not contained in $O$. Therefore, since
$\partial O$ is contained in $K$ and $K$ is connected, it follows that $B$
contains $K$. This contradicts the assumption that $O$ is bounded.

  3. (3)

We first show that $O\setminus H$ is disconnected. Assume to the contrary that
$O\setminus H$ is connected. Let $I$ be a small segment lying in $O$ and
intersecting $H$ transversely. Since $O\setminus H$ is open and assumed to be
connected, it follows that $O\setminus H$ is also path-connected (see the
first assertion of the proposition). Therefore, there exists a path $\rho$
lying in $O\setminus H$ and joining the endpoints of $I$. Then, passing to
isotopic polygonal lines as in the proof of the previous assertion, it is easy
to deduce that there exists a simple closed polygonal line $\gamma$ lying in
$O$ and intersecting $H$ transversely at a single point ($H\cap I$). This
implies that the endpoints of $H$ lie in distinct components of the set
$\mathbb{R}^{2}\setminus\gamma$ (use the Jordan–Schoenflies theorem), which
contradicts the fact that $O$ is simply connected since these endpoints are
both in $\partial{O}$. This contradiction shows that $O\setminus H$ is
disconnected.

To verify that $O\setminus H$ has at most two components, we construct an open
neighborhood $U\subset O$ of the relative interior of $H$ such that $U$ is the
union of open Euclidean disks, each centered at a point of $H$. Then
$U\setminus H$ is clearly a two-component set while each component of
$O\setminus H$ intersects (and hence contains) a component of $U\setminus H$.
∎

######  Definition 6.

We call a nondegenerate segment in $\overline{\mathcal{O}}$, with relative
interior in $\mathcal{O}$ and with endpoints in $\partial{\mathcal{O}}$, a
_chord_. We say that a chord is _good_ if it is contained in a good disk. A
sequence of chords $(H_{k})_{k=1}^{\infty}$ in $\overline{\mathcal{O}}$ is
called _ruled_ if for each $k>1$, the chord $H_{k}$ divides $\mathcal{O}$ in
two parts such that one of these parts contains the relative interior of
$H_{k-1}$ and another one contains the relative interior of $H_{k+1}$.

######  Remark 1.

Definitions readily imply that two points in $\partial{\mathcal{O}}$ are
spherical neighbors with respect to $\partial{\mathcal{O}}$ if and only if
these two points are the endpoints of a good chord.

######  Lemma 4.

If $\mathcal{O}$ is simply connected then there is a ruled sequence of good
chords in $\overline{\mathcal{O}}$.

###### Proof..

Observe that $\overline{\mathcal{O}}$ has at least one good chord because
$\overline{\mathcal{O}}$ contains at least one good disk ([3, 5, 1]) while the
segment joining two good points on the boundary of this good disk is a good
chord. Next observe that in order to prove the lemma it suffices to show that
if $H$ is a good chord then the set $\mathcal{O}\setminus H$ is composed of
two connected components and each of these components contains the relative
interior of a good chord (of $\overline{\mathcal{O}}$). The set
$\mathcal{O}\setminus H$ is composed of two connected components because
$\mathcal{O}$ is simply connected (see Proposition 6). Now, if $R$ is a
component of $\mathcal{O}\setminus H$, let $B$ be that of all good disks
containing $H$ which has maximal intersection with $R$. If $R$ is contained in
$B$ then the part of $\partial\mathcal{O}$ that bounds $R$ is an arc of the
circle $\partial B$, and hence $R$ contains relative interiors of a continuum
of good chords. If $R$ is not contained in $B$, let $x$ be a point in
$R\setminus B$. Then there exists a good disk $D$ containing $x$ (see [3, 5,
1]). Clearly, the relative interior of any good chord contained in $D$, is
contained in $R$. The lemma is thus proved. ∎

######  Definition 7.

Let $\mathbb{S}^{1}$ be a circle, and let
$f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ be a continuous map. If $f$ is
locally injective (see Definition 3) we say that $f$ is a _locally-simple
curve_. If the restriction of $f$ to an arc $\alpha\subset\mathbb{S}^{1}$ is
injective, we say that $\alpha$ is an _$f$ -simple arc_.

The following proposition lists several properties of locally-simple curves.

######  Proposition 7.

If $f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ is a locally-simple curve, then

  1. (1)

For any $y\in f(\mathbb{S}^{1})$, the inverse image $f^{-1}(y)$ is finite;

  2. (2)

Each point $y\in f(\mathbb{S}^{1})$ has a neighborhood $U=U(y)$ such that each
connected component of $f^{-1}(U)$ is an $f$-simple arc;

  3. (3)

$\operatorname{Int}f(\mathbb{S}^{1})=\emptyset$;

  4. (4)

The set of bounded connected components of $\mathbb{R}^{2}\setminus
f(\mathbb{S}^{1})$ is nonempty;

  5. (5)

The set of indices (the winding numbers) of connected components of
$\mathbb{R}^{2}\setminus f(\mathbb{S}^{1})$ is finite.

