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abstract: 'We prove that each superinjective simplicial map of the complex of curves of a compact, connected, nonorientable surface is induced by a homeomorphism of the surface, if $g + n \leq 3$ or $g + n \geq 5$, where $g$ is the genus of the surface and $n$ is the number of the boundary components.'
author:
- Elmas Irmak
title: Superinjective Simplicial Maps of the Complexes of Curves on Nonorientable Surfaces
---

Key words: Mapping class groups, simplicial maps, nonorientable surfaces

MSC: 32G15, 20F38, 30F10, 57M99

Introduction
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Let $N$ be a compact, connected, nonorientable surface of genus $g$ (connected sum of $g$ copies of projective planes) with $n$ boundary components. Mapping class group, $Mod_N$, of $N$ is defined to be the group of isotopy classes of all self-homeomorphisms of $N$. The *complex of curves*, $\mathcal{C}(N)$, on $N$ is an abstract simplicial complex defined as follows: A simple closed curve on $N$ is called [*nontrivial*]{} if it does not bound a disk, a mobius band, and it is not isotopic to a boundary component of $N$. The vertex set, $A$, of $\mathcal{C}(N)$ is the set of isotopy classes of nontrivial simple closed curves on $N$. A set of vertices forms a simplex in $\mathcal{C}(N)$ if they can be represented by pairwise disjoint simple closed curves. The geometric intersection number $i([a], [b])$ of $[a]$, $[b] \in A$ is the minimum number of points of $x \cap y$ where $x \in [a]$ and $y \in [b]$. A simplicial map $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ is called [*superinjective*]{} if the following condition holds: if $[a], [b]$ are two vertices in $\mathcal{C}(N)$ such that $i([a], [b]) \neq 0$, then $i(\lambda([a]), \lambda([b])) \neq 0$. The main result of this paper:

Let $N$ be a compact, connected, nonorientable surface of genus $g$ with $n$ boundary components. Suppose that either $g + n \leq 3$ or $g + n \geq 5$. If $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ is a superinjective simplicial map, then $\lambda$ is induced by a homeomorphism $h : N \rightarrow N$ (i.e $\lambda([a]) = [h(a)]$ for every vertex $[a]$ in $\mathcal{C}(N)$).

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[The author was supported by Faculty Research Incentive Grant, BGSU.]{}

The mapping class groups and complex of curves on orientable surfaces are defined similarly as follows: Let $R$ be a compact, connected, orientable surface. Mapping class group, $Mod_R$, of $R$ is defined to be the group of isotopy classes of orientation preserving homeomorphisms of $R$. Extended mapping class group, $Mod_R^*$, of $R$ is defined to be the group of isotopy classes of all self-homeomorphisms of $R$. The complex of curves, $\mathcal{C}(R)$, on $R$ is defined as an abstract simplicial complex. The vertex set is the set of isotopy classes of nontrivial simple closed curves, where here nontrivial means it does not bound a disk and it is not isotopic to a boundary component of $R$. A set of vertices forms a simplex in $\mathcal{C}(R)$ if they can be represented by pairwise disjoint simple closed curves.

Ivanov proved that the automorphism group of the curve complex is isomorphic to the extended mapping class group on orientable surfaces. As an application he proved that isomorphisms between any two finite index subgroups are geometric. Ivanov’s results were proven by Korkmaz in [@K1] for lower genus cases. Luo gave a different proof of these results for all cases in [@L]. After Ivanov’s work, mapping class group was viewed as the automorphism group of various geometric objects on orientable surfaces. These objects include Schaller’s complex (see [@Sc] by Schaller), the complex of pants decompositions (see [@M] by Margalit), the complex of nonseparating curves (see [@Ir3] by Irmak), the complex of separating curves (see [@BM1] by Brendle-Margalit, and [@MV] by McCarthy-Vautaw), the complex of Torelli geometry (see [@FIv] by Farb-Ivanov), the Hatcher-Thurston complex (see [@IrK] by Irmak-Korkmaz), and complex of arcs (see [@IrM] by Irmak-McCarthy). As applications, Farb-Ivanov proved that the automorphism group of the Torelli subgroup is isomorphic to the mapping class group in [@FIv], and McCarthy-Vautaw extended this result to $g \geq 3$ in [@MV].

