# Automorphisms of some variants of fine graphs

Frédéric Le Roux  Sorbonne Universités, UPMC Univ. Paris 06, Institut de
Mathématiques de Jussieu-Paris Rive Gauche, UMR 7586, CNRS, Univ. Paris
Diderot, Sorbonne Paris Cité, 75005 Paris, France<EMAIL_ADDRESS>and  Maxime Wolff  Sorbonne Universités, UPMC Univ. Paris 06, Institut de
Mathématiques de Jussieu-Paris Rive Gauche, UMR 7586, CNRS, Univ. Paris
Diderot, Sorbonne Paris Cité, 75005 Paris, France<EMAIL_ADDRESS>
###### Abstract.

Recently Bowden, Hensel and Webb defined the fine curve graph for surfaces,
extending the notion of curve graphs for the study of homeomorphism or
diffeomorphism groups of surfaces. Later Long, Margalit, Pham, Verberne and
Yao proved that for a closed surface of genus $g\geqslant 2$, the automorphism
group of the fine graph is naturally isomorphic to the homeomorphism group of
the surface. We extend this result to the torus case $g=1$; in fact our method
works for more general surfaces, compact or not, orientable or not. We also
discuss the case of a smooth version of the fine graph.

##  1\. Introduction

###  1.1. Context and results

For a connected, compact surface $\Sigma_{g}$ of genus $g\geqslant 1$, Bowden,
Hensel and Webb [2] recently introduced the fine curve graph
$\mathcal{C}^{\dagger}(\Sigma)$, as the graph whose vertices are all the
essential closed curves on $\Sigma$, with an edge between two vertices $a$ and
$b$ whenever $a\cap b=\emptyset$, if $g\geqslant 2$, and whenever $|a\cap
b|\leqslant 1$ if $g=1$. They proved that for every $g\geqslant 1$, the graph
$\mathcal{C}^{\dagger}(\Sigma)$ is hyperbolic, and derived a construction of
an infinite dimensional family of quasi-morphisms on
$\operatorname{Homeo}_{0}(\Sigma)$, thereby answering long standing questions
of Burago, Ivanov and Polterovich.

The ancestor of the fine graph is the usual curve complex of a surface
$\Sigma$, i.e., the complex whose vertices are the isotopy classes of
essential curves, with an edge (or a simplex, more generally) between some
vertices if and only if they have disjoint representatives. Since its
introduction by Harvey [5], the curve complex of a surface has been an
extremely useful tool for the study of the mapping class group
$\mathrm{Mod}(\Sigma)$ of that surface, as it acts on it naturally. In
particular, the fact that this complex is hyperbolic, discovered by Masur and
Minsky, has greatly improved the understanding of the mapping class groups
(see [12, 11]). The result of Bowden, Hensel and Webb, promoting the
hyperbolicity of the curve complex to that of the fine curve graph, opens the
door both to the study of what classical properties of usual curve complexes
have counterparts in the fine curve graph, and to the use of this graph to
derive properties of homeomorphism groups. A first step in this direction was
taken by Bowden, Hensel, Mann, Militon and Webb [1], who explored the metric
properties of the action of $\operatorname{Homeo}(\Sigma)$ on this hyperbolic
graph.

A classical theorem by Ivanov [7] states that, when $\Sigma$ is a closed
surface of genus $g\geqslant 2$, the natural map
$\mathrm{Mod}(\Sigma)\to\mathrm{Aut}(\mathcal{C}(\Sigma))$ is an isomorphism.
Recently Long, Margalit, Pham, Verberne and Yao [10] proved the following
natural counterpart of Ivanov’s theorem for fine graphs: provided $\Sigma$ is
a compact orientable surface of genus $g\geqslant 2$, the natural map

$\operatorname{Homeo}(\Sigma)\longrightarrow\mathrm{Aut}(\mathcal{C}^{\dagger}(\Sigma))$

is an isomorphism. They also suggested that this map (with the appropriate
version of $\mathcal{C}^{\dagger}$) may also be an isomorphism when $g=1$, and
conjectured that the automorphism group of the fine curve graph of smooth
curves, should be nothing more than $\mathrm{Diff}(\Sigma)$.

In this article, we address both these questions. Our motivation originates
from the case of the torus: excited by [1], we wanted to understand more
closely the relation between the rotation set of homeomorphisms isotopic to
the identity and the metric properties of their actions on the fine graph.
This subject will be treated in another article, joint with Passeggi and
Sambarino [9]. The methods developed in the present article are valid not only
for the torus but for a large class of surfaces.

We work on nonspherical surfaces (i.e., surfaces not embeddable in the
$2$-sphere, or equivalently, containing at least one nonseparating simple
closed curve), orientable or not, compact or not. We consider the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, whose vertices are the
nonseparating simple closed curves, and with an edge between two vertices $a$
and $b$ whenever they are either disjoint, or have exactly one, topologically
transverse intersection point (see the beginning of Section 2 for more
detail). Our first result answers a problem raised in [10].

######  Theorem 1.1.

Let $\Sigma$ be a connected, nonspherical surface, without boundary. Then the
natural map
$\operatorname{Homeo}(\Sigma)\to\mathrm{Aut}(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$
is an isomorphism.

Our second result concerns the smooth version of fine graphs. We consider the
graph $\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$ whose vertices are
the smooth nonseparating curves in $\Sigma$, with an edge between $a$ and $b$
if they are disjoint or have one, transverse intersection point, in the
differentiable sense (in particular,
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$ is not the subgraph of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ induced by the vertices
corresponding to smooth curves: it has fewer edges). The following result
partially confirms the conjecture of [10]; here we restrict to the case of
orientable surfaces for simplicity.

######  Theorem 1.2.

Let $\Sigma$ be a connected, orientable, nonspherical surface, without
boundary. Then all the automorphisms of
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$ are realized by
homeomorphisms of $\Sigma$.

In other words, if we denote by
$\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)$ the subgroup of
$\operatorname{Homeo}(\Sigma)$ preserving the collection of smooth curves and
preserving transversality, then the natural map

$\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)\longrightarrow\mathrm{Aut}(\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma))$

is an isomorphism. We were surprised to realize however that
$\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)$ is strictly larger than
$\mathrm{Diff}(\Sigma)$.

######  Proposition 1.3.

Every surface $\Sigma$ admits a homeomorphism $f$ such that $f$ and $f^{-1}$
preserve the set of smooth curves, and preserve transversality, but such that
neither $f$ nor $f^{-1}$ is differentiable. In particular, the natural map

$\mathrm{Diff}(\Sigma)\longrightarrow\mathrm{Aut}(\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma))$

is not surjective.

###  1.2. Idea of the proof of Theorem 1.1

The main step in this proof is the following.

######  Proposition 1.4.

If $\\{a,b\\}$, or $\\{a,b,c\\}$ is a $2$-clique or a $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ then, from the graph structure
of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, we can tell the type of the
clique.

If $\\{a_{1},\ldots,a_{n}\\}$ is an $n$-clique in the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, the homeomorphism type of the
subset $\cup_{j=1}^{n}a_{j}$ of $\Sigma$ will be called the type of the
$n$-clique. We will explore this only for $2$ and $3$-cliques. A $2$-clique
$\\{a,b\\}$, i.e., an edge of the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, may have two distinct types:
the intersection $a\cap b$ may be empty or not. For a $3$-clique
$\\{a,b,c\\}$, up to permuting the curves $a$, $b$ and $c$, the cardinals of
the intersections $a\cap b$, $a\cap c$ and $b\cap c$, respectively, may be
$(1,1,1)$, or $(1,1,0)$, or $(1,0,0)$, or $(0,0,0)$. This determines the type
of the $3$-clique, except in the case $(1,1,1)$, where the intersection points
$a\cap b$, $a\cap c$ and $b\cap c$ may be pairwise distinct, in which case we
will speak of a $3$-clique of type necklace, or these intersection points may
be equal, in which case we will speak of a $3$-clique of type bouquet, see
Figure 1.

import geometry;

picture Bouquet, Collier; real L = 30, l = 12; path boucle =
(0,0)dir(30)..L*dir(60)..dir(-90)(0,0); draw (Bouquet,
(0,0)dir(30)..L*dir(60)..dir(-90)(0,0)); draw (Bouquet, rotate(120)*boucle,
(0,0)); draw (Bouquet, rotate(-120)*boucle, (0,0)); add(Bouquet, (0,0));

path cercle = l*dir(0)dir(0)..l*dir(120)dir(300)..cycle; draw (Collier,
cercle); draw (Collier, rotate(120)*cercle, (0,0)); draw (Collier,
rotate(-120)*cercle, (0,0)); add(Collier, (180,0));

Figure 1. A bouquet (left) and a necklace (right) of three circles.

The main bulk of the proof of Proposition 1.4 consists in distinguishing the
$3$-cliques of type necklace from any other $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. Here, the key is that among all
the $3$-cliques, the cliques $\\{a,b,c\\}$ of type necklace are exactly those
such that the union $a\cup b\cup c$ contains nonseparating simple closed
curves other than $a$, $b$ and $c$. In terms of the graph structure, this
leads to the following property, denoted by $N(a,b,c)$, which turns out to
characterize these cliques:

There exists a finite set $F$ of at most $8$ vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, all distinct from $a$, $b$ and
$c$, such that every vertex $d$ connected to $a$, $b$ and $c$ in this graph,
is connected to at least one element of $F$.

From this, we will easily characterize all the configurations of $2$-cliques
and $3$-cliques in terms of similar statements in the first order logic of the
graph $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$.

Now, let $T(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$ denote the set of
edges $\\{a,b\\}$ of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ satisfying
$|a\cap b|=1$. Then we have a map

$\mathrm{Point}\colon\
T(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))\to\Sigma,$

which to each edge $\\{a,b\\}$ of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, associates the intersection
point $a\cap b$. The next step in the the proof of Theorem 1.1 now consists in
characterizing the equality $\mathrm{Point}(a,b)=\mathrm{Point}(c,d)$ in terms
of the structure of the graph. This characterization shows that every
automorphism of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ is realized by
some bijection of $\Sigma$; then we prove that such a bijection is necessarily
a homeomorphism (see Proposition 3.1).

In order to characterize the equality
$\mathrm{Point}(a,b)=\mathrm{Point}(c,d)$, we introduce on
$T(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$ the relation $\diamondvert$
generated, essentially (see section 3.2 for details), by
$(a,b)\diamondvert(b,c)$ if $(a,b,c)$ is a $3$-clique of type bouquet.
Obviously, if $(a,b)\diamondvert(c,d)$ then
$\mathrm{Point}(a,b)=\mathrm{Point}(c,d)$. Interestingly, the converse is
false, but we can still use this idea in order to characterize the points of
$\Sigma$ in terms of the graph structure of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$.

This subtlety between the relation $\diamondvert$ and the equality of points
is related to the non smoothness of the curves involved, and more precisely,
to the fact that a curve may spiral infinitely with respect to another curve
in a neighborhood of a common point. We think that this phenomenon is of
independent interest and we investigate it in Section 4. In particular, we can
easily state, in terms of the graph structure of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, an obstruction for a
homeomorphism to be conjugate to a $C^{1}$-diffeomorphism, see Section 4.6.

###  1.3. Ideas of the proof of Theorem 1.2

In the smooth case, the adaptation of our proof of Theorem 1.1 fails from the
start: indeed, the closed curves contained in the union $a\cup b\cup c$ of a
necklace, and distinct from $a$, $b$ and $c$, are not smooth. This suggests
the idea to use sequences of curves (at the expense of losing the
characterizations of configurations in terms of first order logic).

This time it is easiest to first characterize disjointness of curves (see
Lemma 5.7), and then recover the different types of $3$-cliques. Then the
strategy follows the $C^{0}$ case.

Once we start to work with sequences, it is natural to say that a sequence
$(f_{n})$ of curves not escaping to infinity converges to $a$ in some weak
sense, if for every vertex $d$ such that $\\{a,d\\}$ is an edge of the graph,
$\\{f_{n},d\\}$ is also an edge for all $n$ large enough. As it turns out,
this property implies convergence in $C^{0}$-sense to $a$, and is implied by
convergence in $C^{1}$-sense. But it is not equivalent to the convergence in
$C^{1}$-sense, and it is precisely this default of $C^{1}$-convergence that
enables us to distinguish between disjoint or transverse pairs of curves.

Interestingly, this simple criterion for disjointness has no counterpart in
the $C^{0}$-setting. Indeed, in that setting, no sequence of curves converges
in this weak sense: given a curve $a$, and a sequence $(f_{n})$ of curves
with, say, some accumulation point in $a$, we can build a curve $d$
intersecting $a$ once transversally (topologically), but oscillating so much
that it itersects every $f_{n}$ several times. From this perspective, none of
our approaches in the $C^{0}$-setting and in the $C^{\infty}$-setting are
directly adaptable to the other.

###  1.4. Further comments

We can imagine many variants of fine graphs. For example, in the arXiv version
of [2], for the case of the torus they worked with the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ on which we are working here,
whereas in the published version, they changed to a fine graph in which two
curves $a$ and $b$ are still related by an edge when they have one
intersection, not necessarily transverse.

More generally, in the spirit of Ivanov’s metaconjecture, we expect that the
group of automorphisms should not change from any reasonable variant to
another. And indeed, using the ideas of [10, Section 2] and those presented
here, we can navigate between various versions of fine graphs, and recover,
from elementary properties of one version, the configurations defining the
edges in another version, thus proving that their automorphism groups are
naturally isomorphic. From this perspective, it seems satisfying to recover
the group of homeomorphisms of the surface as the automorphism group of any
reasonable variant of the fine graph. In this vein, we should mention that the
results of [10, Section 2] directly yield a natural map
$\mathrm{Aut}(\mathcal{C}^{\dagger}(\Sigma))\to\mathrm{Aut}(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$,
and from there, our proof of Theorem 1.1 may be used as an alternative proof
of their main result.

All reasonable variants of the fine graphs should be quasi-isometric, and a
unifying theorem (yet out of reach today, as it seems to us) would certainly
be a counterpart of the theorem by Rafi and Schleimer [13], which states that
every quasi-isometry of the usual curve graph is bounded distance from an
isometry.

###  1.5. Organization of the article

Section 2 is devoted to the recognition of the $3$-cliques in the
$C^{0}$-setting, and of some other configurations regarding the nonorientable
case. We encourage the reader to skip, at first reading, everything that
concerns the nonorientable case: these points shoud be easily identified, and
this halves the length of the proof. In Section 3 we prove Theorem 1.1. In
Section 4 we characterize, from the topological viewpoint, the relation
$\diamondvert$ introduced above in terms of the graph structure, and deduce
our obstruction to differentiability. Finally in Section 5 we prove Theorem
1.2 and Proposition 1.3.

### Acknowledgments

We thank Kathryn Mann for encouraging discussions, and Dan Margalit for his
extensive feedback on a preliminary version of this manuscript.

##  2\. Recognizing configurations of curves

###  2.1. Standard facts and notation

We will use, often without mention, the following easy or standard facts for
curves on surfaces.

The first is the classification of connected, topological surfaces with
boundary (not necessarily compact). In particular, every topological surface
admits a smooth structure. Given a closed curve $a$ in a surface $\Sigma$, we
can apply this classification to $\Sigma\smallsetminus a$ and understand all
possible configurations of simple curves; this is the so-called change of
coordinates principle in the vocabulary of the book of Farb and Margalit [3].

In particular, every closed curve $a$ has a neighborhood homeomorphic to an
annulus or a Möbius strip in which $a$ is the “central curve”. Very often in
this article, we will consider the curves $a^{\prime}$ obtained by deforming
$a$ in such a neighborhood, so that $a^{\prime}$ is disjoint from $a$ in the
first case, or intersects it once, transversely, in the second, as in Figure
2. We will say that $a^{\prime}$ is obtained by pushing $a$ aside.

The change of coordinates principle also applies to finite graphs embedded in
$\Sigma$: there is a homeomorphism of $\Sigma$ that sends any given graph to a
smooth graph, such that all edges connected to a given vertex leave it in
distinct directions. In the simple case when the graph is the union of two or
three simple closed curves that pairwise intersect at most once, this
observation justifies the description of the possible configurations of
cliques in the introduction. This also enables, provided two curves $a$ and
$b$ intersect at a single point (or more generally at a finite number of
points), to speak of a transverse (also called essential), or to the contrary
inessential, intersection point, as we did in the introduction.

Here are two other useful facts.

######  Fact 2.1.

A simple closed curve $a$ in a surface, is nonseparating if and only if there
exists a closed curve $b$, such that $a\cap b$ is a single point and this
intersection is transverse.