###### Proof..

  1. (1)

This is a particular case of assertion (1) of Proposition 4.

  2. (2)

This follows from assertion (2) of Proposition 4. (If $f$ is injective and we
have $f^{-1}(U)=\mathbb{S}^{1}$, we pass to a smaller neighborhood.)

  3. (3)

This follows, for example, from the domain invariance theorem.

  4. (4)

The statement follows, for example, from Theorem 2.1 in [7].

  5. (5)

Since all of the components of $\mathbb{R}^{2}\setminus f(\mathbb{S}^{1})$
having nonzero index are contained in a compact domain, it follows by standard
compactness arguments that in order to prove the statement it suffices to show
that an arbitrary point $z\in\mathbb{R}^{2}$ has a neighborhood $U(z)$ such
that the set of indices (the winding numbers) of points in $U(z)\setminus
f(\mathbb{S}^{1})$ is finite. If $z\notin f(\mathbb{S}^{1})$ then $z$ has a
neighborhood where all points are of the same index. Let $z$ be a point of
$f(\mathbb{S}^{1})$. By assertion (2) there exists $\varepsilon_{0}>0$ such
that all connected components of
$V_{\varepsilon_{0}}:=f^{-1}(D_{\varepsilon_{0}}(z))$, where
$D_{\varepsilon_{0}}(z)$ is the open disk of radius $\varepsilon_{0}$ centered
at $z$, are (open) $f$-simple arcs. By assertion (1), only a finite number of
these $f$-simple arcs of $V_{\varepsilon_{0}}$ intersect $f^{-1}(z)$. Let
$\gamma_{1}$, …, $\gamma_{k}$ be the closures of the $f$-simple arcs of
$V_{\varepsilon_{0}}$ intersecting $f^{-1}(z)$. Passing to a smaller
$\varepsilon_{0}$ if necessary, we may assume that for each
$i\in\\{1,\dots,k\\}$, the closed arc $\gamma_{i}$ is $f$-simple as well and
$f(\gamma_{i})$ is a closed simple arc (not a loop) with two distinct
endpoints on the circle $\partial D_{\varepsilon_{0}}(z)$.

Let $f^{\prime}\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ be a curve (not
necessarily locally-simple) of the following form:

     * –

for each $i\in\\{1,\dots,k\\}$, the restriction of $f^{\prime}$ to
$\gamma_{i}$ is injective and the image $f^{\prime}(\gamma_{i})$ is an arc of
$\partial D_{\varepsilon_{0}}(z)$ with the same pair of endpoints as that of
$f(\gamma_{i})$;

     * –

for each $t\in\mathbb{S}^{1}\setminus(\gamma_{1}\cup\dots\cup\gamma_{k})$ we
set $f^{\prime}(t)=f(t)$.

Observe that for each $i\in\\{1,\dots,k\\}$, the union
$f^{\prime}(\gamma_{i})\cup f(\gamma_{i})$ is a Jordan curve. By the
Jordan–Schoenflies theorem this curve bounds an open topological disk
$D_{i}\subset D_{\varepsilon_{0}}(z)$. By construction, for any point $y\in
D_{\varepsilon_{0}}(z)\setminus f(\mathbb{S}^{1})$ the index of $y$ with
respect to $f$ differs from the index of $y$ with respect to $f^{\prime}$ by
at most $k$.

Next, observe that $z\notin f^{\prime}(\mathbb{S}^{1})$ so that by compactness
of $f^{\prime}(\mathbb{S}^{1})$ there exists
$\varepsilon_{1}\in(0,\varepsilon_{0})$ such that the open disk
$D_{\varepsilon_{1}}(z)$ of radius $\varepsilon_{1}$ centered at $z$ does not
intersect $f^{\prime}(\mathbb{S}^{1})$. Consequently, all of the points in
$D_{\varepsilon_{1}}(z)\setminus f(\mathbb{S}^{1})$ have the same index with
respect to $f^{\prime}$ and the set of their indices with respect to $f$ has
cardinality at most $k+1$. It remains to set $U(z):=D_{\varepsilon_{1}}(z)$,
which completes the proof. ∎

######  Theorem 4.

Let $\mathbb{S}^{1}$ be a Euclidean circle in $\mathbb{R}^{2}$, let $d$ be the
angular distance on $\mathbb{S}^{1}$, and let
$f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ be a continuous map. Then
$\Omega_{f}^{sph}$ for $(\mathbb{S}^{1},d)$ is infinite.