On orientable surfaces: Irmak proved that superinjective simplicial maps of the curve complex are induced by homeomorphisms of the surface to classify injective homomorphisms from finite index subgroups of the mapping class group to the whole group (they are geometric except for closed genus two surface) for genus at least two in [@Ir1], [@Ir2], [@Ir3]. Behrstock-Margalit and Bell-Margalit proved these results for lower genus cases in [@BhM] and in [@BeM]. Brendle-Margalit proved that superinjective simplicial maps of separating curve complex are induced by homeomorphisms, and using this they proved that an injection from a finite index subgroup of $K$ to the Torelli group, where $K$ is the subgroup of mapping class group generated by Dehn twists about separating curves, is induced by a homeomorphism in [@BM1]. Shackleton proved that injective simplicial maps of the curve complex are induced by homeomorphisms in [@Sc] (he also considers maps between different surfaces), and he obtained strong local co-Hopfian results for mapping class groups.

On nonorientable surfaces: For odd genus cases, Atalan proved that the automorphism group of the curve complex is isomorphic to the mapping class group if $g + r \geq 6$ in [@A]. Irmak proved that each injective simplicial map from the arc complex of a compact, connected, nonorientable surface with nonempty boundary to itself is induced by a homeomorphism of the surface in [@Ir4]. She also proved that the automorphism group of the arc complex is isomorphic to the quotient of the mapping class group of the surface by its center. Atalan-Korkmaz proved that the automorphism group of the curve complex is isomorphic to the mapping class group in [@AK] if $g + r \geq 5$. They also proved that two curve complexes are isomorphic if and only if the underlying surfaces are homeomorphic. In this paper we use some results from [@AK]. Our techniques give simpler proofs of the some of the results in [@AK]. Since an automorphism of $\mathcal{C}(N)$ is a superinjective simplicial map, our result implies that automorphisms of $\mathcal{C}(N)$ are induced by homeomorphisms of $N$ which was proved in [@AK].

Some small genus cases
======================

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In this section we will prove our main results for $(g, n) \in \{(1, 0), (1, 1), (1, 2),$ $ (2, 0), (2, 1)\}$. We note that since simplicial maps preserve geometric intersection zero, superinjective simplicial maps preserve geometric intersection zero and nonzero properties.

\[A\] Let $N$ be a compact, connected, nonorientable surface of genus $g$ with $n$ boundary components. Suppose that either $(g, n) \in \{(1, 0), (1, 1), (1, 2),$ $(2, 0), (2, 1)\}$. If $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ is a superinjective simplicial map, then $\lambda$ is induced by a homeomorphism $h : N \rightarrow N$ (i.e $\lambda([a]) = [h(a)]$ for every vertex $[a]$ in $\mathcal{C}(N)$).

If $(g, n) = (1, 0)$, $N$ is the projective plane. There is only one element (isotopy class of a 1-sided curve) in the curve complex. Hence, any superinjective simplicial map is induced by the identity homeomorphism. If $(g, n) = (1, 1)$, $N$ is Mobius band. There is only one element (isotopy class of a 1-sided curve) in the curve complex. Hence, any superinjective simplicial map is induced by the identity homeomorphism.

If $(g, n) = (1, 2)$, there are only two elements in the curve complex (see [@Sc]). They are the isotopy classes of $a$ and $b$ as shown in Figure \[figure1\]. We see that $i([a], [b]) = 1$. So, the superinjective simplicial map cannot send both of these elements to the same element. If this simplicial map fixes each of them, then it is induced by the identity homeomorphism, if it switches them then it is induced by a homeomorphism that switches the curves $a$ and $b$.