######  Fact 2.2.

Let $p,q$ be two distinct points, and $x,x^{\prime},x^{\prime\prime}$ three
simple arcs, each with end-points $p$ and $q$, such that

$x\cap x^{\prime}=x^{\prime}\cap x^{\prime\prime}=x\cap
x^{\prime\prime}=\\{p,q\\}.$

If two of the three curves $x\cup x^{\prime},x^{\prime}\cup
x^{\prime\prime},x^{\prime\prime}\cup x$ are separating, then the third one is
also separating.

###### Proof.

Denote $y=x\smallsetminus\\{p,q\\}$, the arc $x$ without its ends, and
similarly, define $y^{\prime}$ and $y^{\prime\prime}$. Suppose $x\cup
x^{\prime}$ and $x\cup x^{\prime\prime}$ are separating. Denote by
$\Sigma_{1},\Sigma_{2}$, resp. $\Sigma_{3},\Sigma_{4}$, the components of
$\Sigma\smallsetminus(x\cup x^{\prime})$, resp. $\Sigma\smallsetminus(x\cup
x^{\prime\prime})$, where $\Sigma_{2}$ contains $y^{\prime\prime}$ and
$\Sigma_{4}$ contains $y^{\prime}$. By looking at neighborhoods of $p$ and $q$
(see Figure 2, left), we see that $\Sigma^{\prime}=\Sigma_{2}\cap\Sigma_{4}$
is non empty, and that the arc $y$ bounds $\Sigma_{1}$ on one side, and
$\Sigma_{3}$ on the other, so $\Sigma^{\prime\prime}=\Sigma_{1}\cup
y\cup\Sigma_{3}$ is a surface. Now, $\Sigma\smallsetminus(x^{\prime}\cup
x^{\prime\prime})=\Sigma^{\prime}\cup\Sigma^{\prime\prime}$, and
$\Sigma^{\prime}$ and $\Sigma^{\prime\prime}$ are disjoint by construction. ∎

import geometry;

real R = 25;

dot ((0,0)); label("$p$", (0,0), WNW); draw ((0,0)–R*dir(100));
draw(R*dir(205)–(0,0)–R*dir(-20)); label("$y^{\prime\prime}$", R*dir(100), N);
label("$y$", R*dir(205), W); label("$y^{\prime}$", R*dir(-20), E);

label("$\Sigma_{1}$", 0.8*R*dir(-80)); label("$\Sigma_{3}$", R*dir(150));
label("$\Sigma_{2}\cap\Sigma_{4}$", 0.7*R*dir(80), E); import geometry;

real L = 50, h = 11, e = 3; path contour = (0,0)–(L, 0)–(L, h)–(0, h)–cycle;
fill (contour, lightgray); draw(contour, lightgray+dotted); draw
((0,h/2)–(L,h/2), blue); label("$a$", (0, h/2), W, blue); draw ((0, h/2+e)–(L,
h/2+e), red); label("$a^{\prime}$", (L, h/2+e), E, red); import geometry;

real L = 50, h = 11, e = 3; path contour = (0,0)–(L, 0)–(L, h)–(0, h)–cycle;
fill (contour, lightgray); draw(contour, lightgray+dotted); draw
((0,h/2)–(L,h/2), blue); label("$a$", (0, h/2), W, blue); draw ((0,
h/2+e)..(1/3*L, h/2+e)right..(2/3*L, h/2-e)right..(L, h/2-e), red);
label("$a^{\prime}$", (L, h/2-e), E, red);

Figure 2. Left: a neighborhood of $p$. Center: pushing a two-sided curve $a$.
Right: pushing a one-sided curve $a$.

###  2.2. Properties characterizing geometric configurations

Now we list the properties, in terms of the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, that will be used as
characterizations of certain configurations of curves. This allows us to
specify the statement of Proposition 1.4, which will be proved in the next
paragraph, and define the relation $\diamondvert$ in terms of the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$.

In the following, the letters $N,D,T,B$ respectively stand for necklace,
disjoint, transverse, and bouquet. If $a,b,c$ are vertices of this graph, we
will denote by:

  * $\bullet$

$N(a,b,c)$ the property that $\\{a,b,c\\}$ is a $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ and there exists a finite set
$F$ of at most $8$ vertices of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$,
all distinct from $a$, $b$ and $c$, such that for every vertex $d$ such that
$\\{a,b,c,d\\}$ is a $4$-clique, there is an edge from $d$ to at least one
element of $F$,

  * $\bullet$

$D(a,b)$ the property that $\\{a,b\\}$ is an edge of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ and there does not exist a
vertex $d$ such that $N(a,b,d)$ holds,

  * $\bullet$

$T(a,b)$ the property that $\\{a,b\\}$ is an edge of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ and $D(a,b)$ does not hold,

  * $\bullet$

$B(a,b,c)$ the property that $T(a,b)$, $T(a,c)$, $T(b,c)$ all hold but
$N(a,b,c)$ does not.

The following proposition is the main part of Proposition 1.4.

######  Proposition 2.3.

Let $a$, $b$ and $c$ be vertices of the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. Property $N(a,b,c)$ holds if
and only if $\\{a,b,c\\}$ is a $3$-clique of type necklace of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$.

The proof of Proposition 2.3 will occupy the next paragraph. The following
corollary complements Proposition 2.3 and provides a precise version of
Proposition 1.4.

######  Corollary 2.4.

  * •

Property $D(a,b)$ holds if and only if the curves $a$ and $b$ are disjoint.

  * •

Property $T(a,b)$ holds if and only if $a$ and $b$ have a unique intersection
point and the intersection is transverse.

  * •

Property $B(a,b,c)$ holds if and only if $\\{a,b,c\\}$ is a $3$-clique of type
bouquet.

###### Proof.

Let $a$ and $b$ be neighbors in the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. Of course, if $a$ and $b$ are
disjoint, then $D(a,b)$ holds: there does not exist a curve $d$ such that
$N(a,b,d)$, since this would mean that $\\{a,b,d\\}$ is of type necklace and
then, by definition, $a$ and $b$ would intersect. Conversely, suppose that $a$
and $b$ are not disjoint, and let us prove that $D(a,b)$ does not hold, i.e.,
let us find a curve $c$ such that $\\{a,b,c\\}$ is a $3$-clique of type
necklace. In the case when one of $a$ or $b$ is one-sided, up to exchanging
the two, suppose $a$ is one-sided. Then we may push $a$ in order to find a
curve $c$ which makes a $3$-clique of type necklace with $a$ and $b$, see
Figure 3 (left). In the case when both $a$ and $b$ are two-sided, then by the
change of coordinates principle, a regular neighborhood of $a\cup b$ is
homeomorphic to a one-holed torus, embedded in $\Sigma$, with a choice of
meridian and longitude coming from $a$ and $b$. In this torus, a curve $c$
with slope $1$ will form a $3$-clique of type necklace with $a$ and $b$, see
Figure 3, right.

import geometry;

picture commun, gauche, droite;

path a = (-40, 0)–(40, 0), b = (0, -32)–(0, 32); path c1 = (-40,
7)–(8,7)right..right(28,-7)–(40,-7); path c21 = (-40, 7)–(0, 7)right..up(8,
15)–(8, 32); path c22 = (8, -32)–(8, -4)up..right(20,7)–(40,7);

draw(commun, a, blue); label(commun, "$a$", (40,0),E, blue); draw(commun, b,
red); label(commun, "$b$", (0,32), N, red); draw(gauche, c1, green);
label(gauche,"$c$", (40,-7),SE, green); draw(droite, c21, green); draw(droite,
c22, green); label(droite,"$c$", (40,7),NE, green);

add(gauche, (-90,0)); add(droite, (90,0)); add(commun, (-90,0)); add(commun,
(90,0));

Figure 3. Completing $(a,b)$ to a $3$-clique of type necklace

This proves the first point.

The second point is a straightforward consequence of the first, and the third
simply follows from the second point together with Proposition 2.3. ∎

###  2.3. Proof of Proposition 2.3

The following lemma is a key step in the proof of the direct implication in
Proposition 2.3.

######  Lemma 2.5.

Let $\\{a,b,c\\}$ be a $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ which is not of type necklace.
Then there exists a vertex $d$ of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ such that $\\{a,b,c,d\\}$ is a
$4$-clique of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, and such that $d$
meets every connected component of $\Sigma\smallsetminus(a\cup b\cup c)$.

Before entering the proof, we note that we cannot remove the hypothesis that
$\\{a,b,c\\}$ is not of type necklace. Indeed, in the flat torus
$\Sigma=\mathbb{R}^{2}/\mathbb{Z}^{2}$, consider three closed geodesics
$a,b,c$ respectively directed by $(1,0)$, $(0,1)$, and $(1,1)$. By pushing $c$
aside if necessary, we obtain a $3$-clique of type necklace. The complement of
$a\cup b\cup c$ in $\Sigma$ has three connected components, and there is no
curve $d$ satisfying the conclusion of the lemma.

###### Proof.

In all the proof, we will denote $\Sigma^{\prime}=\Sigma\smallsetminus(a\cup
b\cup c)$. Up to permuting the curves $a$, $b$ and $c$, we may suppose that
the triple of cardinals of intersections, $(|a\cap b|,|a\cap c|,|b\cap c|)$,
equals $(1,1,1)$, or $(1,1,0)$, or $(1,0,0)$, or $(0,0,0)$. We will deal with
these cases separately.

Let us begin with the case $(0,0,0)$. If $\Sigma^{\prime}$ is connected, then
any curve $d$ making a $4$-clique with $(a,b,c)$ satisfies the Lemma. Such a
curve can be found, for example, by pushing $a$ aside. If $\Sigma^{\prime}$
has two connected components, denote them by $\Sigma_{1}$ and $\Sigma_{2}$.
Since $a$, $b$ and $c$ are each nonseparating, at least two of the curves
$a,b,c$ (say, $a$ and $b$) correspond to boundary components of both
$\Sigma_{1}^{\prime}$ and $\Sigma_{2}^{\prime}$. Choose one point $x_{a}$ in
$a$ and one point $x_{b}$ in $b$. For $i=1,2$, there is an arc $\gamma_{i}$
connecting $x_{a}$ to $x_{b}$ in $\Sigma_{i}$, and disjoint from the boundary
of $\Sigma_{i}$ except at its ends. Then the curve
$d=\gamma_{1}\cup\gamma_{2}$ satisfies the lemma (see Figure 4, left). It may
also happen that $\Sigma^{\prime}$ has three connected components, in which
case we find a curve $d$ exactly in the same way, see Figure 4, right.

import geometry;

real r = 1.5;

picture trou, gauche, droite; path batrou = (-7,4)..(0,0)..(7,4); path hautrou
= point(batrou,0.35)..(0,3)..point(batrou,1.65); draw (trou, batrou);
draw(trou, hautrou); add(gauche, scale(r)*trou, r*(0,17)); add(gauche,
scale(r)*trou, r*(0, 33)); add(gauche, scale(r)*trou, r*(-55, 14)); path
contourg = (0,0)left..(-50,-6)..(-80,20)up..(-65, 32).. (0,50)dir(10)..(60,
25)down..cycle; point A = (-8, -28), B = (-28,18), C = (25, 22); path contourd
= Adir(160)..(-65, -14)up..Bdir(70)..Cdir(-45)..(55,-17)down..cycle;

path ga1 = (0,0)dir(135)..(-3.2, 8)..(0,17)dir(45); path ga2 =
(0,0)dir(45)..(3.2, 8)..(0,17)dir(135); path gb1 = (0,20)dir(135)..(-3.2,
27)..(0,33)dir(45); path gb2 = (0,20)dir(45)..(3.2, 27)..(0,33)dir(135); path
gc1 = (0,36)dir(135)..(-3.2,43)..(0,50)dir(45); path gc2 =
(0,36)dir(45)..(3.2,43)..(0,50)dir(135);

draw (gauche, scale(r)*contourg); draw (droite, contourd); draw (gauche,
scale(r)*ga1, blue); draw(gauche, scale(r)*ga2, dotted+blue); label(gauche,
"$a$", r*(-3,8), W, blue); draw (gauche, scale(r)*gb1, red); draw(gauche,
scale(r)*gb2, dotted+red); label(gauche, "$b$", r*(-3,27), W, red); draw
(gauche, scale(r)*gc1, green); draw(gauche, scale(r)*gc2, dotted+green);
label(gauche, "$c$", r*(-3,43), W, green); label(gauche, "$\Sigma_{1}$",
r*(-35, 20)); label(gauche, "$\Sigma_{2}$", r*(33,37)); path gd =
(0,25)dir(200)..(-12,17)..(0,12)dir(20)..(12,20)..cycle; draw(gauche,
scale(r)*gd); label(gauche, "$d$", r*(12,20), dir(-75));

real R = 1.7; path da1 = Adir(135)..(0,-5)dir(10); path da2 =
Adir(10)..(0,-5)dir(135); point PB = (0,-5)+R*point(hautrou, 0.6); point PC =
(0,-5)+R*point(hautrou, 1.4); path db1 = Bdir(30)..PBdir(-75); path db2 =
Bdir(-60)..PBdir(-40); path dc1 = PCdir(75)..Cdir(20); path dc2 =
PCdir(20)..Cdir(80); path dd =
(0,-15)right..(18,0)up..(0,15)left..(-20,0)down..cycle; draw (droite, da1,
blue); draw(droite, da2, dotted+blue); draw (droite, db1, red); draw(droite,
db2, dotted+red); draw (droite, dc1, green); draw(droite, dc2, dotted+green);
draw (droite, dd); label (droite, "$a$", A, S, blue); label(droite, "$b$", B,
NW, red); label (droite, "$c$", C, NE, green); label(droite, "$d$", (18,0),
SE); label (droite, "$\Sigma_{1}$", (-38, -15)); label(droite, "$\Sigma_{2}$",
(0, 26)); label (droite, "$\Sigma_{3}$", (34, -22));

add (droite, scale(R)*trou, (0,-5)); add(gauche, (-80,0)); add(droite,
(120,30));

Figure 4. Finding $d$ in the case $(0,0,0)$

Next we deal with the case of intersections $(1,0,0)$. In this case, $a$ and
$b$ intersect transversely, once, and $c$ is disjoint from $a\cup b$. By
hypothesis, the curve $c$ is (globally) nonseparating. Consider the union
$a\cup b$. If $a$ or $b$ is two-sided, then $a\cup b$ does not disconnect its
regular neighborhoods. This is seen by travelling along a small band on one
side of $a\cup b$, (see Figure 5, left). In this case, $\Sigma^{\prime}$
cannot have more connected components than $\Sigma\smallsetminus c$, hence
$\Sigma^{\prime}$ is connected, and any curve $d$ obtained by pushing $c$, as
in the preceding case, satisfies the lemma. If both $a$ and $b$ are one-sided,
then $a\cup b$ is locally disconnecting, so $\Sigma^{\prime}$ may have up to
two connected components. In this case, a curve $d$ obtained by pushing $a$
satisfies the lemma (See Figure 5, right).

import geometry; picture commun, gauche, droite;

path a = (-40, 0)–(40, 0), b = (0, -32)–(0, 32); path region1 = (-40,
0)–(0,0)–(0,32)–(-8,32)–(-8,7)–(-40,7)–cycle; path region2 =
reflect((0,0),(0,1))*region1; path region3 = rotate(180)*region1, region4 =
rotate(180)*region2;

path d = (-40, 7)–(12,7)right..right(20,-7)–(40,-7);

fill(gauche, region1, mediumgray); fill(gauche, region3, mediumgray);
fill(droite, region1, mediumgray); fill(droite, region3, mediumgray);
fill(gauche, region2, mediumgray); fill(droite, region2, lightred);
fill(gauche, region4, mediumgray); fill(droite, region4, lightred);
draw(commun, a, 1pt+blue); label(commun, "$a$", (40,0),E, blue); draw(commun,
b, 1pt+red); label(commun, "$b$", (0,32), N, red); draw(droite, d,
1.1pt+black); label(droite,"$d$", (-15,7),N);

add(gauche, (-90,0)); add(droite, (90,0)); add(commun, (-90,0)); add(commun,
(90,0));

Figure 5. Finding $d$ in case $(1,0,0)$

Now assume we are in the case $(1,1,1)$ or $(1,1,0)$. Since $\\{a,b,c\\}$ is
not of type necklace, note that in any case $b$ and $c$ do not meet outside
$a$. We first treat the sub-case when $a$ is two-sided. For this we consider
any curve $d$ obtained by pushing $a$ aside, and we claim that $d$ meets every
connected component of $\Sigma^{\prime}$. Indeed, let $C$ be such a component.
Of course the closure of $C$ meets $a$ or $b$ or $c$. Since both $b$ and $c$
meet $a$, it actually has to meet $a$, as we can see by traveling along $b$ or
$c$ in $C$. More precisely, by following $b$ or $c$ in both directions, we see
that $C$ meets any neighborhood of $a$ from both sides. Thus it meets $d$.