###### Proof..

Our proof is a case-by-case analysis. On its first level we distinguish two
basic cases, the case where $f$ is locally-simple and the case where it is
not. If $f$ is not locally-simple then $\Omega_{f}^{sph}$ is infinite by
Proposition 5 because $f$-neighbors are spherical $f$-neighbors. The rest of
the proof addresses subcases of the case

  1. (C2)

$f$ is locally-simple.

The following exposition consists of two parts. First we give some
preliminaries and introduce the notion of _positive points_ , and then we pass
to the core construction of the proof, which involves studying sequences of
pairs of positive points.

We will use the classical notion of _winding number_ (or _winding index_) of a
closed curve in $\mathbb{R}^{2}$ around a given point (which we also refer to
as the index of a point with respect to a curve). The index of a point $q$
with respect to a curve $\gamma\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ will be
denoted by $\operatorname{wind}(q,\gamma)$. The index depends on orientations.
We fix orientations on $\mathbb{S}^{1}$ and on $\mathbb{R}^{2}$. For
convenience of description we assume that $\mathbb{R}^{2}$ is located in front
of us in $\mathbb{R}^{3}$ in such a way that any Euclidean circle (in our
$\mathbb{R}^{2}$) oriented counterclockwise has winding index $1$ with respect
to the points inside it. Under this assumption, if we have, say, a smooth
closed curve in general position then the indices of any two adjacent
components of the curve complement differ by $1$: the component with the
larger index is on the _left_ side of the curve when we move along the curve
according to its orientation.

Next we introduce the notion of positive points. Let $H$ be a simple closed
arc in $\mathbb{R}^{2}$ such that the relative interior $H^{\circ}$ of $H$
does not intersect $f(\mathbb{S}^{1})$ and an endpoint $\xi$ of $H$ is in
$f(\mathbb{S}^{1})$. Let $x$ be a point in $f^{-1}(\xi)$. Since $f$ is
locally-simple, it follows that there exists a closed $f$-simple arc $\gamma$
containing $x$ in its interior. The Jordan–Schoenflies theorem implies [6]
that there exists an orientation preserving homeomorphism
$\Psi\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that $\Psi(f(\gamma))$ is a
straight segment. We say that $x$ is a _positive point_ for $H$ if the arc
$\Psi(H)$ adjoins our segment $\Psi(f(\gamma))$ on the left side when we move
along $\Psi(f(\gamma))$ according to its orientation (induced by that of
$f(\gamma)$). It is straightforward to check by studying compositions of
homeomorphism of $\mathbb{R}^{2}$ that, for $x$ and $H$, this definition is
independent of choices of $\Psi$ and $\gamma$.

In the above notation, let ${O}_{H}$ be the component of
$\mathbb{R}^{2}\setminus f(\mathbb{S}^{1})$ containing $H^{\circ}$. We claim
that if ${O}_{H}$ is a component where the index attains its largest value
then at least one point in $f^{-1}(\xi)$ is positive for $H$.

To prove this claim, we need an auxiliary construction. Assertion (2) of
Proposition 7 implies that there exists $\varepsilon>0$ such that the inverse
image $V_{\varepsilon,\xi}=f^{-1}(D_{\varepsilon}(\xi))$, where
$D_{\varepsilon}(\xi)$ is an open disk of radius $\varepsilon$ centered at
$\xi$, is an (at most countable) collection of open $f$-simple arcs. By
assertion (1) of Proposition 7, $f^{-1}(\xi)$ is finite so that only a finite
number of these arcs intersect $f^{-1}(\xi)$. Let $k$ be the cardinality of
$f^{-1}(\xi)$, let $x_{1}$, $\dots$, $x_{k}$ be the points of $f^{-1}(\xi)$,
and let $\gamma_{1}$, …, $\gamma_{k}$, where $\gamma_{i}\ni x_{i}$, be the
closures of $f$-simple arcs of $V_{\varepsilon,\xi}$ intersecting
$f^{-1}(\xi)$. Passing to a smaller $\varepsilon$ if necessary, we may assume
that for each $i\in\\{1,\dots,k\\}$, the closed arc $\gamma_{i}$ is $f$-simple
as well and $f(\gamma_{i})$ is a closed simple arc (not a loop) with two
distinct endpoints on the circle $\partial D_{\varepsilon}(\xi)$. Observe that
for each $i\in\\{1,\dots,k\\}$, $f(\gamma_{i})$ splits $D_{\varepsilon}(\xi)$
into two parts, each of which is homeomorphic to $D_{\varepsilon}(\xi)$ by the
Jordan–Schoenflies theorem. Let $D_{i}$ denote that one of these two parts
which does not intersect $H$, and let $J_{i}$ be the closed arc $\partial
D_{i}\cap\partial D_{\varepsilon}(\xi)$ ($J_{i}$ is an arc of $\partial
D_{\varepsilon}(\xi)$ and has the same endpoints as $f(\gamma_{i})$).