If $(g, n) = (2, 0)$, there are only three elements in the curve complex (see [@Sc]). They are the isotopy classes of $a$, $b$ and $c$ as shown in Figure \[figure1\] (ii). We see that $i([a], [b]) = 1$, $i([a], [c]) = 1$ and $i([b], [c]) = 0$. Since superinjective simplicial maps preserve geometric intersection zero and nonzero properties, $\lambda$ fixes $[a]$. If it also fixes each of $[b]$ and $[c]$ then it is induced by the identity homeomorphism, if it switches $[b]$ and $[c]$ then it is induced by a homeomorphism that switches the one sided curves $b$ and $c$, while fixing $a$ up to isotopy.

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If $(g, n) = (2, 1)$, then the curve complex is given by Scharlemann in [@Sc] as follows: Let $a$ and $b$ be as in Figure \[figure2\]. We see that $i([a], [b]) = 1$. The vertex set of the curve complex is $\{[a], [b], t_a^m ([b]) : m \in \mathbb{Z} \}$, where $t_a$ is the Dehn twist about the 2-sided curve $a$. The complex is shown in Figure \[figure2\]. Since $i([a], t_a^m ([b])) \neq 0$ for any $m$, we have $i(\lambda([a]), \lambda(t_a^m ([b])) \neq 0$ for any $m$, i.e. $\lambda([a])$ will not be connected by an edge to any other vertex. For any $m$, the elements $t_a^m ([b])$ and $t_a^{m+1} ([b])$ cannot be send to the same element by $\lambda$ since they have different intersection information with $t_a^{m-1} ([b])$, and $\lambda$ preserves geometric intersection zero and nonzero by definition. Similarly, $t_a^m ([b])$ and $t_a^{n} ([b])$ cannot be send to the same element by $\lambda$ for any $m, n$ with $m \neq n$. For any $m$, $t_a^m ([b])$ cannot be send to $[a]$, since $i(t_a^{m-1} ([b]), t_a^m ([b])) = 0$ and $i(t_a^{m-1} ([b]), [a]) \neq 0$, and so we will have $i(\lambda(t_a^{m-1} ([b])), \lambda(t_a^m ([b]))) = 0$ and $i(\lambda(t_a^{m-1} ([b])), \lambda([a]) \neq 0$. So, either $\lambda(t_a^m ([b])) = t_a^k ([b])$, $\lambda(t_a^{m-1} ([b]))= t_a^{k-1} ([b])$, $\lambda(t_a^{m+1} ([b]))= t_a^{k+1} ([b])$ or $\lambda(t_a^m ([b])) = t_a^k ([b])$, $\lambda(t_a^{m-1} ([b]))= t_a^{k+1} ([b])$, $\lambda(t_a^{m+1} ([b]))= t_a^{k-1} ([b])$ for some $k \in \mathbb{Z}$. In either case, we also have that $\lambda$ fixes $[a]$. By cutting $N$ along $a$ we get a cylinder with one puncture as shown in Figure \[figure2a\]. There is a reflection of the cylinder interchanging front face with the back face which fixes $b$ pointwise. This gives us a homeomorphism $r$ of $N$ such that $r(b) = b$ and $r(a) = a^{-1}$. So, $(r)_\#([b]) = [b]$ and $(r)_\#(t_a([b])) = t_{a^{-1}}([b]) = {t_a}^{-1}([b])$. The map $r_\#$ reflects the graph in Figure \[figure2\] at $[b]$ and fixes $[a]$. Now it is easy to see that $\lambda$ is induced by $t_a^k$ or $t_a^k \circ r$ for some $k \in \mathbb{Z}$.

Properties of Superinjective Simplicial Maps
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Let $N$ be a compact, connected, nonorientable surface of genus $g$ with $n$ boundary components. We will list some properties of superinjective simplicial maps. First we give some definitions.

Let $P$ be a set of pairwise disjoint, nonisotopic, nontrivial simple closed curves on $N$. $P$ is called a [*pair of pants decomposition*]{} of $N$ if each component $\Delta$ of the surface $N_P$, obtained from $N$ by cutting $N$ along $P$, is a pair of pants. A pair of pants of a pants decomposition is the image of one of these connected components under the quotient map $q: N_P \rightarrow N$. Let $a$ and $b$ be two distinct elements in a pair of pants decomposition $P$. Then $a$ is called [*adjacent*]{} to $b$ w.r.t. $P$ iff there exists a pair of pants in $P$ which has $a$ and $b$ on its boundary.