It remains to treat the sub-case when $a$ is one-sided, first for the
$(1,1,1)$ case, and then for the $(1,1,0)$ case. In the $(1,1,1)$ case, the
curves $a$, $b$, $c$ play symmetric roles, and by the above argument it just
remains to consider the case when they are all one-sided. Then the situation
is depicted on Figure 6, left: $a\cup b\cup c$ disconnects its regular
neighborhoods into three connected components, and the figure shows a curve
$d$, obtained by pushing $a$ aside, which intersects all three components and
such that $\\{a,b,c,d\\}$ is a 4-clique. In the remaining case the curves $b$
and $c$ play symmetric roles, and there are three different cases to consider,
regarding whether $b$ and $c$ are one or two-sided. These three cases are
pictured in Figure 6, and in each case, we obtain $d$ by pushing $a$ aside.

import geometry;

picture gauche; path a = (-40, 0)–(40, 0), b = (0, -32)–(0, 32), c =
(-40,-32)–(40,32); path region1 = (-40,
0)–(0,0)–(0,32)–(-8,32)–(-8,7)–(-40,7)–cycle; path region2 =
(-40,0)–(-40,-7)–(-17.5,-7)–(-40,-25)–(-40,-32)–(0,0)–cycle; path region3 =
(-40,-32)–(-32,-32)–(-8,-12.5)–(-8,-32)–(0,-32)–(0,0)–cycle; path region4 =
rotate(180)*region1; path region5 = rotate(180)*region2, region6 =
rotate(180)*region3;

path d1 = (-40,7)–(40,7); path d2 = (-40,
7)–(12,7)right..right(20,-7)–(40,-7);

fill(gauche, region1, mediumgray); fill(gauche, region2, lightred);
fill(gauche, region3, lightblue); fill(gauche, region4, mediumgray);
fill(gauche, region5, lightred); fill(gauche, region6, lightblue);
draw(gauche, a, blue); label(gauche, "$a$", (40,0),E, blue); draw(gauche, b,
red); label(gauche, "$b$", (0,32), N, red); draw(gauche, c, green);
label(gauche, "$c$", (40,32), NE, green);

draw(gauche, d2); label(gauche,"$d$", (40,-7),SE); label(gauche, "Case
$(1,1,1)$", (0,-40));

add(gauche, (-150,0)); add(gauche, (-150,0));

//

picture commun, un, deux, trois, quatre, cinq, six; path a = (-50,0)–(50,0), b
= (-20,-25)–(-20,25), c = (20, -25)–(20,25); path d1 = (-50,7)–(50,7); path d2
= (-50, 7)–(-9,7)right..right(9,-7)–(50,-7); path region1 =
(-50,0)–(-20,0)–(-20,25)–(-28,25)–(-28,7)–(-50,7)–cycle; path region2 =
reflect((0,0),(1,0))*region1; path region3 =
(-20,-25)–(-12,-25)–(-12,-7)–(12,-7)–(12,-25)–(20,-25)–(20,0)–(-20,0)–cycle;
path region4 = rotate(180)*region1, region5 = rotate(180)*region2; path
region6 = rotate(180)*region3;

//////////////////fill(quatre, region1, mediumgray); fill(quatre, region2,
mediumgray); fill(quatre, region3, lightred); fill(quatre, region4,
mediumgray); fill(quatre, region5, mediumgray); fill(quatre, region6,
lightred); fill(cinq, region1, mediumgray); fill(cinq, region2, mediumgray);
fill(cinq, region3, mediumgray); fill(cinq, region4, mediumgray); fill(cinq,
region5, mediumgray); fill(cinq, region6, mediumgray); fill(six, region1,
mediumgray); fill(six, region2, mediumgray); fill(six, region3, mediumgray);
fill(six, region4, mediumgray); fill(six, region5, mediumgray); fill(six,
region6, mediumgray);

draw(commun, a, blue); label(commun, "$a$", (50,0), E, blue); draw(commun, b,
red); label(commun, "$b$", (-20,25), N, red); draw(commun, c, green);
label(commun, "$c$", (20,25), N, green); //////draw(quatre, d2); label(quatre,
"$d$", (50,-7), SE); draw(cinq, d2); label(cinq, "$d$", (50,-7), SE);
draw(six, d2); label(six, "$d$", (50,-7), SE); //////label(quatre, "$b$ and
$c$ two-sided", (0,-36)); label(cinq, "$b$ two-sided, $c$ one-sided",
(0,-36)); label(six, "$b$ and $c$ one-sided", (0,-36));
//add(rotate(90)*quatre, (-50, 0)); add(rotate(90)*cinq, (50, 0));
add(rotate(90)*six, (150, 0)); //add(rotate(90)*commun, (-50, 0));
add(rotate(90)*commun, (50, 0)); add(rotate(90)*commun, (150, 0));

Figure 6. Finding $d$ in cases $(1,1,1)$ and $(1,1,0)$

∎

We deduce the following.

######  Lemma 2.6.

Let $\\{a,b,c\\}$ be a $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, not of type necklace. Let
$(\alpha_{1},\ldots,\alpha_{n})$ be a finite family of vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, all distinct from $a$, $b$ and
$c$. Then there exists a vertex $d$ of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, such that

  * $\bullet$

$\\{a,b,c,d\\}$ is a $4$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$,

  * $\bullet$

for all $j\in\\{1,\ldots,n\\}$, the intersection $d\cap\alpha_{j}$ is
infinite; in particular, $\\{d,\alpha_{j}\\}$ is not an edge of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$.

As a corollary, we get the direct implication in Proposition 2.3.

######  Corollary 2.7.

If $\\{a,b,c\\}$ is a $3$-clique not of type necklace, then $N(a,b,c)$ does
not hold.

###### Proof of Lemma 2.6.

The hypotheses that $\\{a,b,c\\}$ is not of type necklace, and
$\alpha_{j}\not\in\\{a,b,c\\}$, impose that for every $j$, the curve
$\alpha_{j}$ is not contained in the union $a\cup b\cup c$. Hence, there
exists a small subarc $\beta_{j}\subset\alpha_{j}$ lying in the complement of
$a\cup b\cup c$, and we may further suppose that these $n$ arcs are pairwise
disjoint, and choose a point $x_{j}$ in $\beta_{j}$ for each $j$.

Now, let $d_{0}$ be a vertex of $\mathcal{NC}_{\leqslant 1}^{\dagger}(\Sigma)$
as from Lemma 2.5. Since $d_{0}$ meets every component of
$\Sigma^{\prime}=\Sigma\smallsetminus(a\cup b\cup c)$, we may perform a
surgery on $d_{0}$, far from $a\cup b\cup c$, to obtain a new curve $d_{1}$
such that $\\{a,b,c,d_{1}\\}$ is still a $4$-clique, and $d_{1}$ still meets
every component of $\Sigma^{\prime}$, and $d_{1}$ passes through $x_{1}$. We
may iterate this process, to get a curve $d_{n}$ which passes through $x_{j}$
for every $j$, and such that $\\{a,b,c,d_{n}\\}$ is a $4$-clique. Finally, we
may perform a last surgery on $d_{n}$, in the neighborhood of all $x_{j}$, in
order to obtain a curve $d$ such that for each $j$, $|\beta_{j}\cap d|$ is
infinite. ∎

It remains to prove the converse implication in Proposition 2.3, which we
restate as Lemma 2.8.

######  Lemma 2.8.

Let $\\{a,b,c\\}$ be a $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ of type necklace. Then there
exists a finite set $F$, of at most 8 vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ all distinct from $a$, $b$ and
$c$, and such that every $d$ such that $\\{a,b,c,d\\}$ is a $4$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ is connected by an edge to some
element of $F$.

In fact this set $F$ can be chosen explicitely, with cardinal at most 8, as
follows. If $\\{a,b,c\\}$ is a $3$-clique of type necklace, then there exists
a family of arcs $(x,X,y,Y,z,Z)$, all embedded in $\Sigma$, such that $a=x\cup
X$, $b=y\cup Y$, $c=z\cup Z$, and such that these six arcs pairwise intersect
at most at their ends. The union $a\cup b\cup c$ may be viewed as a graph
embedded in $\Sigma$, and these six arcs are the edges of this embedded graph.
We let $F$ be the set of nonseparating curves, among the 8 curves $X\cup Y\cup
Z$, $X\cup Y\cup z$, $X\cup y\cup Z$ etc., (there is one choice of upper/lower
case for each letter). In the course of the proof of Lemma 2.8, we will see
that $F$ is nonempty, and satisfies the Lemma.

###### Proof of Lemma 2.8.

Let $\\{a,b,c\\}$ be a $3$-clique of type necklace. Let $F$ be the set of
nonseparating curves, as above, among all the 8 curves $x\cup y\cup z$, $X\cup
y\cup z$, $X\cup Y\cup z$, etc. Let $d$ be such that $\\{a,b,c,d\\}$ is a
$4$-clique. Up to permuting the curves $a$, $b$ and $c$, we may suppose that
$(|a\cap d|,|b\cap d|,|c\cap d|)$ equals $(1,1,1)$, or $(1,1,0)$, or $(1,0,0)$
or $(0,0,0)$ ; our proof proceeds case by case.

The easiest case is $(1,0,0)$. In this case, up to exchanging the arcs $X$ and
$x$, we may suppose that $d$ intersects $a$ at an interior point of $X$, and
is disjoint from all the other arcs. Consider $f=X\cup y\cup z$. This curve
intersects $d$ at a unique point, transversely. It follows that $f$ is
nonseparating. Hence $f\in F$ and $f$ satisfies the conclusion of the lemma.

Now let us deal simultaneously with the cases $(1,1,1)$ and $(1,1,0)$. Suppose
first that the intersections of $d$ with $a\cup b\cup c$ do not occur at the
intersection points $a\cap b$, $a\cap c$ or $b\cap c$. Up to exchanging $x$
with $X$, $y$ with $Y$ and $z$ with $Z$, we may suppose that the intersections
occur in the interior of the arcs $X$, $Y$, and $Z$ in the case $(1,1,1)$, and
in the interior of the arcs $X$ and $Y$ in the case $(1,1,0)$. Now the curve
$f=X\cup y\cup z$, for instance, satisfies the conclusion of the lemma.

Now suppose that $d$ contains one of the points $a\cap b$, $a\cap c$ or $b\cap
c$. In case $(1,1,1)$ we may suppose, up to permuting $a$, $b$ and $c$, that
$d$ contains the point $a\cap b$, and in the case $(1,1,0)$, this is
automatic, as $d$ is disjoint from $c$. Now in any case, $d$ cannot contain
$a\cap c$ nor $b\cap c$, because it intersects $a$ and $b$ only once. Hence,
up to exchanging $z$ with $Z$, we may suppose $d\cap z=\emptyset$. In the
neighborhood of the point $a\cap b$, up to homeomorphism, the configuration of
our curves is as depicted in Figure 7, because all the intersections are
supposed to be transverse. Then, up to exchanging $X$ with $x$ or $Y$ with
$y$, we can suppose that the arc $X\cup Y$ has a transverse intersection with
$d$, and then the arc $f=X\cup Y\cup z$ satisfies the conclusion of the lemma.

import geometry;

path a1 = (-40, 0)–(0, 0), a2 = (0,0)–(40,0); path b1 = (0, -32)–(0, 0), b2 =
(0,0)–(0,32); path d = (-40, -32)–(40, 32); draw (a1, blue); draw(a2,
1.9pt+blue); draw(b1, red); draw(b2, 1.9pt+red); draw(d); label("$a$", (40,
0), E, blue); label("$b$", (0, 32), N, red); label("$d$", (40, 32), NE);
label("$x$", (-30,0), S, blue); label("$X$", (30,0),S, blue); label("$y$",
(0,-19), E, red); label("$Y$", (0,19),E, red);

Figure 7. A suitable choice of $X$ and $Y$

We are left with the case $(0,0,0)$. In this case, any curve in $F$ will
satisfy the conclusion of the lemma, and hence all we have to do is to prove
that $F$ is nonempty. If $X\cup Y\cup Z$ and $X\cup Y\cup z$ were both
separating, then so would be $c=z\cup Z$, by Fact 2.2. Hence, among these two
curves, at least one is nonseparating, and $F$ is nonempty (in fact, it
contains at least 4 elements). ∎

###  2.4. One or two-sided curves, and extra bouquets

In this last paragraph of this section, we will see how to recognize, from the
graph strucutre of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, some
additional configurations. We insist that the work in this paragraph is useful
only in the case when $\Sigma$ is non orientable; it is needed in order to
make our proof of Theorem 1.1 work in that case (see Remark 3.3 below).

We start with a simple characterization of one-sided and two-sided curves.

  * $\bullet$

$\mathrm{Two}(a)$ the property that for all $b$ such that $T(a,b)$ holds,
there exists $c$ such that $T(b,c)$ and $D(a,c)$ both hold,

  * $\bullet$

$\mathrm{One}(a)$ the negation of $\mathrm{Two}(a)$: there exists $b$ such
that $T(a,b)$ and such that there does not exist $c$ satisfying $T(b,c)$ and
$D(a,c)$.

######  Observation 2.9.

Let $a$ be a vertex of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. Then the
curve $a$ is one-sided, if and only if $\mathrm{One}(a)$ holds.

###### Proof.

If $a$ is two-sided, and $b$ satisfying $T(a,b)$, by pushing $a$ aside we find
another curve $c$ as in the definition of $\mathrm{Two}(a)$. This proves the
revers implication.

If $a$ is one-sided, let $b$ be a curve obtained by pushing $a$ aside. We have
$T(a,b)$, and $a$ and $b$ bound a disk. Any curve $c$ disjoint from $a$, and
with $T(b,c)$, has to enter this disk: but then, it has to get out, which is
impossible without touching $a$ and without intersecting $b$ another time. ∎

Our next objective is to characterize when two one-sided curves $a$ and $b$
meet exactly once, non-transversely. We will do this in several steps.

######  Lemma 2.10.

Let $a$, $b$ be one-sided simple curves of $\Sigma$. Suppose the intersection
$a\cap b$ is not connected. Then there exists a vertex $c$ of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, distinct from $a$ and $b$, such
that for every neighbor $d$ of both $a$ and $b$ in this graph, and such that
$D(a,d)$ or $D(b,d)$ (or both), the vertices $c$ and $d$ are neighbors in this
fine graph.

###### Proof.

Let $x$ be a subarc of $b$, whose endpoints lie in $a$, and disjoint from $a$
otherwise. Since $a\cap b$ is disconnected, such an arc exists, and has two
distinct end points, $p$ and $q$. Let $x^{\prime}$ and $x^{\prime\prime}$ be
the two subarcs of $a$ whose ends are $p$ and $q$. From Fact 2.2, we know that
$x\cup x^{\prime}$ or $x\cup x^{\prime\prime}$ (or both) is a nonseparating
curve; denote it by $c$. By construction, we have $c\neq a$ and $c\neq b$.

Now let $d$ be a curve satisfying the hypothesis of the lemma. If $d$ is
disjoint from $a$ and $b$, then it is disjoint from $c$; otherwise $d$
intersects exactly one of $a$, $b$, far away from the other. So the
intersection between $d$ and $c$, if any, is still transverse, and $d$ is a
neighbor of $c$ in $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. ∎

This contrasts with the situation we want to characterize, as we see now.

######  Lemma 2.11.

Let $a$ and $b$ be two one-sided curves, and suppose that $a\cap b$ consists
in one, inessential intersection point. Then, for every nonseparating curve
$c\not\in\\{a,b\\}$, there exists $d$ such that $T(a,d)$ and $D(b,d)$ hold but
such that the intersection $c\cap d$ is infinite.

###### Proof.

We first observe that $\Sigma^{\prime}=\Sigma\smallsetminus(a\cup b)$ is
connected. This is seen by following the curves $a$ and $b$ in both
directions: the union $a\cup b$ does not disconnect its small neighborhoods.
Let $c$ be a curve as above. Then, we may consider a first curve $d_{0}$,
obtained by pushing $a$ aside, in such a way that $d_{0}$ is disjoint from $b$
(this is possible since the intersection $a\cap b$ is inessential). Since
$c\not\in\\{a,b\\}$, the curve $c$ intersects $\Sigma^{\prime}$. Since $d_{0}$
meets every component of $\Sigma^{\prime}$ (there is only one), we may deform
it into a curve $d$ which intersects $c$ infinitely many times, exactly as in
the proof of Lemma 2.6. ∎

After these two lemmas, we have a simple sentence in terms of the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, which holds when $a\cap b$ is a
single inessential intersection point, and which guarantees that $a\cap b$ is
connected. In order to upgrade this into a characterization of the first
situation, we need to be able to exclude as well the cases when $a\cap b$ is a
non degenerate arc. These cases fall into two subcases: the intersection arc
$a\cap b$ can be essential or inessential, exactly as an intersection point.
One way to formalize this, is to say that the intersection $a\cap b$ is
essential if $a$ cuts a regular neighborhood of $a\cap b$ into two regions
both containing a subarc of $b$, and inessential otherwise.