Set $\delta_{i}=-1$ if $x_{i}$ is positive for $H$ and $\delta_{i}=1$
otherwise. It is straightforward to check that if
$f_{1}\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ is a curve such that
$f_{1}(\gamma_{1})=J_{1}$ and $f_{1}(t)=f(t)$ for
$t\in\mathbb{S}^{1}\setminus\gamma_{1}$, then for an arbitrary point
$q\in\mathbb{R}^{2}\setminus(f(\mathbb{S}^{1})\cup f_{1}(\mathbb{S}^{1}))$ we
have

$\operatorname{wind}(q,f_{1})=\begin{cases}\operatorname{wind}(q,f)+\delta_{1}&\text{if}\leavevmode\nobreak\
q\in D_{1},\\\ \operatorname{wind}(q,f)&\text{if}\leavevmode\nobreak\ q\notin
D_{1}.\end{cases}$

Applying this argument to all arcs $f(\gamma_{i})$ we see that if $p$ is a
point in ${O}_{H}$ and $q$ is a point in $\mathbb{R}^{2}\setminus
f(\mathbb{S}^{1})$ such that $p$ and $q$ are joined by a path that intersects
$f(\mathbb{S}^{1})$ only at points of $f(\gamma_{1}\cup\dots\cup\gamma_{k})$
then

(1) $\operatorname{wind}(q,f)=\operatorname{wind}(p,f)+\sum_{\\{i\mid q\in
D_{i}\\}}\delta_{i}.$

Observe that if $\varepsilon_{1}>0$ is sufficiently small then the open disk
$D_{\varepsilon_{1}}(\xi)$ of radius $\varepsilon_{1}$ centered at $\xi$
intersects $O_{H}$ and $D_{1}$ and does not intersect
$f(\mathbb{S}^{1})\setminus f(\gamma_{1}\cup\dots\cup\gamma_{k})$. Then it
follows by (1) in an obvious way that if ${O}_{H}$ is a component where the
index attains its largest value then at least one of $\delta_{i}$’s is
negative so that at least one of $x_{i}$’s is positive for $H$. The claim is
thus proved.

We pass to the second part of the proof for case (C2).

Assertions (4) and (5) of Proposition 7 imply that there exists a bounded
connected component of $\mathbb{R}^{2}\setminus f(\mathbb{S}^{1})$ where the
index attains its extreme value. Let $\mathcal{O}$ be such a component. By
reversing the orientation of $\mathbb{S}^{1}$ if necessary we get that in
$\mathcal{O}$ the index attains the largest value.222Notice that if our curve
with its initial orientation is, e. g., a Euclidean circle oriented clockwise,
then there is no bounded connected component of $\mathbb{R}^{2}\setminus
f(\mathbb{S}^{1})$ where the index attains its largest value (which is $0$ in
this case).

Observe that $\mathcal{O}$ is simply connected because it is a bounded
connected component of $\mathbb{R}^{2}\setminus f(\mathbb{S}^{1})$ while
$f(\mathbb{S}^{1})$ is compact and connected (see Proposition 6). Then Lemma 4
implies that there exists a ruled sequence of good chords in
$\overline{\mathcal{O}}$. Let $(H_{i})_{i=1}^{\infty}$ be such a sequence. For
each $i$, let $\xi_{i}$ and $\zeta_{i}$ be the endpoints of $H_{i}$ listed in
an arbitrary order. The claim in the above part of the proof implies that if
$H$ is a chord for $\overline{\mathcal{O}}$ and $\xi$ is an endpoint of $H$,
then at least one point in $f^{-1}(\xi)$ is positive for $H$. For each $i$,
let $a_{i}\in f^{-1}(\xi_{i})$ and $b_{i}\in f^{-1}(\zeta_{i})$ be any points
positive for $H_{i}$. (Note that $a_{i}$ and $b_{i}$ are spherical
$f$-neighbors because $\xi_{i}$ and $\zeta_{i}$ are the endpoints of a good
chord.)

Passing to subsequences if necessary, we may and will assume that each of the
sequences $(a_{i})_{i=1}^{\infty}$ and $(b_{i})_{i=1}^{\infty}$ is

  * (A1)

convergent (since $\mathbb{S}^{1}$ is compact),

  * (A2)

contained in a closed $f$-simple arc (since $f$ is assumed to be locally-
simple),

  * (A3)

monotone in a containing $f$-simple arc (since each infinite numerical
sequence contains an infinite monotone subsequence).