Let $P$ be a pair of pants decomposition of $N$. Let $[P]$ be the set of isotopy classes of elements of $P$. Note that $[P]$ is a maximal simplex of $\mathcal{C}(N)$. Every maximal simplex $\sigma$ of $\mathcal{C}(N)$ is equal to $[P]$ for some pair of pants decomposition $P$ of $N$.

On orientable surfaces all maximal simplices have the same dimension $3g + n -4$ where $g$ is the genus and $n$ is the number of boundary components of the orientable surface. On nonorientable surfaces this is not the case. There are different dimensional maximal simplices. In Figure \[Fig0\] we see some of them in $g=7$ case. In the figure we see cross sings. This means that the interiors of the disks with cross signs inside are removed and the antipodal points on the resulting boundary components are identified.

The following lemma is given in [@A] and [@AK]:

\[dim\] Let $N$ be a nonorientable surface of genus $g \geq 2$ with $n$ boundary components. Let $a_r= 3r+n-2$ and $b_r = 4r +n -2$ if $g = 2r+1$, and let $a_r= 3r+n-4$ and $b_r= 4r +n -4$ if $g =2r$. Then there is a maximal simplex of dimension $q$ in $\mathcal{C}(N)$ if and only if $a_r \leq q \leq b_r$.

Now we prove some properties of superinjective simplicial maps.

\[inj\] Suppose that $(g, n)= (3, 0)$ or $g + n \geq 4$. A superinjective simplicial map $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ is injective.

Let $[a]$ and $[b]$ be two distinct vertices in $\mathcal{C}(N)$. If $i([a], [b]) \neq 0$, then $i(\lambda([a]), \lambda([b])) \neq 0$, since $\lambda$ preserves nondisjointness. So, $\lambda([a]) \neq \lambda([b])$. If $i([a], [b]) = 0$, we choose a vertex $[c]$ of $\mathcal{C}(N)$ such that $i([a], [c]) = 0$, and $i([b], [c]) \neq 0$. Then $i(\lambda([a]), \lambda([b])) = 0$, $i(\lambda([b]), \lambda([c])) \neq 0$. So, $\lambda([a]) \neq \lambda([b])$. Hence, $\lambda$ is injective.

\
If a maximal simplex has dimension $b_r = 4r +n -2$ if $g = 2r+1$, or $b_r = 4r +n -4$ if $g=2r$, we will call it a [*top dimensional maximal simplex*]{}. Since a superinjective map $\lambda$ is injective, $\lambda$ sends top dimensional maximal simplices to top dimensional maximal simplices. In the following lemmas we will see that adjacency and nonadjacency are preserved w.r.t. top dimensional maximal simplices.

\[adjacent\] Suppose that $(g, n)= (3, 0)$ or $g + n \geq 4$. Let $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ be a superinjective simplicial map. Let $P$ be a pair of pants decomposition on $N$ which corresponds to a top dimensional maximal simplex in $\mathcal{C}(N)$. Let $a, b \in P$ such that $a$ is adjacent to $b$ w.r.t. $P$. There exists $a'  \in \lambda([a])$ and $b'  \in \lambda([b])$ such that $a'$ is adjacent to $b'$ w.r.t. $P'$ where $P'$ is a set of pairwise disjoint elements of $\lambda([P])$ containing $a', b'$.

Suppose that $(g, n)= (3, 0)$ or $g + n \geq 4$. The statement is easy to see in $(g, n)= (3, 0)$ and $(g, n) = (1,3)$ cases as there are only two curves in a pair of corresponding to a top dimensional maximal simplex. Assume that $(g, n) \neq (3, 0)$ and $(g, n) \neq (1,3)$.