######  Lemma 2.12.

Let $a$ and $b$ be one-sided curves. Suppose that $a\cap b$ is a non
degenerate arc, and suppose this intersection is essential. Then there exist
curves $\alpha,\alpha^{\prime},\beta,\beta^{\prime}$ obtained by pushing $a$
aside, such that $B(a,\alpha,\beta)$, $B(b,\alpha,\beta)$,
$B(a,\alpha^{\prime},\beta^{\prime})$, $B(b,\alpha^{\prime},\beta^{\prime})$,
and $N(a,\alpha,\beta^{\prime})$.

###### Proof of Lemma 2.12.

The curves $\alpha$, $\beta$, $\alpha^{\prime}$ and $\beta^{\prime}$ may be
taken in a neighborhood of $a\cup b$, as pictured in Figure 8.

import geometry;

point A = (0,70), B = (-60,35), C = (-60,-35), D = (0, -70), E = (90, -70), F
= (150, -35), G = (150, 35), H = (90, 70); point O1 = (-40/3,0), O2 =
(90+40/3,0); point U = (-40,0), V = (130, 0); path contour =
A–B–(-40,0)–C–D–(20,-35)–(70,-35)– E–F–(130,0)–G–H–(70, 35)–(20,35)–cycle;

path alpha = ((5*C+4*D)/9)..((5*U+4*D)/9)dir(60)..(20,0)dir(25)..
((5*V+4*H)/9)dir(60)..((5*G+4*H)/9); path beta =
((4*C+5*D)/9)..((4*U+5*D)/9)dir(60)..(20,0)dir(50)..
((4*V+5*H)/9)dir(60)..((4*G+5*H)/9);

path alphaprim = ((2*C+7*D)/9)..((2*U+7*D)/9+10*dir(60))dir(60)..
(60,0)dir(40)..((2*V+7*H)/9)dir(60)..((2*G+7*H)/9); path betaprim =
((1*C+8*D)/9)..((1*U+8*D)/9+20*dir(60))dir(60)..
(60,0)dir(65)..((1*V+8*H)/9)dir(60)..((1*G+8*H)/9);

fill (contour, lightgray); draw (contour, gray+dotted);

draw (O1–O2, magenta+1.2pt); draw ((A+2*B)/3–O1, red+1.2pt); label("$b$",
(A+2*B)/3, NW, red); draw ((2*C+D)/3–O1, blue+1.2pt); label("$a$", (2*C+D)/3,
SW, blue); draw ((E+2*F)/3–O2, red+1.2pt); draw ((2*G+H)/3–O2, blue+1.2pt);

draw (alpha, cyan); label("$\alpha$", (5*G+4*H)/9, NE, cyan); draw(beta,
green); label("$\beta$", (4*G+5*H)/9, NNE, green); draw(alphaprim, cyan);
label("$\alpha^{\prime}$", (2*C+7*D)/9, SW, cyan); draw(betaprim, green);
label("$\beta^{\prime}$", (1*C+8*D)/9, SSW, green);

Figure 8. The curves $\alpha,\beta,\alpha^{\prime},\beta^{\prime}$

∎

Finally, we deal with inessential arcs.

######  Lemma 2.13.

Let $a$ and $b$ be one-sided curves, such that $a\cap b$ is an inessential arc
or intersection point. Let $c$ be such that $T(a,c)$. Then there exists $d$
such that $B(a,c,d)$ and $D(b,d)$, if and only if the intersection point
$a\cap c$ does not belong to $b$.

###### Proof.

If $a\cap c$ belongs to $b$, then obviously, any curve $d$ satisfying
$B(a,b,d)$ will meet $b$. Otherwise, we may push $a$ aside, in order to obtain
a curve $d$, as in Figure 9, left. ∎

Putting all together, this yields the following characterization of
inessential intersection points between one-sided curves.

######  Corollary 2.14.

Let $a$ and $b$ be one-sided curves. Then, $a\cap b$ consists of one,
inessential intersection point, if and only if the following conditions are
satisfied:

  1. (1)

$a$ and $b$ are not neighbors in
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$,

  2. (2)

for every $c\not\in\\{a,b\\}$, there exists $d$ such that $d$ is a neighbor of
$a$ and $b$ in $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, and $D(a,d)$ or
$D(b,d)$ (or both), but $d$ is not a neighbor of $c$ in that graph,

  3. (3)

there do not exist $\alpha,\beta,\alpha^{\prime},\beta^{\prime}$ such that
$B(a,\alpha,\beta)$, $B(b,\alpha,\beta)$,
$B(a,\alpha^{\prime},\beta^{\prime})$, $B(b,\alpha^{\prime},\beta^{\prime})$
and $N(a,\alpha,\beta^{\prime})$ all hold,

  4. (4)

there do not exist $c_{1},c_{2}$ with $D(c_{1},c_{2})$ and with the following
property: for $i=1,2$ we have: $T(a,c_{i})$ and for all $d$, $B(a,c_{i},d)$
and $D(b,d)$ do not both hold.

This enumeration of conditions expressed only in terms of the graph structure
of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, with the addition of the
conditions $\mathrm{One}(a)$ and $\mathrm{One}(b)$, will be also denoted by
$I(a,b)$, for inessential intersection (of one-sided curves).

###### Proof.

First, let us check that if $a$ and $b$ have one, inessential intersection
point then $I(a,b)$ holds. Condition (1) holds by definition, and (2) follows
from Lemma 2.11. The negation of condition (3) would imply that the cardinal
of $a\cap b$ is at least 2. Indeed, $B(a,\alpha,\beta)$ implies that
$\alpha\cap\beta$ is a point lying in $a$. Thus, the bouquet conditions imply
that both $\alpha\cap\beta$ and $\alpha^{\prime}\cap\beta^{\prime}$ lie in
$a\cap b$. And the condition $N(a,\alpha,\beta^{\prime})$ then implies that
$a\cap\alpha$ and $a\cap\beta^{\prime}$ are disjoint, hence the two points
$\alpha\cap\beta$ and $\alpha^{\prime}\cap\beta^{\prime}$ are distinct.
Finally, condition (4) follows from Lemma 2.13. Indeed, this lemma implies
that the two curves $c_{1}$ and $c_{2}$ should both contain a point of $a\cap
b$, hence they cannot be disjoint.

Now, let $a$ and $b$ be any two nonseparating curves and suppose that
$I(a,b)$. By conditions (1) and (2), the intersection $a\cap b$ is non empty,
and connected. Along the lines of the proof of Lemma 2.12, we can see that
condition (3) implies that $a\neq b$, so $a\cap b$ is an inessential
intersection point, or an arc. Suppose for contradiction that it is a
nondegenerate arc. By Lemma 2.12 and condition (3), this intersection arc
cannot be essential. Now Figure 9, right, shows the desired contradiction with
condition (4).

import geometry;

point A = 28*dir(120), B = 14*dir(120), C = (0,0), D = (40,0), E =
D+28*dir(60); path d = A+(-4,0)..tension 3 ..(B-4*dir(-20))..B..
(B+4*dir(-20))..tension 3 ..(C+4*dir(60))..tension 4 ..
(D+4*dir(120))..tension 3 ..(E+(-4,0)); draw (C–D, magenta); draw (A–C, blue);
draw(D–E, blue); draw (C–(28*dir(-120)), red); draw(D–((40,0)+28*dir(-60)),
red); draw ((-25,3)..B..(7,24), green);

draw (d, cyan);

label ("$b$", -28*dir(60), W, red); label("$a$", E, SE, blue); label("$c$",
(-25,3), W, green); label("$d$", 4*dir(60), NE, cyan); import geometry;

point A = (0,70)/2, B = (-60,35)/2, C = (-60,-35)/2, D = (0, -70)/2, E = (90,
-70)/2, F = (150, -35)/2, G = (150, 35)/2, H = (90, 70)/2; point O1 = (0,0)/2,
O2 = (90,0)/2; point U = (-40,0)/2, V = (130, 0)/2; path contour =
A–B–U–C–D–((20,-35)/2)–((70,-35)/2)– E–F–V–G–H–((70, 35)/2)–((20,35)/2)–cycle;

fill (contour, lightgray); draw (contour, gray+dotted);

draw (O1–O2, magenta+1pt); draw ((A+B)/2–O1, red+1pt); label("$b$", (A+B)/2,
NW, red); draw ((C+D)/2–O1, blue+1pt); label("$a$", (C+D)/2, SW, blue); draw
((E+F)/2–O2, red+1pt); draw ((G+H)/2–O2, blue+1pt);

draw (((4*A+2*B)/6)..tension 2 ..(5,0)..tension 2 .. ((4*D+2*C)/6),
green+0.9pt); draw (((5*A+B)/6)..tension 2 ..(10,0)..tension 2 .. ((5*D+C)/6),
cyan+0.9pt);

draw (((4*F+2*E)/6)..tension 2 ..(55,0)..tension 2 .. ((4*G+2*H)/6),
green+0.9pt); draw (((5*F+E)/6)..tension 2 ..(60,0)..tension 2 .. ((5*G+H)/6),
cyan+0.9pt);

label("$c_{1}$", ((4*F+2*E)/6), SSE, green); label("$c_{2}$", ((5*G+H)/6),
ENE, cyan);

Figure 9. Some configurations of curves for properties $I$ and $xB$

∎

Finally, we deal with extra bouquets. We denote by $xB(a,b,c)$ the property
that $T(a,b)$, $T(a,c)$, $I(b,c)$ all hold and moreover: for all $d$ such that
$B(a,b,d)$ holds, $D(c,d)$ does not.

######  Lemma 2.15.

Let $a$, $b$ and $c$ be such that $T(a,b)$, $T(a,c)$ and $I(b,c)$, with $b$
and $c$ one-sided. Then $xB(a,b,c)$ holds if and only if the intersection
points $a\cap b$, $a\cap c$ and $b\cap c$ coincide.

###### Proof.

Of course if these points coincide, then property $xB(a,b,c)$ holds: every
curve $d$ such that $B(a,b,d)$ holds, must contain this point and hence cannot
be disjoint from $c$.

Now suppose that these points do not coincide, hence, are three pairwise
distinct points. Then, we may push $b$ aside, in order to find a curve $d$
which does not intersect $c$ any more, as the intersection $b\cap c$ is not
essential. This curve $d$, obtained by pushing $b$, can be made to satisfy
$T(b,d)$, while crossing $b$ precisely at the point $a\cap b$, and this
intersection can be made transverse; the illustration of this situation is
similar to Figure 9, left, and this time we leave it to the reader. This
yields a curve $d$ such that $B(a,b,d)$ holds and $d$ disjoint from $c$. ∎

##  3\. Proof of Theorem 1.1

Here as above, $\Sigma$ is a connected surface admitting a nonseparating
closed curve.

###  3.1. From bijections to homeomorphisms

In order to prove Theorem 1.1, it suffices to prove that every automorphism of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ is supported by a bijection of
the surface, in virtue of the following observation.

######  Proposition 3.1.

Let $f\colon\Sigma\to\Sigma$ be a bijection. We suppose that for every
nonseparating simple closed curve $\alpha\subset\Sigma$, the sets $f(\alpha)$
and $f^{-1}(\alpha)$ are also nonseparating simple closed curves in $\Sigma$.
Then $f$ is a homeomorphism.

###### Proof.

If our hypothesis was that $f$ and $f^{-1}$ are closed (i.e., send closed sets
to closed sets), then $f$ would be a homeomorphism. So our strategy is to use
our hypothesis here in a similar fashion. We need only prove that $f$ is
continuous, the argument for $f^{-1}$ is symmetric.

Let $x\in\Sigma$ and suppose that $f$ is not continuous at $x$. Then there
exists a sequence $(x_{n})_{n\geqslant 0}$ of distinct points converging to
$x$, and a neighborhood $V$ of $f(x)$, such that for all $n$, we have
$f(x_{n})\not\in V$. Notice that in the open unit disk of the plane, up to
homeomorphism, there is only one sequence of distinct points converging to the
origin. With this in head, we may construct an embedded arc, in $\Sigma$, with
one end at $x$, and which contains all the points $x_{n}$. Then we may
construct a nonseparating simple closed curve $\alpha$ containing this arc. By
hypothesis, $f(\alpha)$ is a nonseparating closed curve in $\Sigma$, which
contains $f(x)$. We may perform a surgery of $f(\alpha)$ inside $V$, to obtain
a nonseparating simple closed curve $\beta$, which coincides with $f(\alpha)$
outside $V$ but which does not contain $f(x)$. Now $f^{-1}(\beta)$ is, by
hypothesis, a closed subset of $\Sigma$, which contains all the points $x_{n}$
but not $x$. This is a contradiction. ∎

###  3.2. The adjacency relation $\diamondvert$

Let $E_{T}(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$ denote the set of
edges $\\{a,b\\}$ of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ satisfying
$T(a,b)$. Then we have a map

$\mathrm{Point}\colon\
E_{T}(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))\to\Sigma,$

which to each edge $\\{a,b\\}$, associates the intersection point $a\cap b$.
The main part of proof of Theorem 1.1 consists in showing that we can express
the equality

$\mathrm{Point}(a,b)=\mathrm{Point}(\alpha,\beta)$

in terms of the graph. For this we introduce the equivalence relation
$\diamondvert$ on $E_{T}(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$ as
follows. Let $\\{a,b\\}$ in
$E_{T}(\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma))$. If $\\{a,b,c\\}$ is a
$3$-clique of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ of type bouquet we
set $\\{a,b\\}\diamondvert\\{a,c\\}$. We also set
$\\{a,b\\}\diamondvert\\{a,c\\}$ if $a$, $b$ and $c$ satisfy the “extra
bouquet” condition, denoted above by $xB(a,b,c)$, see paragraph 2.4 (this is
void when $\Sigma$ is orientable). Then $\diamondvert$ is defined as the
equivalence relation generated by these relations. When $\Sigma$ is
orientable, the relation $\diamondvert$ corresponds to the equivalence
relation on triangles, generated by adjacency, in the subgraphs of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ induced by curves passing
through a common point. This is what motivates our notation.

The relation $\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$ obviously
implies $\mathrm{Point}(a,b)=\mathrm{Point}(a^{\prime},b^{\prime})$. We will
see that the converse is not true, and describe geometrically the equivalence
classes in Section 4, but for now we will only need the following partial
statement.

######  Proposition 3.2.

Let $a$, $b$, $a^{\prime}$, $b^{\prime}$ be such that $T(a,b)$ and
$T(a^{\prime},b^{\prime})$. Suppose that they have the same intersection
point, $x=\mathrm{Point}(a,b)=\mathrm{Point}(a^{\prime},b^{\prime})$, and
suppose that the germs of $a$ and $a^{\prime}$ coincide, i.e., there exists a
neighborhood $V$ of $x$ such that $a\cap V=a^{\prime}\cap V$. Then
$\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$.

######  Remark 3.3.

If $\Sigma$ is a Klein bottle, there are no couples $\\{a,b\\}$ of two-sided
curves such that $T(a,b)$, and for every one-sided curve $b$, the curves $c$
such that $T(b,c)$ holds fall into only two isotopy classes: that of $b$ and
that of a two-sided curve, prescribed by $b$. It follows that, without the
extra bouquets in the definition of $\diamondvert$, there would have been too
many classes of $\diamondvert$, as such a class would remember the isotopy
class of a one-sided curve, and Proposition 3.2 would not be true in this
special case. These extra bouquets will be used in the proof of Lemma 3.9
below.

We postpone the proof of the proposition to the end of this section; for now
we will explain how it implies Theorem 1.1.

###  3.3. Proof of Theorem 1.1

If $a,b,c$ are vertices of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, we
denote by $F(a,b,c)$ the property that $T(a,b)$ holds, and there exists an
edge $\\{a^{\prime},b^{\prime}\\}$ with
$\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$ such that
$\\{a^{\prime},b^{\prime},c\\}$ is a $3$-clique which is not of type bouquet.
Note that this property $F(a,b,c)$ implies that $c$ does not contain the point
$\mathrm{Point}(a^{\prime},b^{\prime})=\mathrm{Point}(a,b)$. The next lemma
asserts that $F(a,b,c)$ actually characterises this geometric property, and
the letter $F$ stands for: “$a\cap b$ is far from $c$”.