(The correctness of transition to subsequences is ensured by the fact that
subsequences of ruled sequences are ruled as well.)

Let $J_{a}\subset\mathbb{S}^{1}$ and $J_{b}\subset\mathbb{S}^{1}$ be closed
$f$-simple arcs containing $(a_{i})_{i=1}^{\infty}$ and
$(b_{i})_{i=1}^{\infty}$ respectively and such that $(a_{i})_{i=1}^{\infty}$
is monotone in $J_{a}$ and $(b_{i})_{i=1}^{\infty}$ is monotone in $J_{b}$.

We distinguish two subcases of (C2):

  1. (C2.1)

Either $(a_{i})_{i=1}^{\infty}$ or $(b_{i})_{i=1}^{\infty}$ is eventually
constant.

  2. (C2.2)

None of $(a_{i})_{i=1}^{\infty}$ and $(b_{i})_{i=1}^{\infty}$ is eventually
constant.

If one of $(a_{i})_{i=1}^{\infty}$ and $(b_{i})_{i=1}^{\infty}$ is eventually
constant then the second one is not (because $(H_{i})_{i=1}^{\infty}$ is not
eventually constant by construction), whence it follows that
$\\{d(a_{i},b_{i})\\}_{i=1}^{\infty}$ is infinite. Then $\Omega_{f}^{sph}$ is
infinite since $\Omega_{f}^{sph}$ contains
$\\{d(a_{i},b_{i})\\}_{i=1}^{\infty}$ because $a_{i}$ and $b_{i}$ are
spherical $f$-neighbors.

The rest of the proof deals with subcase (C2.2).

In this subcase, since any monotone sequence is either eventually constant or
contains a strictly monotone subsequence, passing again to subsequences if
necessary we may and will assume in addition that

  * (A4a)

$(a_{i})_{i=1}^{\infty}$ is strictly monotone in $J_{a}$,

  * (A4b)

$(b_{i})_{i=1}^{\infty}$ is strictly monotone in $J_{b}$.

Now, we have two subsubcases of (C2.2):

  1. (C2.2.1)

either $(a_{i})_{i=1}^{\infty}$ is strictly increasing (in $J_{a}$ with
respect to the fixed orientation of $\mathbb{S}^{1}$) and
$(b_{i})_{i=1}^{\infty}$ is strictly decreasing (in $J_{b}$ with respect to
the fixed orientation of $\mathbb{S}^{1}$) or vice versa,

  2. (C2.2.2)

$(a_{i})_{i=1}^{\infty}$ and $(b_{i})_{i=1}^{\infty}$ are either both strictly
increasing or both strictly decreasing.

In case (C2.2.1) a straightforward geometric analysis shows that
$\\{d(a_{i},b_{i})\\}_{i=1}^{\infty}$ is infinite. Then $\Omega_{f}^{sph}$ is
infinite since $\Omega_{f}^{sph}$ contains
$\\{d(a_{i},b_{i})\\}_{i=1}^{\infty}$ because $a_{i}$ and $b_{i}$ are
spherical $f$-neighbors.

Thus our proof boils down to the very specific subcase (C2.2.2). We will show
that this subcase is empty, i. e., no continuous map
$f\colon\mathbb{S}^{1}\to\mathbb{R}^{2}$ satisfies all conditions determining
this subcase. Denote by $\mathcal{A}$ the closed subarc of the simple arc
$f(J_{a})$ with endpoints at $f(a_{1})$ and $f(a_{3})$, and let $\mathcal{B}$
be the closed subarc of $f(J_{b})$ with endpoints at $f(b_{1})$ and
$f(b_{3})$.

We split (C2.2.2) into two subcases:

  1. (C2.2.2.1)

$\mathcal{A}\cap\mathcal{B}=\emptyset$,

  2. (C2.2.2.2)

$\mathcal{A}\cap\mathcal{B}\neq\emptyset$.

In the first subcase (C2.2.2.1) observe that the set

$Q(\mathcal{A},\mathcal{B})=\mathcal{A}\cup\mathcal{B}\cup H_{1}\cup H_{3}$

is a Jordan curve and hence by the Jordan–Schoenflies theorem there is a
homeomorphism $\psi\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ that maps
$Q(\mathcal{A},\mathcal{B})$ to a Euclidean circle $\mathbb{S}_{0}^{1}$.

Let $r_{a}$ and $r_{b}$ denote the orientations of $\psi(\mathcal{A})$ and
$\psi(\mathcal{B})$ determined by the orderings/directions
$\psi(f(a_{1}))\to\psi(f(a_{2}))\to\psi(f(a_{3}))$ and
$\psi(f(b_{1}))\to\psi(f(b_{2}))\to\psi(f(b_{3}))$ respectively.