Let $P$ be a pair of pants decomposition on $N$ which corresponds to a top dimensional maximal simplex in $\mathcal{C}(N)$. Let $a, b \in P$ such that $a$ is adjacent to $b$ w.r.t. $P$. We can find a simple closed curve $c$ on $N$ such that $c$ intersects only $a$ and $b$ nontrivially (with nonzero geometric intersection) and $c$ is disjoint from all the other curves in $P$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$. Since $\lambda$ is injective by Lemma \[inj\], $\lambda$ sends top dimensional maximal simplices to top dimensional maximal simplices. So, $P'$ corresponds to a top dimensional maximal simplex.

Assume that $\lambda([a])$ and $\lambda([b])$ do not have adjacent representatives w.r.t. $P'$. Since $i([c], [a]) \neq 0$ and $i([c], [b]) \neq 0$, we have $i(\lambda([c]), \lambda([a])) \neq 0$ and $i(\lambda([c]), \lambda([b])) \neq 0$ by superinjectivity. Since $i([c], [e]) = 0$ for all $e \in P \setminus \{a, b\}$, we have $i(\lambda([c]), \lambda([e])) = 0$ for all $e \in P \setminus \{a, b\}$. But this is not possible because $\lambda([c])$ has to intersect geometrically essentially with some isotopy class other than $\lambda([a])$ and $\lambda([b])$ in $\lambda([P])$ to be able to make essential intersections with $\lambda([a])$ and $\lambda([b])$ since $\lambda([P])$ is a top dimensional maximal simplex. This gives a contradiction to the assumption that $\lambda([a])$ and $\lambda([b])$ do not have adjacent representatives.

\[nonadjacent\] Suppose that $g + n \geq 4$. Let $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ be a superinjective simplicial map. Let $P$ be a pair of pants decomposition on $N$ which corresponds to a top dimensional maximal simplex in $\mathcal{C}(N)$. Let $a, b \in P$ such that $a$ is not adjacent to $b$ w.r.t. $P$. There exists $a'  \in \lambda([a])$ and $b'  \in \lambda([b])$ such that $a'$ is not adjacent to $b'$ w.r.t. $P'$ where $P'$ is a set of pairwise disjoint elements of $\lambda([P])$ containing $a', b'$.

Suppose that $g + n \geq 4$. Let $P$ be a pair of pants decomposition on $N$ which corresponds to a top dimensional maximal simplex in $\mathcal{C}(N)$. Let $a, b \in P$ such that $a$ is not adjacent to $b$ w.r.t. $P$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$. We can find simple closed curves $c$ and $d$ on $N$ such that $c$ intersects only $a$ nontrivially and is disjoint from all the other curves in $P$, $d$ intersects only $b$ nontrivially and is disjoint from all the other curves in $P$, and $c$ and $d$ are disjoint. Since $\lambda$ is injective by Lemma \[inj\], $\lambda$ sends top dimensional maximal simplices to top dimensional maximal simplices. We have $i(\lambda([a]), \lambda([c]) \neq 0$, $i(\lambda([a]), \lambda([x]) = 0$ for all $x \in P \setminus \{a\}$, $i(\lambda([b]), \lambda([d]) \neq 0$, $i(\lambda([d]), \lambda([x]) = 0$ for all $x \in P \setminus \{b\}$, and $i(\lambda([c]), \lambda([d]) = 0$ by superinjectivity. This is possible only when $\lambda([a])$ and $\lambda([b])$ have representatives which are not adjacent w.r.t. $P'$.

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A simple closed curve is called [*1-sided*]{} if its regular neighborhood is a Mobius band. It is called [*2-sided*]{} if its regular neighborhood is an annulus. There are 1-sided simple closed curves as well as 2-sided simple closed curves on nonorientable surfaces. Some 1-sided simple closed curves have orientable complements and some have nonorientable complements. We will first prove some properties of $\lambda$.

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\[1-sided-co\] Let $g=1$. Suppose that $n \geq 3$. Let $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ be a superinjective simplicial map. If $a$ is a 1-sided simple closed curve on $N$, then $\lambda(a)$ is the isotopy class of a 1-sided simple closed curve on $N$.