######  Lemma 3.4.

Let $a,b,c$ be vertices of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, and
suppose $T(a,b)$ holds. Then

$F(a,b,c)\Leftrightarrow\mathrm{Point}(a,b)\not\in c.$

###### Proof.

The direct implication follows directly from the definitions; we have to prove
the converse implication. Suppose $\mathrm{Point}(a,b)\not\in c$. Since
$\Sigma$ is connected, there exists a regular neighborhood of $c$ containing
the point $\mathrm{Point}(a,b)$. Depending on whether $c$ is one-sided or two-
sided, up to homeomorphism, this leads to only two distinct situations. In
Figure 10, we represent in bold the germs of the curves $a$ and $b$ near the
point $a\cap b$, and show how to complete these germs to new curves
$a^{\prime}$, $b^{\prime}$ such that $\\{a^{\prime},b^{\prime},c\\}$ is a
$3$-clique not of type bouquet. When $c$ is one-sided (see Figure 10, left),
we may use two curves $a^{\prime}$, $b^{\prime}$ obtained by pushing $c$,
while when $c$ is two-sided (see Figure 10, right), we have to use a curve $d$
which meets $c$ once transversely.

Now, by Proposition 3.2, we have
$\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$, and
$\\{a^{\prime},b^{\prime},c\\}$ is a non-bouquet $3$-clique, so we have
$F(a,b,c)$ by definition.

import geometry;

picture commun, gauche, droite; path agros = (0, 10)–(10, 20), bgros = (0,
20)–(10, 10), c = (-30, 0)–(40,0);

draw (commun, c, 1pt+black); draw(commun, agros, 2pt+blue); draw (commun,
bgros, 2pt+red); add(gauche, commun); add(droite, commun);

path b1fingauche = (-30, 6)dir(0)..(-20, 6)dir(0)..(0,20)dir(-45); path
b2fingauche = (10, 10)dir(-45)..dir(0)(25, -6)–(40, -6); path a1fingauche =
(-30, -6)–(-15,-6)dir(0)..(0,10)dir(45); path a2fingauche = (10,
20)dir(45)..right(30, 6)–(40,6); draw (gauche, b1fingauche, red); draw(gauche,
b2fingauche, red); draw (gauche, a1fingauche, blue); draw(gauche, a2fingauche,
blue); label(gauche, "$a^{\prime}$", (-30,-6), SW, blue); label(gauche, "$c$",
(-30,0), W); label(gauche, "$b^{\prime}$", (-30,6), NW, red);

path d = (-10, -20)–(-10, 40); draw (droite, d, 1pt+black); draw (droite,
b1fingauche, red); path b2findroite = (10,10)dir(-45)..right(30, 6)–(40,6);
draw (droite, b2findroite, red); path a1findroite =
(-10,-20)–(-10,-10)up..(0,10)dir(45); path a2findroite =
(10,20)dir(45)..up(-10,35)–(-10,40); draw (droite, a1findroite, blue);
draw(droite, a2findroite, blue); label(droite, "$c$", (-30, 0), W);
label(droite, "$b^{\prime}$", (-30,6), NW, red); label(droite, "$a^{\prime}$",
(10,20), E, blue); label(droite, "$d$", (-10,40), SW);

add(gauche, (-50, 0)); add(droite, (60,0));

Figure 10. Completing the germs of $a$ and $b$ to form a non-bouquet
$3$-clique $(a^{\prime},b^{\prime},c)$

∎

######  Corollary 3.5.

Suppose $T(a,b)$ and $T(\alpha,\beta)$ hold. Then
$\mathrm{Point}(a,b)\neq\mathrm{Point}(\alpha,\beta)$ if and only if there
exists a nonseparating closed curve $c$ such that $F(\alpha,\beta,c)$ holds
but not $F(a,b,c)$.

The corollary is a direct consequence of Lemma 3.4. It follows that the
equality $\mathrm{Point}(a,b)=\mathrm{Point}(\alpha,\beta)$ can be expressed
in terms of the graph structure of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, because, as a consequence of
Corollary 2.4, being a $3$-clique not of type bouquet is also characterized in
terms of this graph structure. Now we can conclude the proof of Theorem 1.1,
provided Proposition 3.2 holds.

###### Proof of Theorem 1.1.

Let $\varphi$ be an automorphism of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. Given a point $x$ in $\Sigma$,
we choose two nonseparating simple closed curves $a$, $b$ intersecting exactly
once, transversely, at $x$, and set
$\varphi_{\Sigma}(x)=\mathrm{Point}(\varphi(a),\varphi(b))$. This formula is
valid because, by Proposition 1.4, $\varphi(a)$ and $\varphi(b)$ are still
nonseparating simple closed curves intersecting exactly once. The point
$\varphi_{\Sigma}(x)$ does not depend on the choice of $(a,b)$, because if
$(\alpha,\beta)$ is another choice, the equalities
$\mathrm{Point}(a,b)=\mathrm{Point}(\alpha,\beta)$ and
$\mathrm{Point}(\varphi(a),\varphi(b))=\mathrm{Point}(\varphi(\alpha),\varphi(\beta))$
can be all expressed in terms of the graph structure of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$, thanks to Corollary 3.5. Thus,
the map $\varphi_{\Sigma}$ is well-defined, and by following the definitions
we observe that the map $(\varphi^{-1})_{\Sigma}$ is its inverse: hence
$\varphi_{\Sigma}$ is a bijection of $\Sigma$. Finally, it follows from Lemma
3.4 that for any nonseparating simple closed curve $\alpha$, the curve
$\varphi(\alpha)$ coincides with the set of points $\varphi_{\Sigma}(x)$ as
$x$ describes $\alpha$. In other words, the automorphism $\varphi$ is realized
by the bijection $\varphi_{\Sigma}$. Proposition 3.1 concludes. ∎

###  3.4. Connectedness of some arc graphs

In order to finally prove Proposition 3.2, we will first need a couple of
elementary results on fine arc graphs.

######  Lemma 3.6.

Let $S$ be a connected topological surface, with boundary, and let $x,y$ be
two distinct points of $\partial S$. Let
$\mathcal{E}\mathcal{A}^{\dagger}(S,x,y)$ be the graph whose vertices are the
simple arcs joining $x$ and $y$ and which meet $\partial S$ only at their
ends, with an edge between two such arcs if and only if they are disjoint
except at $x$ and $y$. Then the graph
$\mathcal{E}\mathcal{A}^{\dagger}(S,x,y)$ is connected.

Note that, when $x$ and $y$ are taken in the same connected component, we are
not requiring that the arcs be nonseparating; this is the reason why we use
the letter $\mathcal{E}$, for extended, in the same fashion as in [10].

###### Proof.

Let $a$, $b$ be two vertices of this graph. As a first case we suppose that
$a\cap b$ is made of a finite number of transverse intersection points: we
will prove by induction on the cardinal of $a\cap b$ that, in this case, $a$
and $b$ are connected in $\mathcal{E}\mathcal{A}^{\dagger}(S,x,y)$. If $a\cap
b$ is as small as possible, i.e., is equal to $\\{x,y\\}$, then $a$ and $b$
are neighbors in this graph. Otherwise, if $a$ and $b$ intersect at other
points than $x$ and $y$, we may, in the spirit of [6], pick a unicorn path
$c$, made of one subarc of $a$ beginning at $x$, and one subarc of $b$ ending
at $y$. (For example, we may follow $a$ until it first meets $b$ after $x$,
and then continue along $b$). Now we may push $c$ aside while fixing its ends,
at the appropriate side of $c$, to obtain a new arc $c^{\prime}$ with both
$c^{\prime}\cap a$ and $c^{\prime}\cap b$ of cardinal strictly lower than that
of $a\cap b$.

This proves that arcs intersecting at finitely many points are connected in
$\mathcal{E}\mathcal{A}^{\dagger}(S,x,y)$.

Now if the intersection $a\cap b$ is infinite, fix a differentiable structure
on the surface $S$. We may consider a smooth curve $a^{\prime}$, neighbor of
$a$, by pushing $a$ aside while fixing its ends, and similarly, a smooth
neighbor $b^{\prime}$ of $b$ similarly. Up to perturbing $b^{\prime}$, we may
suppose that $b^{\prime}$ is transverse to $a^{\prime}$. By the step above,
$a^{\prime}$ and $b^{\prime}$ are connected in
$\mathcal{E}\mathcal{A}^{\dagger}(S,x,y)$ and the Lemma is proved. ∎

We will also need a version for nonseparating arcs.

######  Lemma 3.7.

Let $S$ be a connected surface containing a nonseparating curve. Suppose $S$
has boundary, and let $x$ and $y$ be two distinct points in a same boundary
component of $S$. Let $\mathcal{NA}^{\dagger}(S,x,y)$ be the set of
nonseparating simple arcs connecting $x$ to $y$, with an edge when they are
disjoint away from $x$ and $y$. Then $\mathcal{NA}^{\dagger}(S,x,y)$ is
connected.

The points $x$ and $y$ add some technicality; let us first prove the following
simpler statement.

######  Lemma 3.8.

Let $S$ be a connected surface containing a nonseparating curve, and with at
least one boundary component, denoted $C$. Let $\mathcal{NA}^{\dagger}(S,C)$
be the graph whose vertices are the nonseparating arcs joining two distinct
points of $C$, and with an edge between two such vertices whenever they are
disjoint. Then this graph is connected.

This lemma is a variation on [10, Corollary 3.2]; here we additionnaly require
that the arcs end at $C$. In fact, in [10], Corollary 3.3 is stated for
surfaces with $b>0$ boundary components, but proved only in the case $b=1$,
which is the case needed in the proof of their main theorem. Lemma 3.8 may be
used to extend this corollary to any $b>0$.

###### Proof.

We begin with the observation that the graph $\mathcal{NA}^{\dagger}(S,C)$ has
no isolated point. Indeed, if $\gamma$ is a vertex of
$\mathcal{NA}^{\dagger}(S,C)$, by definition it is nonseparating. So we may
consider a simple closed curve $u$ with one, transverse intersection point
with $\gamma$. Obviously, this curve $u$ is nonseparating; this follows from
Fact 2.1, applied to $u$, and a curve $v$ obtained by concatenation of
$\gamma$ with some arc of $C$. Now we can perform a surgery on $u$, and push
its intersection point towards one end of $\gamma$ until we hit $C$. This
constructs an arc $\alpha$, which is now disjoint from $\gamma$, and which is
also nonseparating.

Next, we claim that we can suppose, without loss of generality, that the
surface $S$ is compact. Indeed, if $\gamma_{1},\gamma_{2}$ are vertices of
$\mathcal{NA}^{\dagger}(S,C)$, and if $u$ is a nonseparating curve
intersecting $\gamma_{1}$ as above, consider the set
$K=C\cup\gamma_{1}\cup\gamma_{2}\cup u$. This set is compact, hence there
exists a compact topological subsurface $S^{\prime}$ of $S$ containing $K$.
This surface $S^{\prime}$ contains nonseparating curves, as it contains $u$
and $\gamma_{1}\cup C$, which may be used as above to find two simple closed
curves $u$, $v$ with one, essential intersection. Now a path joining
$\gamma_{1}$ to $\gamma_{2}$ in $S^{\prime}$ is also a path joining
$\gamma_{1}$ to $\gamma_{2}$ in $S$. So, until the end of the proof, $S$ is
now supposed to be compact.

Next, observe that if two vertices $\gamma_{1},\gamma_{2}$ of
$\mathcal{NA}^{\dagger}(S,C)$ are isotopic (i.e., there exists a continuous
map $H\colon[0,1]^{2}\to S$ such that $\gamma_{1}(t)=H(0,t)$ and
$\gamma_{2}(t)=H(1,t)$ for all $t$, $H(s,0),H(s,1)\in C$ for all $s$ and the
curve $H_{s}\colon t\to H(s,t)$ is injective for all $s$), then $\gamma_{1}$
and $\gamma_{2}$ are in the same component of $\mathcal{NC}^{\dagger}(S,C)$.
This argument is borrowed from [2]: for all $s$, the arc $H_{s}$ has at least
a neighbor $\alpha_{s}$ (by the first observation above), and the set of
$s^{\prime}$ such that $H_{s^{\prime}}$ is still a neighbor of $\alpha_{s}$ is
open in $[0,1]$. By compactness of $[0,1]$, there exist a finite number of
arcs $\alpha_{1},\ldots,\alpha_{n}$, and a subdivision
$0=t_{0}<t_{1}<\cdots<t_{n}=1$ such that $\alpha_{j}$ is disjoint from $H_{t}$
for all $t\in[t_{j-1},t_{j}]$ for all $j$, and now
$(\gamma_{1},\alpha_{1},\ldots,\alpha_{n},\gamma_{2})$ is a path of
$\mathcal{NC}^{\dagger}(S,C)$ joining $\gamma_{1}$ to $\gamma_{2}$. As a
result of this observation, we need only prove the connectedness of the graph
$\mathrm{NA}(S,C)$, whose vertices are the isotopy classes of arcs between two
distinct points of $C$, and with an edge between two vertices whenever the
corresponding classes admit disjoint representatives.

We proceed with the observation that the graph $\mathrm{A}(S,C)$, defined
exactly as $\mathrm{NA}(S,C)$ excpept we consider essential arcs, which may be
separating, is connected. A simple way to do this is by using the idea of
unicorn arcs exactly as in the proof of the preceding lemma: if two arcs $a$
and $b$ are in minimal position then their unicorn arcs are essential, and
have fewer intersections with both $a$ and $b$ than the number of points of
$a\cap b$.

We will promote the connectedness of $\mathrm{A}(S,C)$ to that of
$\mathrm{NA}(S,C)$, by induction on the number of boundary components of $S$.

First, suppose that $S$ has only one boundary component, $C$. Let
$\gamma_{1},\gamma_{2}$ be two vertices of $\mathrm{NA}(S,C)$. We may connect
them by a path
$(\gamma_{1},\alpha_{1},\alpha_{2},\ldots,\alpha_{n},\gamma_{2})$ in
$\mathrm{A}(S,C)$, where each of the $\alpha_{j}$ may be separating; consider
such a path with minimal number of separating arcs. For contradiction, and up
to some relabeling, suppose $\alpha_{1}$ is separating. Then it cuts $S$ in
two components; denote by $S_{1}$ the one containing $\gamma_{1}$ and $S_{2}$
the other. Then $\alpha_{2}$ is also contained in $S_{1}$, otherwise we may
delete $\alpha_{1}$ from our path. Since $S$ has no boundary component other
than $C$ and since the curve $\alpha_{1}$ is essential, the surface $S_{2}$
contains a nonseparating arc, $\alpha_{1}^{\prime}$. This arc may be used
instead of $\alpha_{1}$ in our initial path from $\gamma_{1}$ to $\gamma_{2}$,
contradicting the minimality of the number of separating arcs. This proves
that $\mathrm{NA}(S,C)$ is connected if $S$ has no other boundary component.

Now, we suppose, for inductive hypothesis, that
$\mathrm{NA}(S^{\prime},C^{\prime})$ is connected for every surface
$S^{\prime}$ with less boundary components than $S$. Let
$\gamma_{1},\gamma_{2}$ be two vertices of $\mathrm{NA}(S,C)$. As before,
consider a path
$(\gamma_{1},\alpha_{1},\alpha_{2},\ldots,\alpha_{n},\gamma_{2})$ in
$\mathrm{A}(S,C)$ between them, with minimal number of separating arcs. For
contradiction, and up to some relabeling, suppose $\alpha_{1}$ is separating:
it cuts $S$ into two subsurfaces, let $S_{1}$ be the one containing
$\gamma_{1}$, and, by hypothesis, must also contain $\alpha_{2}$, and let
$S_{2}$ be the other. If $S_{2}$ contains nonseparating arcs, we conclude as
before. If not, then $S_{2}$ contains some of the boundary components of $S$,
hence the surface with boundary $S^{\prime}=S_{1}\cup\alpha_{1}$ has strictly
less boundary components than $S$. One is $C^{\prime}$, composed by an arc of
$C$ and the arc $\alpha_{1}$, and there may be others.

If $\alpha_{2}$ is nonseparating, then, by the induction hypothesis, there is
a path $(\gamma_{1},\beta_{1},\ldots,\beta_{k},\alpha_{2})$ of
$\mathrm{NA}(S^{\prime},C^{\prime})$ connected them. The arcs
$\beta_{1},\ldots,\beta_{k}$ may have end points in $\alpha_{1}$, but we may
perform a surgery in order to push all these points to $C$, and obtain arcs
$\beta_{1}^{\prime},\ldots,\beta_{k}^{\prime}$ which are also vertices of
$\mathrm{NA}(S,C)$, and we are done in this case.