Then $r_{a}$ and $r_{b}$ induce opposite orientations on $\mathbb{S}^{1}_{0}$.
This means in particular that

  * (P1)

if $X$ is a component of $\mathbb{R}^{2}\setminus\mathbb{S}^{1}_{0}$ then
either $X$ adjoins $\psi(\mathcal{A})$ on the left side when we move along
$\psi(\mathcal{A})$ according to $r_{a}$ and $X$ adjoins $\psi(\mathcal{B})$
on the right side when we move along $\psi(\mathcal{B})$ according to $r_{b}$,
or vice versa.

At the same time, since we address subcase (C2.2.2), where
$(a_{i})_{i=1}^{\infty}$ and $(b_{i})_{i=1}^{\infty}$ are either both strictly
increasing or both strictly decreasing, it follows that $r_{a}$ and $r_{b}$
induce, via the map $\psi\circ f$, one and the same orientation on
$\mathbb{S}^{1}$. Therefore, since $a_{2}$ and $b_{2}$ are positive for
$H_{2}$ by construction, it follows by the definition of positive points that

  * (P2)

either $\psi(H_{2})$ adjoins both $\psi(\mathcal{A})$ and $\psi(\mathcal{B})$
on the left side with respect to $r_{a}$ and $r_{b}$, or $\psi(H_{2})$ adjoins
both $\psi(\mathcal{A})$ and $\psi(\mathcal{B})$ on the right side with
respect to $r_{a}$ and $r_{b}$.

However this is impossible: conditions (P1) and (P2) are incompatible because
$\psi(H_{2})$ intersects only one component of
$\mathbb{R}^{2}\setminus\mathbb{S}^{1}_{0}$.

This shows that subcase (C2.2.2.1) is empty.

In the second subcase (C2.2.2.2) where $\mathcal{A}$ and $\mathcal{B}$
intersect and it is possible that no Jordan curve contains them, we will study
two Jordan curves. Let $m_{i}$ denote the middle point of $H_{i}$, and let
$\tau$ be a simple polygonal path in $\mathcal{O}$ with endpoints $m_{1}$ and
$m_{3}$ and such that $\tau\cap H_{i}=\\{m_{i}\\}$ for $i\in\\{1,2,3\\}$. Let
$H_{i}^{a}$ denote the half of $H_{i}$ joining $f(a_{i})$ and $m_{i}$, and let
$H_{i}^{b}=\overline{H_{i}\setminus H_{i}^{a}}$. Observe that the sets

$Q(\mathcal{A},\tau)=\mathcal{A}\cup\tau\cup H_{1}^{a}\cup H_{3}^{a}$

and

$Q(\tau,\mathcal{B})=\tau\cup\mathcal{B}\cup H_{1}^{b}\cup H_{3}^{b}$

are Jordan curves. Applying argument of subcase (C2.2.2.1) to these Jordan
curves and using their common arc $\tau$ we arrive at a similar set of
incompatible conditions (details are left to the reader), which shows that
subcase (C2.2.2.2) is empty as well. This completes the proof of Theorem 4. ∎

######  Corollary 1.

Let $S^{1}$ be a topological circle, let $d$ be an intrinsic metric on $S^{1}$
compatible with its topology, and let $f\colon S^{1}\to\mathbb{R}^{2}$ be a
continuous map. Then $\Omega_{f}^{sph}$ for $(S^{1},d)$ is infinite.

###### Proof..

This follows from Theorem 4 because any intrinsic metric on $S^{1}$ is
proportional to the metric induced by the angular metric on a Euclidean circle
$\mathbb{S}^{1}\subset\mathbb{R}^{2}$ via a homeomorphism
$S^{1}\to\mathbb{S}^{1}$. ∎

##  4\. A quantitative generalisation of the Hopf theorem

Let $n$ be a positive integer. The Hopf theorem states that if $M$ is a
compact Riemannian manifold of dimension $n$ and $f\colon M\to\mathbb{R}^{n}$
is a continuous map, then for an arbitrary $\delta>0$, there are points $a$
and $b$ in $M$ such that they are joined by a geodesic of length $\delta$ and
$f(a)=f(b)$.

In this section we prove that if in addition to the assumptions of the Hopf
theorem we assume that $n>1$ and no two points in $M$ (not necessarily
distinct) are joined by an infinite number of geodesics of length $\delta>0$,
then $M\times M$ contains uncountably many pairs of points such that $f$ takes
each pair to a singleton and points in each pair are joined by a geodesic of
length $\delta$.