Let $a$ be a 1-sided simple closed curve on $N$. Since $g=1$, the complement of $a$ is orientable. Let $a' \in \lambda(a)$. Since $g=1$, $a'$ cannot be a 2-sided nonseparating simple closed curve on $N$.

Suppose $a'$ is a 2-sided separating simple closed curve on $N$.

Case (i): Suppose that $n=3$. We complete $a$ to a pair of pants decomposition $P=\{a, x\}$ as shown in Figure \[Fig1a\] (i). $P$ corresponds to a top dimensional maximal simplex. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $x' \in \lambda([x]) \cap P'$. Since $P'$ is a pants decomposition $N$, $g=1$ and $a'$ is separating, we see that $x'$ has to be a sided curve as shown in Figure \[Fig1a\] (ii). Consider the curves $z, t$ as shown in \[Fig1a\] (i). We have $[z] \neq [t]$, each of $z$ and $t$ intersects $x$ essentially, and each of them is disjoint from $a$. Let $z' \in \lambda([z]), t' \in \lambda([t])$ such that each of $z'$ and $t'$ intersects each of $a', x'$ minimally. Since $\lambda$ is injective by Lemma \[inj\], $[z'] \neq [t']$. Since $z', t'$ are not in $P'$ and they are disjoint from $a'$, they both have to lie in the subsurface, $M$, which is a projective plane with two boundary components containing $x'$, and having $a'$ as a boundary component as shown in Figure \[Fig1a\] (ii). This gives a contradiction as there are no two nontrivial simple closed curves with distinct isotopy classes such that each of them intersects $x'$ essentially in $M$. By Scharlemann’s result about complex of curves of projective plane with two boundary components in [@Sc], there are only two curves up to isotopy in $M$, (see Figure \[figure1\]). They are $x'$ and a 1-sided curve intersecting $x'$ only once.

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Case (ii): Suppose that $n=4$. We complete $a$ to a pair of pants decomposition $P=\{a, x, y\}$ as shown in Figure \[Fig1\] (i). $P$ corresponds to a top dimensional maximal simplex. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $x' \in \lambda([x]) \cap P'$. Since $a$ is adjacent to only $x$ w.r.t. $P$, by Lemma \[adjacent\] and Lemma \[nonadjacent\], $a'$ should be adjacent to only $x'$ w.r.t $P'$. Since $g=1$, $a'$ is separating and $a'$ is adjacent to only $x'$ w.r.t $P'$, there is a subsurface $T \subseteq N$ such that it is homeomorphic to sphere with four boundary components where three of the boundary components of $T$ are the boundary components of $N$, $x'$ is a boundary component of $T$, and $a'$ divides $T$ into two pair of pants as shown in Figure \[Fig1\] (ii). Let $y' \in \lambda([y]) \cap P'$. Since $x$ is adjacent to $y$, $x'$ is adjacent to $y'$. Since $n=4$ and $g=1$, we have $a', x', y', N$ as shown in Figure \[Fig1\] (ii). But this will give us a contradiction as follows: Consider the curves $z, t$ as shown in \[Fig1\] (i). We see that $[z] \neq [t]$, each of $z$ and $t$ intersects $y$ essentially, and each of them is disjoint from $a$ and $x$. Let $z' \in \lambda([z]), t' \in \lambda([t])$ such that each of $z'$ and $t'$ intersects each of $a', x', y'$ minimally. Since $\lambda$ is injective by Lemma \[inj\], $[z'] \neq [t']$. Since $z', t'$ are not in $P'$ and they are disjoint from $a'$ and $x'$, they both have to lie in the subsurface, $M$, which is a projective plane with two boundary components containing $y'$, and having $x'$ as a boundary component as shown in the figure. This gives a contradiction as there are no two nontrivial simple closed curves with distinct isotopy classes such that each of them intersects $y'$ essentially in $M$, (see [@Sc]).