Finally, if $\alpha_{2}$ is a separating arc (of $S$, or of $S_{1}$,
equivalently), then we may find an arc $\alpha_{2}^{\prime}$ of $S_{1}$ which
is nonseparating and disjoint from $\alpha_{2}$. By following the last case
above, there exists a path
$(\gamma_{1},\beta_{1},\ldots,\beta_{k},\alpha_{2}^{\prime})$ in
$\mathrm{NA}(S,C)$, hence the path
$(\gamma_{1},\beta_{1},\ldots,\beta_{k},\alpha_{2}^{\prime},\alpha_{2},\ldots,\alpha_{n},\gamma_{2})$
of $\mathrm{A}(S,C)$ has one less separating arc than the initial path. This
contradiction ends the proof. ∎

###### Proof of Lemma 3.7.

Let $\gamma_{1},\gamma_{2}$ be two vertices of
$\mathcal{NA}^{\dagger}(S,x,y)$. First, we may construct a neighbor
$\gamma_{2}^{\prime}$ in $\mathcal{NA}^{\dagger}(S,x,y)$ of $\gamma_{2}$,
which, in a neighborhood of $x$ (resp. $y$), touches $\gamma_{1}$ only at $x$
(resp. $y$).

Indeed, there is a neighborhood $U_{x}$ of $x$ homeomorphic to the closed half
unit disk

$\\{z,|z|\leqslant 1\text{ and }\mathrm{Im}(z)\geqslant 0\\},$

where the middle ray ($\mathrm{Re}(z)=0$) corresponds to the points of
$\gamma_{1}$. On either side of this ray, we may find an arc disjoint from
$\gamma_{1}$ and $\gamma_{2}$ except at $0$, arbitrarily close to the boundary
($\mathrm{Im}(z)=0$), and joining $0$ to the unit circle, and then this small
arc may be continued to construct a curve $\gamma_{2}^{\prime}$ which consists
of pushing $\gamma_{2}$ aside.

So we may suppose that $\gamma_{1}$ and $\gamma_{2}$, close to $x$ and $y$,
intersect only at these points, and we may now find neighborhoods $U_{x}$ and
$U_{y}$ as above, such that their intersections with $\gamma_{1}$ and
$\gamma_{2}$ are along rays in this disk, in distinct directions around $0$.
Let $S^{\prime}$ be the surface obtained by removing the interiors of $U_{x}$
and $U_{y}$ from $S$. Then the path given by applying Lemma 3.8 to $S$, yields
a path from $\gamma_{1}$ to $\gamma_{2}$ in $\mathcal{NA}^{\dagger}(S,x,y)$,
just by adding some rays in $U_{x}$ and $U_{y}$ to the corresponding arcs. ∎

###  3.5. Proof of Proposition 3.2

Let us go back to the proof of Proposition 3.2. For the remaining of the
section we fix a point $x\in\Sigma$. Let $X$ denote the set of nonseparating
simple closed curves passing through $x$.

######  Lemma 3.9.

Let $a,b,c\in X$. Suppose that $T(a,b)$ and $T(a,c)$ hold.

Then $(a,b)\diamondvert(a,c)$.

###### Proof.

Let $S$ be the surface obtained by cutting $\Sigma$ along $a$: it is the
surface with boundary obtained by gluing back two copies of the curve $a$ to
$\Sigma\smallsetminus a$. The point $x$ of $\Sigma$ yields two points, $p$ and
$q$, of $\partial S$, and the curves $b$ and $c$ define two arcs of $S$
joining $p$ and $q$. By Lemma 3.6, there exists a finite sequence
$\gamma_{0}=b$, …, $\gamma_{n}=c$, of arcs of $S$ joining $p$ and $q$, with
$\gamma_{i}$ and $\gamma_{i+1}$ disjoint except at $p$ and $q$. For each $i$,
the arc $\gamma_{i}$ defines a closed curve in $\Sigma$, which has precisely
one, transverse intersection with $a$; we will still denote it by
$\gamma_{i}$, abusively.

For every $i$, if $T(\gamma_{i},\gamma_{i+1})$ holds, then we have
$(a,\gamma_{i})\diamondvert(a,\gamma_{i+1})$, by definition. If
$T(\gamma_{i},\gamma_{i+1})$ does not hold, then either $\gamma_{i}$ or
$\gamma_{i+1}$ are both one-sided, or one of them is two-sided. In the first
case, the condition $xB(a,\gamma_{i},\gamma_{i+1})$ holds, by definition, and
hence $(a,\gamma_{i})\diamondvert(a,\gamma_{i+1})$. In the second, up to
reversing the notation suppose $\gamma_{i}$ is two-sided. Figure 11 shows how
to insert a curve $\delta$ such that $B(a,\delta,\gamma_{i})$ and
$B(a,\delta,\gamma_{i+1})$ both hold, and hence we still have
$(a,\gamma_{i})\diamondvert(a,\gamma_{i+1})$ in this case.

import geometry;

picture commun, gauche, droite;

path a = (-30,0)–(30,0), gamma1 = 30*dir(120)–(0,0)–30*dir(-120); path gamma2
= rotate(180)*gamma1;

path dgauche = (-29, 4)right::(0,0)dir(-30)::(29, -4)right; path ddroit1 =
(30*dir(127))dir(-60)::(0,0)dir(-30)::(29,-4)right; path ddroit2 = (-29,
-4)right..tension 3 ..(10*dir(210))dir(-60).. tension 3
..(30*dir(-127))dir(-120);

draw (commun, a, blue+0.9pt); label(commun, "$a$", (-30,0), W, blue);
draw(commun, gamma1, magenta+0.9pt); label(commun, "$\gamma_{i}$",
30*dir(120), N, magenta); draw(commun, gamma2, olive+0.9pt); label(commun,
"$\gamma_{i+1}$", 30*dir(60), N, olive);

draw(gauche, dgauche, cyan+0.9pt); draw(droite, ddroit1, cyan+0.9pt);
draw(droite, ddroit2, cyan+0.9pt); label(commun, "$d$", (29, -4), E, cyan);

label(gauche, "Case when $a$ is one-sided", (0,-40)); label(droite, "Case when
$a$ is two-sided", (0,-40));

add(commun, (-80,0)); add(commun, (80,0)); add(gauche, (-80,0)); add(droite,
(80,0));

Figure 11. Connecting the curves by common adjacency

By transitivity, we deduce that $(a,b)\diamondvert(a,c)$. ∎

The last ingredient for the proof of Proposition 3.2 is the following
observation.

######  Observation 3.10.

Let $a$, $a^{\prime}$ be two nonseparating simple closed curves in $\Sigma$
such that $a\cap a^{\prime}$ is an arc. Then, both sides of this arc lie in
the same connected component of $\Sigma\smallsetminus(a\cup a^{\prime})$.

###### Proof.

A priori, the complement of $\Sigma\smallsetminus(a\cup a^{\prime})$ may have
up to four connected components, as suggested in Figure 12.

import geometry;

real R = 35; point Dec = (20,0); draw ((0,0)–Dec, magenta); draw
(shift(Dec)*(R*dir(50)–(0,0))); label ("$a$ or $a^{\prime}$", Dec+R*dir(50),
E); draw ((0,0)–(-R*dir(50)), red); label ("$a^{\prime}$", -R*dir(50), W,
red); draw (shift(Dec)*(R*dir(-50)–(0,0))); label ("$a$ or $a^{\prime}$",
Dec+R*dir(-50), E); draw ((0,0)–(-R*dir(-50)), blue); label ("$a$",
-R*dir(-50), W, blue); label ("$A$", Dec/2+(0, 20)); label("$B$", Dec/2+(0,
-20)); label ("$C$", (-30, 0)); label ("$D$", Dec+(30, 0));

Figure 12. The arc $a\cap a^{\prime}$ cannot disconnect

Suppose first that the intersection $a\cap a^{\prime}$ is essential. If $a$
(resp. $a^{\prime}$) is one-sided, by following the curve $a$ (resp.
$a^{\prime}$) we see that $A=B$. If both $a$ and $a^{\prime}$ are two-sided,
by following $a$ we see that $A=D$ and $C=B$, while by following $a^{\prime}$
we get $A=C$ and $B=D$, so $A=B$.

Now, suppose the intersection arc $a\cap a^{\prime}$ is inessential. By
following $a$, we see that $C=D$, regardless of $a$ being one or two-sided.
Thus, if $A\neq B$, then one of $A$ or $B$, say $A$, is not connected from any
of $B,C,D$. But this implies that $a$ is separating, a contradiction. ∎

We are now in a position to prove Proposition 3.2, but instead we will prove
the following stronger statement, which will be more convenient later in this
article.

######  Proposition 3.11.

Let $a$, $b$, $a^{\prime}$, $b^{\prime}$ be such that $T(a,b)$ and
$T(a^{\prime},b^{\prime})$, with intersection point
$x=\mathrm{Point}(a,b)=\mathrm{Point}(a^{\prime},b^{\prime})$, and suppose
that $a$ and $a^{\prime}$ locally “half coincide” near $x$, i.e., $a\cap
a^{\prime}$ contains a non degenerate arc with endpoint $x$. Then
$\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$.

###### Proof of Proposition 3.11.

Suppose first that $a$ and $a^{\prime}$ coincide along some arc with $x$ as an
end-point, and are disjoint apart from this arc. By observation 3.10, there
exists a curve $d$ passing through $x$ such that $T(a,d)$ and
$T(a^{\prime},d)$. By Lemma 3.9, this implies
$(a,d)\diamondvert(a^{\prime},d)$, and by the same lemma we also have
$(a,b)\diamondvert(a,d)$ and
$(a^{\prime},b^{\prime})\diamondvert(a^{\prime},d)$. Hence
$(a,b)\diamondvert(a^{\prime},b^{\prime})$.

Now we do not make the assumption any more that $a$ and $a^{\prime}$ meet only
along an arc. Still, thanks to the hypothesis of the proposition, we may
choose a set $V$ homeomorphic to a closed disk, with $x$ on its boundary, and
such that $a\cap V=a^{\prime}\cap V$ is an arc whose endpoints are $x$ and
some other point $y$. Lemma 3.7, applied to the surface
$\Sigma\smallsetminus\ring{V}$, provides a sequence $a_{0}=a$, …,
$a_{n}=a^{\prime}$, of nonseparating curves such that for all $i$, the curves
$a_{i}$ and $a_{i+1}$ intersect only along the arc $a\cap V$, hence we may
conclude by applying iteratively the reasoning above. ∎

##  4\. Local subgraphs

In section 3.2, we considered edges $\\{a,b\\},\\{a^{\prime},b^{\prime}\\}$ in
the graph $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ satisfying $|a\cap
b|=|a^{\prime}\cap b^{\prime}|=1$, and
$\mathrm{Point}(a,b)=\mathrm{Point}(a^{\prime},b^{\prime})$. We defined and
used the equivalence relation $\diamondvert$. The aim of this section is to
provide a geometric interpretation of the equivalence classes. The results
here are not used anywhere else in the paper. In particular, this section is
not used in the proof of our main results. Nevertheless, we think it may help
the reader to get a clear picture of the situation.

###  4.1. The graph of germs

Let $x$ be a marked point in the surface $\Sigma$. In this section the we will
study the local geometry of curves near $x$, so we may assume that
$(\Sigma,x)=(\mathbb{R}^{2},0)$ whenever this is convenient. Given two simple
arcs $a,a^{\prime}:[0,1]\to\Sigma$ with $a(0)=a^{\prime}(0)=x$, we say that
$a$ and $a^{\prime}$ _locally coincide_ at $x$ if there exists a neighborhood
$V$ of $x$ such that $a([0,1])\cap V=a^{\prime}([0,1])\cap V$. This is an
equivalence relation, whose equivalence classes are called _germs of simple
arcs at $x$_. The germ of $a$ is denoted $[a]_{x}$. We say that $a$ and
$a^{\prime}$ _locally intersect only at $x$_ if there exists a neighborhood
$V$ of $x$ such that $a([0,1])\cap a^{\prime}([0,1])\cap V=\\{x\\}$. This
second relation obviously induces a relation on germs. Let us consider the
graph $\mathcal{A}(x)$ whose vertices are the germs of simple arcs at $x$,
with an edge between the germs of $a$ and $a^{\prime}$ whenever $a$ and
$a^{\prime}$ locally intersect only at $x$.

This graph is _not_ connected, in fact it has infinitely (uncountably) many
connected components, as we will see below. We postpone the description of the
connected components to explain the relation with the adjacency relation
defined in section 3.2. We say that two vertices $\alpha,\alpha^{\prime}$ of
the graph $\mathcal{A}(x)$ are _comparable_ if they belong to the same
connected component of the graph.

###  4.2. Germs and adjacency

Given a point $x$ in $\Sigma$ and a simple closed curve $a$ in $\Sigma$ that
contains $x$, we choose any one of the two germs of simple arc at $x$ included
in $a$ and denote it by $\lfloor a\rfloor_{x}$. Which one of the two germs is
chosen will not matter in what follows.

######  Proposition 4.1.

Let $a,b,a^{\prime},b^{\prime}$ be vertices in
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ such that $T(a,b)$ and
$T(a^{\prime},b^{\prime})$ hold, and assume
$\mathrm{Point}(a,b)=\mathrm{Point}(a^{\prime},b^{\prime})$. Then
$\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$ if and only if the germs
$\lfloor a\rfloor_{x}$ and $\lfloor a^{\prime}\rfloor_{x}$ are comparable.

Proposition 4.1 will be proved in section 4.5 below. The aim of the next two
sections is to provide a simple characterization of distance, and connected
components, in the graph of germs; see Proposition 4.3 below.

###  4.3. Distance in local subgraphs

In this section, we give a geometrical interpretation of the distance in three
different graphs, which are very much like the graph of germs
$\mathcal{A}(x)$.

Let $\Sigma$ be one of the following surfaces: (1) the compact annulus
$\mathbb{S}^{1}\times[0,1]$, (2) the open annulus
$\mathbb{S}^{1}\times\mathbb{R}$, or (3) the 2-torus
$\mathbb{T}^{2}=\mathbb{S}^{1}\times\mathbb{S}^{1}$. We consider non-oriented
simple arcs in $\Sigma$, more precisely simple curves connecting both sides of
the annulus in case (1), properly embedded images of the real line connecting
both ends of the open annulus in case (2), or simple closed curves in a fixed
homotopy class, say homotopic to $\\{0\\}\times\mathbb{S}^{1}$ in case (3).
Let $\mathcal{A}$ denote the graph whose vertices are one of the three above
family of curves, with an edge between two curves whenever they are disjoint.

In order to express geometrically the distance in $\mathcal{A}$, let us
consider the cyclic cover $p:\widetilde{\Sigma}\to\Sigma$, respectively in
case (1), (2), (3)

$p:\mathbb{R}\times[0,1]\to\mathbb{R}/\mathbb{Z}\times[0,1],\ \ \ \
p:\mathbb{R}\times\mathbb{R}\to\mathbb{R}/\mathbb{Z}\times\mathbb{R},\ \ \ \
p:\mathbb{R}\times\mathbb{S}^{1}\to\mathbb{R}/\mathbb{Z}\times\mathbb{S}^{1}$

given by the formula $p(x,y)=(x\text{ mod }1,y)$. Let $T$ be the deck
transformation $(x,y)\to(x+1,y)$.

Now consider two curves $a,b$ which are vertices of the graph $\mathcal{A}$.
Let $\widetilde{a},\widetilde{b}$ be respective lifts of $a,b$ under the
covering map $p$. Note that the set

$\\{k\in\mathbb{Z},\ \ T^{k}(\widetilde{a})\cap\widetilde{b}\neq\emptyset\\}$

is an interval of $\mathbb{Z}$, which is finite in the compact cases (1) and
(3) but may be infinite in the open annulus case (2). We define the _relative
width_ $0pt(a,b)$ as the cardinal of this set. This is an element of
$\\{0,1,\dots,+\infty\\}$. The reader may check easily that
$0pt(a,b)=0pt(b,a)$.

######  Proposition 4.2.

For every vertices $a\neq b$ of the graph $\mathcal{A}$, the distance in the
graph is given by

$d(a,b)=0pt(a,b)+1.$

In cases (1) and (3), the graph $\mathcal{A}$ is connected. In case (2), $a$
and $b$ are in the same connected component of $\mathcal{A}$ if and only if
$0pt(a,b)<+\infty$.

###### Proof.

Let $a,b$ be as in the statement, and denote $w=0pt(a,b)$. We first assume
that $w<+\infty$. By Schoenflies’ theorem (in case (2), applied in the two-
point compactification of the annulus, which is a sphere), we may assume that
$a$ is a vertical curve whenever this makes our life easier. We first note
that if $w=0$ then $a$ and $b$ admit lifts that are disjoint from every
$T$-translate of each other, which shows that $a$ and $b$ are disjoint, and
thus $d(a,b)=1$. Let us now assume $w>0$, and prove the two following key
properties.