First we recall Hopf’s original proof of his theorem and then we prove our
theorem using the main idea of Hopf’s proof.

###### Proof..

Without loss of generality we can suppose that in $M$ there are no closed
geodescis of length $\delta$.

Let $p\in M$, and let $T_{p}M$ be the tangent space at $p$. Choose a standard
basis in $T_{p}M$, and let $\mathbb{S}_{p}^{n-1}\subset T_{p}M$ be the unit
sphere. It is well known that for a given tangent vector $\mathfrak{F}\in
T_{p}M$, there exists a unique geodesic with the tangent vector $\mathfrak{F}$
at $p$. Thus for each $\mathfrak{F}\in\mathbb{S}_{p}^{n-1}$, there is a unique
geodesic, which we denote by $\gamma_{\mathfrak{F}}$.

We will denote by $\gamma_{\mathfrak{F}}(s)$, where $s\in\mathbb{R}$, the
geodesic $\gamma_{\mathfrak{F}}$ with a natural parametrization, and we will
suppose that $\gamma_{\mathfrak{F}}(0)=p$ and
$\frac{d}{ds}(\gamma_{\mathfrak{F}}(s))|_{s=0}=\mathfrak{F}$ for each
$\mathfrak{F}\in\mathbb{S}_{p}^{n-1}$. Observe that $\gamma_{\mathfrak{F}}$
and $\gamma_{\mathfrak{-F}}$ are the same geodesics, but
$\gamma_{\mathfrak{F}}(s)$ and $\gamma_{\mathfrak{-F}}(s)$ differ in the
direction of moving when $s$ increases.

Study the vector
$V(\mathfrak{F})_{s,p,\delta}:=f(\gamma_{\mathfrak{F}}(s+\delta))-f(\gamma_{\mathfrak{F}}(s))$.
Suppose that the statement of Theorem 1 is not true, and let
$v(\mathfrak{F})_{s,p,\delta}=\frac{V(\mathfrak{F})_{s,p,\delta}}{|V(\mathfrak{F})_{s,p,\delta}|}$,
where $|V(\mathfrak{F})_{s,p,\delta}|$ denotes the Euclidean norm of
$V(\mathfrak{F})_{s,p,\delta}$ in $\mathbb{R}^{n}$. We get continuous maps
$T_{s,p,\delta}\colon\mathbb{S}_{p}^{n-1}\to\mathbb{S}^{n-1}$ for each $p\in
M$ and $s\in\mathbb{R}$, where $\mathbb{S}^{n-1}$ is the Euclidean sphere.
Note that $T_{-\delta/2,p,\delta}$ is antipodal and hence the modulo 2 degree
of $T_{-\delta/2,p,\delta}$ is 1 for each $p\in M$.

Denote by $f_{1},\dots,f_{n}$ coordinate functions of $f$. By compactness
arguments there is $\xi\in M$ such that $f_{n}(\xi)=\max_{p\in M}{f_{n}(p)}$.
This means that $T_{0,\xi,\delta}(\mathbb{S}_{\xi}^{n-1})$ belongs to the
hemisphere $x_{n}\leq 0$ and hence the modulo 2 degree of $T_{0,\xi,\delta}$
is zero. But there is a homotopy between $T_{0,\xi,\delta}$ and
$T_{-\delta/2,\xi,\delta}$ by $s$. This is a contradiction. ∎

######  Definition 8.

Let $n$ be a positive integer, and let $\delta$ be a positive real number. Let
$M$ be a compact Riemannian manifold of dimension $n$, and let $f\colon
M\to\mathbb{R}^{n}$ be a continuous map. We denote by $\mathcal{F}(\delta)$
the subset of $M\times M$ such that $\\{a,b\\}\in\mathcal{F}(\delta)$ if and
only if $f(a)=f(b)$ and the points $a$ and $b$ are joined by a geodesic of
length $\delta$.

We call points $a,b\in M$ $\delta$-_conjugate_ if they are joined by an
infinite number of geodesics of length $\delta$. We denote the set of such
points by $\mathcal{C}(\delta)$.

######  Theorem 5.

Let $n$ be a positive integer such that $n>1$, and let $\delta$ be a positive
real number. Let $M$ be a compact Riemannian manifold of dimension $n$, and
let $f\colon M\to\mathbb{R}^{n}$ be a continuous map. If $\mathcal{C}(\delta)$
is empty, then $\mathcal{F}(\delta)$ is uncountable.

###### Proof..

If $f(M)$ is a singleton, then the result easily follows. Otherwise
$\partial{f(M)}$ is uncountable (since $f(M)$ is bounded and connected).