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Case (iii): Suppose that $n > 4$. The argument is similar. We complete $a$ to a pair of pants decomposition $P= \{a, x_1, \cdots, x_{n-2}\}$ as shown in Figure \[Fig2\] (i). $P$ corresponds to a top dimensional maximal simplex. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $x_1' \in \lambda([x_1]) \cap P'$. Since $a$ is adjacent to only $x_1$ w.r.t. $P$, by Lemma \[adjacent\] and Lemma \[nonadjacent\], $a'$ should be adjacent to only $x_1'$ w.r.t $P'$. Since $g=1$, $n > 4$, $a'$ is separating and $a'$ is adjacent to only $x_1'$ w.r.t $P'$, we see that there is a subsurface $T \subseteq N$ such that it is homeomorphic to sphere with four boundary components where three of the boundary components of $T$ are the boundary components of $N$, $x_1'$ is a boundary component on $T$, and $a'$ divides $T$ into two pair of pants as shown in Figure \[Fig2\] (ii). Let $x_i' \in \lambda([x_i]) \cap P'$ for each $i= 2, \cdots, n-2$. Since $x_1$ is adjacent to only $x_2$, $x_1'$ is adjacent to only $x_2'$. Similar to the previous case using that adjacency and nonadjacency are preserved, $n > 4$ and $g=1$, we have elements of $P'$ and $N$ as shown in Figure \[Fig2\] (ii). But this will give us a contradiction as in the previous case (consider the curves $z, t$ as shown in Figure \[Fig2\] (i)).

We showed that $a'$ cannot be a 2-sided simple closed curve on $N$. Hence, $a'$ is a 1-sided simple closed curve on $N$.

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\[1-sided-cn\] Let $g \geq 2$. Suppose that $(g, n) = (3, 0)$ or $g+n \geq 4$. Let $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ be a superinjective simplicial map. If $a$ is a 1-sided simple closed curve on $N$ whose complement is nonorientable, then $\lambda(a)$ is the isotopy class of a 1-sided simple closed curve whose complement is nonorientable.

Let $a$ be a 1-sided simple closed curve on $N$ whose complement is nonorientable. Let $a' \in \lambda([a])$.

[**Case 1:**]{} Suppose $a'$ is a 1-sided simple closed curve whose complement is orientable. This case happens only if the genus of $N$ is odd. So, suppose $g =2r +1$, where $r \geq 1$. In this case $a$ can be put into a maximal simplex $\Delta$ of dimension $4r + n -2$. Since $\lambda$ is injective by Lemma \[inj\], $\lambda(\Delta)$ is a simplex of dimension $4r + n-2$. By using Euler characteristic arguments we see that the complement of $a'$ has genus $r$ and $n+1$ boundary components. So, in the complement of $a'$ there can be at most a $3r + n - 3$ dimensional simplex, hence there can be at most a $3r + n - 2$ dimensional simplex containing $a'$ on $N$. Since dim $\lambda(\Delta) = 4r + n - 2 > 3r + n - 2$ as $r \geq 1$, we get a contradiction.

[**Case 2:**]{} Suppose $a'$ is a 2-sided nonseparating simple closed curve on $N$.

[**(i)**]{} Suppose the genus of $N$ is even, $g=2r$ for some $r \geq 1$. We can put $a$ into a maximal simplex $\Delta$ of dim $4r+n-4$. Then $\lambda(\Delta)$ has dimension $4r+n-4$.

\(a) Suppose now that the complement of $a'$ is nonorientable. This case happens when $g \geq 4$. The complement of $a'$ has genus $2r-2$ and it has $n+2$ boundary components. By using Lemma \[dim\] we get the following: In the complement of $a'$ there can be at most a $4(r-1) + (n+ 2) - 4 = 4r + n -6$ dimensional simplex. Hence, there can be at most a $4r + n - 5$ dimensional simplex containing $a'$ on $N$. Since dim $ \lambda(\Delta) = 4r + n -4 > 4r + n - 5$, we get a contradiction.