  1. (i)

For every vertex $a^{\prime}$ of $\mathcal{A}$ such that $d(a,a^{\prime})=1$,

$0pt(a^{\prime},b)\geq\omega-1.$

  2. (ii)

There exists a vertex $a^{\prime}$ of $\mathcal{A}$ such that
$d(a,a^{\prime})=1$ and

$0pt(a^{\prime},b)\leq w-1.$

To prove the first property, consider $a^{\prime}$ such that
$d(a,a^{\prime})=1$. By definition of the width $w$, we may find lifts
$\widetilde{a},\widetilde{b}$ of $a,b$ such that $\widetilde{b}$ is disjoint
from $\widetilde{a},T^{w+1}\widetilde{a}$ but meets
$T(\widetilde{a}),\dots,T^{w}(\widetilde{a})$. Since $a$ and $a^{\prime}$ are
disjoint, there is a lift $\widetilde{a^{\prime}}$ of $a^{\prime}$ which is
between $\widetilde{a}$ and $T(\widetilde{a})$. Then the curves

$T(\widetilde{a^{\prime}}),\dots,T^{w-1}(\widetilde{a^{\prime}})$

are between the two curves $T(\widetilde{a})$ and $T^{w}(\widetilde{a})$, and
those two curves are not in the same connected component of
$\widetilde{\Sigma}\setminus T^{i}(\widetilde{a^{\prime}})$, for
$i=1,\dots,w-1$. Since the curve $\widetilde{b}$ is connected and meets the
two curves $T(\widetilde{a})$ and $T^{w}(\widetilde{a})$, it must meet all the
$T^{i}(\widetilde{a^{\prime}})$. This proves that $0pt(a^{\prime},b)\geq w-1$.

Let us prove the second property. We consider $\widetilde{a},\widetilde{b}$ as
above. Let $S$ denote the compact strip or annulus bounded by
$\widetilde{a}\cup T(\widetilde{a})$. Remember that $\widetilde{b}$ meets
$T(\widetilde{a})$ but not $\widetilde{a}$. Thus $\widetilde{b}\cap S$ is
included in a (maybe infinite) family of _bigons_ , i.e., topological disks
bounded by a simple closed curve made of a segment of the curve
$T(\widetilde{a})$ and a segment of the curve $\widetilde{b}$. Let $S^{+}$
denote the union of these bigons. Symmetrically, the curve
$\widetilde{b^{\prime}}:=T^{-w}(\widetilde{b})$ meets $\widetilde{a}$ but not
$T(\widetilde{a})$. Thus $T^{-w}(\widetilde{b})\cap S$ is included in a union
$S^{-}$ of bigons formed by the curves $\widetilde{a}$ and
$\widetilde{b^{\prime}}$. A key point is that the sets $S^{-}$ and $S^{+}$ are
disjoint, because the curves $\widetilde{b}$ and $\widetilde{b}^{\prime}$ are
disjoint, since $b$ is simple. Thus we may construct a homeomorphism $H$
supported in $S$ such that $H(S^{-})$ is included in an arbitrarily small
neighborhood of $\widetilde{a}$, and $H(S^{+})$ is included in an arbitrarily
small neighborhood of $T(\widetilde{a})$. In particular, we may find a curve
$\widetilde{a^{\prime}}$, which is a lift of some element $a^{\prime}$ of
$\mathcal{A}$, included in the interior of $S$ and disjoint from both $S^{-}$
and $S^{+}$ (to be more explicit, take
$\widetilde{a}^{\prime}=H^{-1}(\\{1/2\\}\times[0,1])$ in the annulus case, in
coordinates for which $a$ is the vertical curve $\\{0\\}\times[0,1]$). Note
that $\widetilde{a^{\prime}}$ is disjoint from $\widetilde{b}$ and
$T^{-w}(\widetilde{b})$, and separate both curves, i.e., the first one is on
the right-hand side of $\widetilde{a^{\prime}}$, and the second one is on the
left-hand side. Thus the set

$\\{k\in\mathbb{Z},\ \
T^{k}(\widetilde{b})\cap\widetilde{a^{\prime}}\neq\emptyset\\}$

has cardinality at most $w-1$. Which proves that $0pt(a^{\prime},b)\leq w-1$,
as wanted.

Using (i) and (ii), an induction on $n$ shows that $d(a,b)=n$ if and only if
$0pt(a,b)+1=n$, which completes the proof in the case when $0pt(a,b)$ is
finite. When $0pt(a,b)=+\infty$, an argument analogous to property (i) above
shows that $0pt(a^{\prime},b)=+\infty$ for every $a^{\prime}$ such that
$d(a,a^{\prime})=1$. This shows that $a$ and $b$ are not in the same connected
component of the graph. This completes the proof of the proposition. ∎

###  4.4. Distance in the graph of germs

Let us go back to the graph of germs $\mathcal{A}(x)$. Assume
$(\Sigma,x)=(\mathbb{R}^{2},0)$. Given two vertices $a,b$ of $\mathcal{A}(x)$,
we define their _local relative width_ $0pt(a,b)$ as follows. The plane minus
the origin is identified with the open annulus
$\mathbb{S}^{1}\times\mathbb{R}$, and we consider the graph $\mathcal{A}$ from
the previous section in the open annulus case. Then $0pt(a,b)$ is defined as
the infimum of the quantity $0pt(A,B)$, where $A$ and $B$ are vertices of
$\mathcal{A}$ whose germs respectively equal $a$ and $b$. Here is a more
practical definition, which is easily seen to be equivalent. Consider the
universal cover $p:\widetilde{\Sigma}\to\Sigma$ as above. Abuse the definition
by still denoting $a,b:[0,1]\to\Sigma$ two curves with $a(0)=b(0)=0$ whose
germs respectively equal $a,b$. Let $\widetilde{a},\widetilde{b}$ denote lifts
of (the restrictions to $(0,1]$ of) $a,b$ in $\widetilde{\Sigma}$. Then the
number $0pt(a,b)=w$ is characterized by the two following properties:

  * (i)

for every $t_{0}\in(0,1]$, the restriction of $\widetilde{a}$ to $(0,t_{0}]$
meets at least $w$ integer translates of $\widetilde{b}$;

  * (ii)

there exists $t_{0}\in(0,1]$ such that the restriction of $\widetilde{a}$ to
$(0,t_{0}]$ meets exactly $w$ integer translates of $\widetilde{b}$;

Analogously to the previous section, the distance in the graph of germs is
characterized by the local relative width.

######  Proposition 4.3.

Let $a\neq b$ be two vertices of the graph $\mathcal{A}(x)$. Then $a$ and $b$
are in the same connected component of $\mathcal{A}(x)$ if and only if
$0pt(a,b)<+\infty$. In this case, the distance in the graph is given by

$d(a,b)=0pt(a,b)+1.$

The proof is very similar to the proof in the previous section. Details are
left to the reader.

###  4.5. Proof of Proposition 4.1

Let $a,b,a^{\prime},b^{\prime}$ be vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ such that $T(a,b)$ and
$T(a^{\prime},b^{\prime})$ hold, and assume
$\mathrm{Point}(a,b)=\mathrm{Point}(a^{\prime},b^{\prime})$. Denote $x$ the
common intersection point.

If $c$ is another vertex such that $\\{a,b,c\\}$ is a 3-clique of type bouquet
or extra bouquet, then the germs $\lfloor a\rfloor_{x}$ and $\lfloor
c\rfloor_{x}$ are disjoint, thus obviously comparable. This entails the direct
implication in Proposition 4.1.

Let us prove the converse implication. We assume that the germs $\lfloor
a\rfloor_{x}$ and $\lfloor a^{\prime}\rfloor_{x}$ are comparable. In other
words, there exists arcs $\alpha_{0},\dots,\alpha_{n}$ with $\alpha_{i}(0)=x$
and whose sequence of corresponding germs is a path from $\lfloor
a\rfloor_{x}$ to $\lfloor a^{\prime}\rfloor_{x}$ in the graph of germs. Note
that each germ $\alpha_{i}$ may be extended to a non separating curve $a_{i}$,
and we can find another non separating curve $b_{i}$ such that
$T(a_{i},b_{i})$ holds. Thus the end of the proof is a direct consequence of
the following lemma.

######  Lemma 4.4.

Let $a,b,a^{\prime},b^{\prime}$ be vertices in
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ such that $T(a,b)$ and
$T(a^{\prime},b^{\prime})$ hold. Assume that for some choices $\lfloor
a\rfloor_{x},\lfloor a^{\prime}\rfloor_{x}$ of arcs at $x$ included
respectively in $a$ and $a^{\prime}$, the germs $\lfloor a\rfloor_{x},\lfloor
a^{\prime}\rfloor_{x}$ intersect only at $x$. Then
$\\{a,b\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$.

###### Proof of the lemma.

Let $c$ be an arc that contains $x$ in its interior and locally coincides with
$\lfloor a\rfloor_{x}\cup\lfloor a^{\prime}\rfloor_{x}$. Extend $c$ into a non
separating closed curve, still denoted $c$, and consider any other non
separating curve $d$ such that $T(c,d)$ holds. Since $c$ locally “half
coincides” near $x$ with both $a$ and $a^{\prime}$, we may apply Proposition
3.11 twice, and get that
$\\{a,b\\}\diamondvert\\{c,d\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$. ∎

###  4.6. Curves and diffeomorphisms

In this short subsection we explain how one can use the fine curve graph to
detect fundamental non differentiability.

Let $\Phi$ be an automorphism of
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$. We introduce the following
property $D(\Phi)$:

_For every vertices $a$, $b$ of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$
such that $T(a,b)$ holds, if
$\mathrm{Point}(\Phi(a),\Phi(b))=\mathrm{Point}(a,b)$ then there exists
$a^{\prime},b^{\prime}$ such that $T(a^{\prime},b^{\prime})$ holds,
$\mathrm{Point}(a^{\prime},b^{\prime})=\mathrm{Point}(a,b)$ and
$\\{\Phi(a^{\prime}),\Phi(b^{\prime})\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$._

Note that this property is clearly invariant under conjugacy in the group of
automorphisms. Let $h$ be a homemorphism of $\Sigma$, and denote
$\Phi=\Phi_{h}$ the action of $h$ on the graph
$\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$.

######  Observation 4.5.

If $h$ is differentiable everywhere, then property $D(\Phi_{h})$ holds.

Indeed, hypothesis $\mathrm{Point}(\Phi(a),\Phi(b))=\mathrm{Point}(a,b)$ is
equivalent to the fact that the point $x=\mathrm{Point}(a,b)$ is a fixed point
of $h$. Since $h$ is differentiable at $x$, it is easy to check that every
germ of smooth arc at $x$ is comparable to its image. Take any two smooth
curves $a^{\prime},b^{\prime}$ such that $T(a^{\prime},b^{\prime})$ and
$\mathrm{Point}(a^{\prime},b^{\prime})=\mathrm{Point}(a,b)$, then Proposition
4.1 tells us that
$\\{\Phi(a^{\prime}),\Phi(b^{\prime})\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$.

Now consider a particular homeomorphism $h$ of $\Sigma$ and assume that $h$
admits a fixed point where, for some local polar coordinates, $h$ writes

$(r,\theta)\mapsto(r,\theta+\frac{1}{r}).$

######  Observation 4.6.

Property $D(\Phi_{h})$ does not hold.

An easy proof of this is obtained by considering the _local rotation interval_
of $h$ at $x$, as defined in [8], section 2.3. Indeed, the local rotation
interval of $h$ at $x$ equals $\\{+\infty\\}$, which accounts for the fact
that orbits turn faster and faster around $x$, in the positive direction, as
we get nearer and nearer to $x$ (the quickest way to check this is to show
that the _local rotation set_ of $h$ at $x$ is $\\{+\infty\\}$, and then to
apply Théorème 3.9 of [8] that relates the local rotation set and the local
rotation interval). We argue by contradiction to show that property
$D(\Phi_{h})$ does not hold. Assuming property $D(\Phi_{h})$ holds, consider
curves $a,b$ such that $T(a,b)$ holds and $\mathrm{Point}(a,b)=x$. Let
$a^{\prime},b^{\prime}$ be given by property $D(\Phi_{h})$, such that
$\\{\Phi(a^{\prime}),\Phi(b^{\prime})\\}\diamondvert\\{a^{\prime},b^{\prime}\\}$.
The reverse direction of Proposition 4.1 tells us that the germs of
$h(a^{\prime})$ and $a^{\prime}$ are comparable at $x$. This entails easily,
from the definition, that the local rotation interval of $h$ at $s$ is a
bounded interval, a contradiction.

##  5\. Fine graph of smooth curves

In this section we address the case of smooth curves, and prove Theorem 1.2
and Proposition 1.3. In all the section, $\Sigma$ will be a connected,
nonspherical surface without boundary, endowed with a smooth structure. In
Section 5.3 we will restrict to the orientable case.

###  5.1. From bijections to higher regularity

One step in the proof of Theorem 1.1 was Proposition 3.1, in which we proved
that if an automorphism of $\mathcal{NC}_{\pitchfork}^{\dagger}(\Sigma)$ is
supported by a bijection of $\Sigma$, then that bijection is a homeomorphism
of $\Sigma$.

We may ask the same question about automorphisms of
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$, and this paragraph is
devoted to the proof of the following two statements. We denote by
$\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)$ the group of bijections of
$\Sigma$ which preserve the family of smooth, nonseparating closed curves, and
preserve transversality between such curves. The first statment below
justifies this notation. Here, for simplicity we restrict to the case of
orientable surfaces.

######  Proposition 5.1.

Let $\Sigma$ be a connected, non spherical orientable surface. The group
$\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)$ is contained in
$\operatorname{Homeo}(\Sigma)$.

###### Proof of Proposition 5.1.

Let $h\in\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)$. We will prove that
the image under $h$ of any open set is an open set. This is the continuity of
$h^{-1}$, and by applying the argument to $h$ we also get the continuity of
$h$.

To do this, we only need to consider the images of a family of sets that
generates the topology. Given three non separating curves $a,b,c$, we denote
$V(a;b,c)$ the union of all the non separating curves $d$ that meet $a$ and
are disjoint from $b$ and $c$.

######  Observation 5.2.

The set $V(a;b,c)$ is the union of some of the connected components of the
complement of $b\cup c$ that meet $a$. In particular, it is an open set.

Indeed, let $x$ be a point of $V(a;b,c)$. By definition there is a non
separating curve $d$ passing through $x$ and meeting $a$ but not $b$ nor $c$.
Consider another point $y$ that belongs to the connected component $V_{x}$ of
the complement of $b\cup c$ that contains $x$. By modifying $d$ using an arc
connecting $x$ to $y$ in $V_{c}$, we find another curve $d^{\prime}$, isotopic
to $d$, still meeting $a$ but not $b$ nor $c$, and passing through $y$. This
proves that $V(a;b,c)$ contains $V_{x}$, and the observation follows.

Now let $a$ be a nonseparating curve. Let $a^{+},a^{-}$ be obtained by pushing
$a$ to both sides. Then $V(a;a^{+},a^{-})$ is a neighborhood of $a$, and by
making $a^{+}$ and $a^{-}$ vary we get a basis of neighborhoods
${\mathcal{B}}(a)$ of the curve $a$. The union of all these families
${\mathcal{B}}(a)$ clearly generates the topology of $\Sigma$.

Thus it suffices to check that the image under $h$ of each set $V(a;b,c)$ is
an open set. But since $h$ is a bijection, we have

$h(V(a;b,c))=V(h(a);h(b),h(c)).$

By hypothesis $h(a),h(b),h(c)$ are nonseparating closed curves, and by the
observation this set is open. ∎

We now prove Proposition 1.3 stated in the introduction, namely the existence
of elements of $\operatorname{Homeo}_{\infty\pitchfork}(\Sigma)$ that are not
smooth.

###### Proof of Proposition 1.3.

We will construct a homeomorphism $F\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$,
which is not differentiable at the origin, but such that both $F$ and $F^{-1}$
send smooth curves to smooth curves. The construction can easily be modified
to make $F$ compactly supported, and then be transported on our surface
$\Sigma$. It will be clear from the constuction that this map preserves
transversality.

Let $h\colon\mathbb{R}\to\mathbb{R}$ be a smooth diffeomorphism supported in
the segment $[1/2,2]$. That is to say, $h(x)=x$ for all $x$ outside $[1/2,2]$;
we suppose however that $h(1)\neq 1$. We consider the map $F$ defined by
$F(x,y)=(x,xh(y/x))$ if $x\neq 0$, and $F(x,y)=(x,y)$ otherwise. We claim that
this map has the desired property.