If no $\xi$ in $\partial f(M)$ yields $(f^{-1}(\xi)\times f^{-1}(\xi))\cap
F(\delta)=\emptyset$, then the statement easily follows.

Otherwise take $\xi\in\partial f(M)$ such that $(f^{-1}(\xi)\times
f^{-1}(\xi))\cap F(\delta)=\emptyset$ and let $q$ be a point in $f^{-1}(\xi)$.

Let $G_{\nu,q}$ be the radial projection with center at $\nu\in\mathbb{R}^{n}$
on a Euclidean sphere $\mathbb{S}^{n-1}$ centered at $\nu$. Denote by
$\mathcal{S}_{q}^{n-1}$ the sphere of geodesic radius $\delta$ centered at
$q$, and let $\mathcal{O}$ be the connected component of
$\mathbb{R}^{n}\setminus f(\mathcal{S}_{q}^{n-1})$ such that
$f(q)\in\mathcal{O}$ (such a component exists since otherwise
$(f^{-1}(\xi)\times f^{-1}(\xi))\cap F(\delta)\neq\emptyset$).

There exists $\varepsilon>0$ such that
$U_{\varepsilon}(f(q))\subset\mathcal{O}$, where $U_{\varepsilon}(f(q))$ is
the $\varepsilon$-neighborhood of $f(q)$ in $\mathbb{R}^{n}$. Since $f(q)$ is
in $\partial{f(M)}$, it follows that there exists a point $\mathcal{B}\in
U_{\varepsilon}(f(q))$ such that $\mathcal{B}\notin f(M)$. Since
$f(\mathcal{S}_{q}^{n-1})$ is contractible in $f(M)$, it is contractible in
$\mathbb{R}^{n}\setminus\mathcal{B}$ as well. Hence the degree of
$G_{\mathcal{B},q}\circ f\colon\mathcal{S}_{q}^{n-1}\to\mathbb{S}^{n-1}$ is
zero as well as the degree of $G_{\xi,q}\circ
f\colon\mathcal{S}_{q}^{n-1}\to\mathbb{S}^{n-1}$ and hence of
${T}_{0,q,\delta}$ (see the proof of the Hopf theorem for notation).

From the continuity of $f$ it follows that there is $\varepsilon_{1}>0$ such
that for each $p\in U_{\varepsilon_{1}}(q)$, the degree of ${T}_{0,p,\delta}$
is zero, where $U_{\varepsilon_{1}}(q)$ is the $\varepsilon_{1}$-neighborhood
of $q$ in $M$. First observe that no at most countable collection of geodesics
in $M$ covers $U_{\varepsilon_{1}}(q)$, since any geodesic has zero
$n$-dimensional Lebesgue measure. Second observe that through each point $p$
in $U_{\varepsilon_{1}}(q)$ passes a geodesic $\gamma_{p}$ containing points
in some pair in $\mathcal{F}(\delta)$ (the distance between points in this
pair along $\gamma_{p}$ is $\delta$). Indeed, otherwise we could use the Hopf
trick and get a contradiction.

Now if $\mathcal{F}(\delta)$ is at most countable, then there is a pair of
points $\\{\mu_{1},\mu_{2}\\}\in\mathcal{F}(\delta)$ such that $\mu_{1}$ and
$\mu_{2}$ are joined by uncountably many geodesics of length $\delta$ and
hence $\mathcal{C}(\delta)$ is not empty. This completes the proof. ∎

## References

  * [1] D. Attali, J-D. Boissonnat, H. Edelsbrunner, Stability and computation of medial axes: a state-of-the-art report, In: Mathematical Foundations of Scientific Visualization, Computer Graphics, and Massive Data Exploration, Springer, Berlin, 2009, pp. 109–125.
  * [2] H. Hopf, Eine Verallgemeinerung bekannter Abbildungs- und Überdeckungssätze, Portugal. Math., 4 (1944), 129–139.
  * [3] A. Lieutier, Any open bounded subset of $\mathrm{R}^{n}$ has the same homotopy type as its medial axis, Computer-Aided Design, 36 (2004), 1029–1046.
  * [4] A. V. Malyutin, O. R. Musin, Neighboring mapping points theorem, arXiv:1812.10895, 2022, 1–21.
  * [5] G. Matheron, Examples of topological properties of skeletons, In: Image Analysis and Mathematical Morphology. Vol. 2: Theoretical Advances, Academic Press, London, 1988, pp. 217–238.
  * [6] W. Thurston, A comment on https://mathoverflow.net/questions/57766/why-are-there-no-wild-arcs-in-the-plane. Accessed August 15, 2022.
  * [7] G. T. Whyburn, On locally simple curves, Bull. Amer. Math. Soc., 53 (1947), 986–992.