This map, as well as its inverse, is obviouly smooth in restriction to
$\mathbb{R}^{2}\smallsetminus\\{(0,0)\\}$. Direct computation shows that $F$
has directional derivatives in all directions around the origin, but the
“differential” fails to be linear: both partial derivatives are those of the
identity, while the directional derivative in the direction $(1,1)$ is not. So
$F$ is not differentiable at the origin.

Now, let $\gamma\colon\mathbb{R}\to\mathbb{R}^{2}$ be a smooth, proper
embedding. If $(0,0)$ is not in the image of $\gamma$, then of course,
$F\circ\gamma$ is still smooth. So suppose, say, that $\gamma(0)=(0,0)$. If
$\gamma^{\prime}(0)$ lies outside the two (opposite) sectors of vectors of
slopes between $1/2$ and $2$, then $F\circ\gamma$ and $\gamma$ have the same
germ at $0$. Otherwise, and up to reparameterization, we can write, near $0$,
$\gamma(t)=(t,\alpha(t))$ where $\alpha$ is a smooth map (satisfying
$\alpha(0)=0)$), from a neighborhood of $0$, to $\mathbb{R}$. This yields the
formula

$F\circ\gamma(t)=\left(t,th\left(\frac{\alpha(t)}{t}\right)\right).$

Now, the smoothness of $F\circ\gamma$ follows from the following elementary
observation.

Claim. Let $\alpha\colon\mathbb{R}\to\mathbb{R}$ be a smooth map satisfying
$\alpha(0)=0$. Then the map $t\mapsto\frac{\alpha(t)}{t}$ when $t\neq 0$, and
$\alpha^{\prime}(0)$ when $t=0$, is smooth.

Indeed, by the fundamental theorem of Calculus, for all
$t\in\mathbb{R}^{\ast}$ we have

$\frac{\alpha(t)}{t}=\int_{0}^{1}\alpha^{\prime}(ts)ds,$

and this integral with parameter can be differentiated indefinitely.111We
borrow this elegant argument from [4]. ∎

###  5.2. A weak convergence for sequences of curves

In order to prove Theorem 1.2, we now explain how to recognize configurations
of smooth curves. Given two vertices $a$ and $b$ of the graph
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$, we will denote by $a-b$
the property that they are neighbors in the graph. If
$(f_{n})_{n\in\mathbb{N}}$ is a sequence of vertices, we denote by
$(f_{n})_{n\in\mathbb{N}}-a$ the property that for all $n$ large enough,
$f_{n}-a$. The first property of sequences we may recover from the graph is
the distinction of what curves go to infinity.

######  Lemma 5.3.

Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$. The following are
equivalent.

  * -

for all $d$, we have $(f_{n})_{n\in\mathbb{N}}-d$;

  * -

for every compact subset $K$ of $\Sigma$, for every $n$ large enough, $K\cap
f_{n}=\emptyset$.

###### Proof.

The second statement obviously implies the first, as we may just take $K=d$.
Let us prove the converse implication by contraposition. Suppose $K$
intersects infinitely many $f_{n}$. Since $K$ is compact, there is a point
$x\in K$, such that every neighborhood of $x$ intersects infinitely many
$f_{n}$. We consider two open sets $B_{1}$ and $B_{2}$ with $x\in B_{1}$ and
$\overline{B_{1}}\subset B_{2}$, and three bottle-shaped arcs, as in Figure
13. These arcs may be continued to form three nonseparating closed curves,
$d_{1}$, $d_{2}$ and $d_{3}$. Now, let $n$ be such that $f_{n}$ enters
$B_{1}$. If $f_{n}$ does not enter nor leave $B_{2}$ through the neck of the
bottle corresponding to $d_{1}$, then we cannot have $f_{n}-d_{1}$, since
$f_{n}$ and $d_{1}$ have to intersect at least twice. Hence, $f_{n}$ passes
through the neck of $d_{1}$, and in order to impose that $f_{n}-d_{2}$,
another arc of $f_{n}$ has to get out of the bottle corresponding of $d_{2}$
through its neck. But then $f_{n}$ has to meet $d_{3}$ twice, and we cannot
have $f_{n}-d_{3}$. In other words, for all $n$ such that $f_{n}$ enters
$B_{1}$, we can’t have $f_{n}-d_{1}$ and $f_{n}-d_{2}$ and $f_{n}-d_{3}$,
hence the first statement is not true, and our implication is proved.

import geometry;

fill(circle((0,0),35), lightgray);

fill(circle((0,0),17), lightgreen); label("$B_{1}$", (0,0), NW);
label("$B_{2}$", 35*dir(0), E);

dot((0,0)); label("$x$", (0,0), SE);

real r1 = 29, r2 = 25, r3 = 21; path d1 = (-40,
8)–(-32,8)right..(r1*dir(135))dir(45)..(r1*dir(90))right..
(r1*dir(0))down..(r1*dir(-90))left..(r1*dir(-135))dir(135)..
left(-32,-8)–(-40,-8);

path d2 = (-40, 8)–(-32,8)right..(r2*dir(135))dir(45)..(r2*dir(90))right..
(r2*dir(0))down..(r2*dir(-90))left..(r2*dir(-135))dir(135)..
left(-32,-8)–(-40,-8); path d2tourne = rotate(120)*d2;

path d3 = (-40, 8)–(-32,8)right..(r3*dir(135))dir(45)..(r3*dir(90))right..
(r3*dir(0))down..(r3*dir(-90))left..(r3*dir(-135))dir(135)..
left(-32,-8)–(-40,-8); path d3tourne = rotate(-120)*d3;

draw(d1, blue); draw(rotate(120)*d2, red); draw(rotate(-120)*d3, green);

label("Any arc from $B_{1}$ to the outside of $B_{2}$ and", (0,-50));
label("disjoint from $d_{1}$ must meet $d_{2}$ and $d_{3}$.", (0,-64));

Figure 13. Three bottles

∎

Thus, we will say here that a sequence $(f_{n})_{n\in\mathbb{N}}$ is relevant
if it has no subsequence $(f_{\varphi(n)})_{n\in\mathbb{N}}$ such that for all
$d$, $(f_{\varphi(n)})_{n\in\mathbb{N}}-d$. We now explore, for such
sequences, the following notion of convergence. We say that a relevant
sequence $(f_{n})$ converges in a weak sense to a curve $a$ if for every $d$
such that $a-d$, we have $(f_{n})-d$. We denote this property by
$W((f_{n}),a)$.

######  Lemma 5.4.

Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of vertices, and $a$ be a vertex
of $\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$.

  * $\bullet$

If $W((f_{n}),a)$, then the sequence $(f_{n})_{n\in\mathbb{N}}$ converges in
the Hausdorff topology to $a$: for every neighborhood $V$ of the curve $a$,
for all $n$ large enough, we have $f_{n}\subset V$.

  * $\bullet$

If the sequence $(f_{n})_{n\in\mathbb{N}}$ of curves, with some appropriate
parameterization, converges in $C^{1}$-topology to $a$, then
$W((f_{n})_{n\in\mathbb{N}},a)$.

###### Proof.

We first prove the first point. Let us first mention that the statement we
wrote is indeed equivalent to the Hausdorff convergence, because any essential
curve in a small enough neighbourhood of $a$ must pass close to every point of
$a$, and thus is Hausdorff-close to $a$. Suppose for contradiction that for
some neighborhood $V$ of $a$, we have $f_{n}\not\subset V$ infinitely often.
Then we may find a point $x$, not in $\overline{V}$, such that every
neighborhood of $x$ is visited by infinitely many $f_{n}$. We may construct
three bottle-shaped arcs around $x$ exactly as in the proof of the preceding
lemma, and complete these arcs to non separating simple closed curves $d_{1}$,
$d_{2}$, $d_{3}$, which can be requested to satisfy $a-d_{i}$ for $i=1,2,3$.
Then we cannot have $f_{n}-d_{i}$ for all $i\in\\{1,2,3\\}$ for the same
reason as in this preceding proof, and this contradicts the hypothesis that
$W((f_{n}),a)$ holds.

The second point is the well known stability of transversality in the
$C^{1}$-topology. ∎

######  Remark 5.5.

In fact, the condition $W((f_{n}),a)$ implies $C^{0}$-convergence, in the
following sense. Given a parameterization $\alpha$ of $a$, we can choose the
parameterizations of the $f_{n}$’s yielding a sequence of parmaetrized curves
converging uniformly to $\alpha$.

As we will not use this fact, we only sketch a quick argument. Let $V$ be a
small tubular neighborhood of $a$, and let $d_{1},\ldots,d_{N}$ be simple
closed nonseparating curves, each meeting $a$ transversely at one point, and
cutting $V$ in small chunks $V_{1},\ldots,V_{N}$ that are met by $\alpha$ in
that cyclic order. Let $n$ be large enough so that $f_{n}\subset V$ and
$f_{n}-d_{i}$ for each $i$. Then $f_{n}\cap V_{i}$ is connected, and $f_{n}$
visits the pieces $V_{1},\ldots,V_{N}$ in that order. Hence we may choose a
parameterization of $f_{n}$, say, $F_{n}$, in such a way that for all
$i\in\\{1,\ldots,N\\}$ and all $t$, $F_{n}(t)\in V_{i}$ if and only if
$\alpha(t)\in V_{i}$. This implies that $F_{n}$ is uniformly close to
$\alpha$.

This notion is actually somewhere strictly in between $C^{0}$-convergence and
$C^{1}$-convergence, as we remark in the following example. This construction
will play a crucial role below in the proof of Theorem 1.2.

######  Example 5.6.

Let $a$ be a smooth nonseparating curve in $\Sigma$. We choose a point $p$ in
$a$, and a chart around one of its point, diffeomorphic to $\mathbb{R}^{2}$,
in such a way that $a$ corresponds to the axis of equation $y=0$ in that
plane. In this chart, we consider the functions $f_{1}\colon
x\mapsto\frac{2}{1+x^{2}}$, and for all $n\geqslant 2$, $f_{n}\colon
x\mapsto\frac{f_{1}(nx)}{n}$. Abusively, we still denote their graphs by the
same letters, and then, we may extend these arcs, viewed in $\Sigma$, to
simple closed curves (consisting of pushing $a$ aside), that converges $C^{1}$
to $a$ outside of the point $p$. Abusively we still use the same letters
$f_{n}$ to denote these closed curves.

Obviously, the sequence $(f_{n})_{n\geqslant 1}$ does not converge $C^{1}$ to
$a$, because $f_{n}$ has slope $-1$ at the point $(1/n,1/n)$.

Nonetheless, we claim that $W((f_{n}),a)$ holds. Indeed, let $d$ be such that
$a-d$. Since the sequence $(f_{n})$ converges $C^{1}$ to $a$ everywhere except
at the origin of this $\mathbb{R}^{2}$ chart, the only case in which it is not
already clear that $(f_{n})-d$ is when $d$ meets $a$ transversely at the
origin. If $d$ has a strictly positive slope there, then for $n$ large enough,
the intersection $f_{n}\cap d$ will be transverse because the slopes of
$f_{n}$ are all negative in the region $x>0$. The case when $d$ has negative
slope is symmetric, and if $d$ has vertical slope, it will be transverse with
$f_{n}$ since these have bounded slopes.

###  5.3. Recognizing configurations of smooth curves

In this last section we assume that our surface $\Sigma$ is orientable.

If $a$, $b$ are vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$, we will denote by
$D_{\infty}(a,b)$ the condition that $a-b$ and for all sequences $(f_{n})$ and
$(g_{m})$ such that $W((f_{n}),a)$ and $W((g_{m}),b)$, we have $f_{n}-g_{m}$
for all $m,n$ large enough.

######  Lemma 5.7.

Let $a$, $b$ be smooth nonseparating curves. Then $D_{\infty}(a,b)$ holds if
and only if $a$ and $b$ are disjoint.

###### Proof.

Suppose first that $a$ and $b$ are disjoint. Then they admit disjoint
neighborhoods, $V_{1}$ and $V_{2}$. For any sequences $(f_{n})$ and $(g_{m})$
with $W((f_{n}),a)$ and $W((g_{m}),b)$, for all $m,n$ large enough we have
$f_{n}\subset V_{1}$ and $g_{m}\subset V_{2}$, by Lemma 5.4. Hence,
$f_{n}-g_{m}$ for all $m,n$ large enough, and $D_{\infty}(a,b)$ holds indeed.

Now, suppose that $a$ and $b$ are not disjoint. Since $a-b$, the curves $a$
and $b$ have a transverse intersection, and in an appropriate chart
diffeomorphic to $\mathbb{R}^{2}$, the curves $a$ and $b$ correspond
respectively to the axes $y=0$ and $x=0$.

Then we may form a sequence $(f_{n})$ such that $W((f_{n}),a)$ exactly as in
example 5.6, and for $(g_{n})$ we just exchange coordinates $x$ and $y$. For
all $n$, the curves $f_{n}$ and $g_{n}$ have a non transverse intersection
point (at $(1/n,1/n)$ in the chart of Example 5.6), hence the condition
$D_{\infty}(a,b)$ does not hold. ∎

In the end of the proof, the curves $f_{n}$ and $g_{n}$ were tangent at their
intersection point, hence not neighbors in the graph
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$. This may look accidental,
but upon changing the formula of $f_{1}$ in Example 5.6 to
$x\mapsto\frac{3}{2+x^{2}}$, for example, we get three intersection points.

From now on, we restrict ourselves to the case of orientable surfaces. One
reason is that it would take more work to recover the extra bouquets and not
only the bouquets; one other reason is that the next lemma works best when at
least one of $a$, $b$ or $c$ is two-sided.

######  Lemma 5.8.

Suppose $\Sigma$ is orientable. Let $\\{a,b,c\\}$ be a $3$-clique of
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$, and suppose that these
three curves pairwise intersect. Then the following are equivalent.

  1. (1)

This $3$-clique is of type bouquet.

  2. (2)

There exists a relevant sequence $(f_{n})$ of vertices of
$\mathcal{NC}_{\pitchfork}^{\dagger\infty}(\Sigma)$, which are all disjoint
from $a$, and such that for all $d$ disjoint from $b$ and satisfying $c-d$, we
have $(f_{n})-d$.

###### Proof.

Suppose $\\{a,b,c\\}$ is of type bouquet. Then the sequence
$(f_{n})_{n\in\mathbb{N}}$ can be constructed explicitly. Let $p=b\cap c$. Fix
a (smooth) metric on $\Sigma$, we remove all points of the ball
$B(p,\frac{1}{n})$ off the curves $b$ and $c$, this gives two arcs. There is a
natural way of adding smooth subarcs of $B(p,\frac{1}{n})$ in order to extend
this union of two arcs, to a curve $f_{n}$ which does not intersect $a$. In a
one-holed torus neighborhood of $b\cup c$, with a choice of meridian and
longitude coming from $b$ and $c$, these curves $f_{n}$ have slope $1$, or
$-1$; these are indeed nonseparating simple closed curves. Now if $d$ is
disjoint from $b$ and satisfies $c-d$, then either $d$ is disjoint from $c$,
and then $f_{n}$ is disjoint from $d$ for all $n$ large enough, or $d$ has a
transverse intersection with $c$ at a point distinct from $p$, and we also
have $f_{n}-d$ for all $n$ large enough. Thus, (1) implies (2).

Conversely, suppose (2). We first claim that the sequence $(f_{n})$ then
concentrates into neighborhoods of $b\cup c$. For contradiction, suppose that
we can find a neighborhood $V$ of $b\cup c$, such that $f_{n}\not\subset V$
for infinitely many $n$. Then, there exists a point $x$, with $x\not\in b\cup
c$, and such that every neighborhood of $x$ meets infinitely many $f_{n}$.
Then we may choose three bottle-shaped arcs around $x$, and complete them into
curves $d_{1}$, $d_{2}$ and $d_{3}$ disjoint from $b$ and satisfying $d_{j}-c$
for $j=1,2,3$. Indeed, we may start with a curve $d_{0}$ obtained by pushing
$b$ aside, and then perform surgeries on $d_{0}$. The same reasoning as in the
proof of Lemma 5.3 shows that $f_{n}\not-d_{j}$ for some $j\in\\{1,2,3\\}$ and
for infinitely many $n$, contradicting the hypothesis (2).

Now, suppose for contradiction that $\\{a,b,c\\}$ is a necklace. Then, for a
sufficiently small regular neighborhood $V$ of $b\cup c$, we may observe that
$V\smallsetminus a$ is contractible. Hence it cannot contain any nonseparating
simple closed curve $f_{n}$, and the hypothesis (2) cannot be fullfilled. This
proves that (2) implies (1). ∎

Now the proof of Theorem 1.2 is a straightforward adaptation of the proof of
Theorem 1.1. The statements about connectedness of complexes of arcs, for
example, are equivalent to their counterparts with regularity, because of the
argument of homotopy recalled in the proof of Lemma 3.8 and borrowed from [2].

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