\(b) Suppose that the complement of $a'$ is orientable. The genus of the complement is $r-1$ and the number of boundary component is $n + 2$. By using Lemma \[dim\] we get the following: In the complement of $a'$ there can be at most a $3(r-1) + (n+ 2) - 4 = 3r + n -5$ dimensional simplex. Hence, there can be at most a $3r + n - 4$ dimensional simplex containing $a'$ on $N$. Since dim $\lambda(\Delta) = 4r + n -4 > 3r + n - 4$, we get a contradiction (since $r \geq 1$).

[**(ii)**]{} Suppose the genus of $N$ is odd, $g =2r + 1 $ for some $r \in \mathbb{Z}$. In this case complement of $a'$ is nonorientable. It has genus $2r-1$, and it has $n+2$ boundary components. By using Lemma \[dim\] we get the following: We can put $a$ into a maximal simplex $\Delta$ of dim $4r+n-2$. Then $\lambda(\Delta)$ has dimension $4r+n-2$. In the complement of $a'$ there can be at most a $4(r-1) + (n+ 2) - 2 = 4r + n - 4$ dimensional simplex by Lemma \[dim\]. Hence, there can be at most a $4r + n - 3$ dimensional simplex containing $a'$ on $N$. Since dim $ \lambda(\Delta) = 4r + n - 2 > 4r + n - 3$, we get a contradiction.

=1.8in =1.55in

[**Case 3:**]{} Suppose $a'$ is a separating simple closed curve on $N$. This case does not occur when $(g, n)=(3, 0)$ as there are no such nontrivial simple closed curves. So, we assume that $g \geq 2$ and $g+n\geq 4$.

[**(i)**]{} Suppose $g \geq 4$. If $(g,n) = (4, 0)$, we complete $a$ to a pants decomposition $P$ as shown in Figure \[Fig4a\]. Let $P'$ be a set of pairwise disjoint elements in $\lambda([P])$ containing $a'$. Let $c' \in \lambda([c]) \cap P'$. If $a'$ is a separating simple closed curve on $N$, we get a contradiction by Lemma \[adjacent\], as $c'$ will be in one of the connected components in the complement of $a'$, and hence $c'$ cannot be adjacent to four curves w.r.t. $P'$ even though $c$ is adjacent to four curves w.r.t. $P$.

Assume $(g, n) \neq (4, 0)$. We can choose a 1-sided curve $b$ on $N$ intersecting $a$ exactly once such that a regular neighborhood $R$ of $a \cup b$ will be a projective plane with two boundary components $x, y$ such that $N \setminus R$ is connected and there is a curve $z$ as shown in Figure \[Fig3\] such that $a, x, y, z$ can be completed to a pair of pants decomposition $P$, which corresponds to a top dimensional maximal simplex in $\mathcal{C}(N)$, on $N$, and $a$ is adjacent to $x$ and $y$ and $x$ and $y$ are both adjacent to $z$ as shown in the figure. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Since $a$ is only adjacent to two curves $x, y$ w.r.t. $P$, there exist $x' \in \lambda([x]) \cap P'$ and $y' \in \lambda([y]) \cap P'$ such that $a'$ is only adjacent to $x'$ and $y'$ w.r.t. $P'$ by Lemma \[adjacent\] and Lemma \[nonadjacent\]. Since $x$ and $y$ are also adjacent to $z$, there exists $z' \in \lambda([z]) \cap P'$ such that $x'$ and $y'$ are adjacent to $z'$. These imply that the only case that $a'$ could be a 2-sided separating curve is that $a'$ and two boundary components (if exists) would form a pair of pants on $N$ and $a', x', y'$ would be as shown in Figure \[Fig3\].

\(a) Suppose $g = 2r$ for some $r \in \mathbb{Z}$. Then the complement of $R$ (the projective plane with two boundary components $x, y$) in $N$ (see Figure \[Fig3\]) is a nonorientable surface with genus $2r -1$ and $n+2$ boundary components. By using Lemma \[dim\] we get the following: In this complement there is a $4(r-1) + (n+ 2) - 4 = 4r + n -6$ dimensional simplex. Hence, there is a $4r + n - 3$ dimensional simplex $\Delta$ containing $a, x, y$ on $N$. The complement of the four holed sphere shown in Figure \[Fig3\] is a connected nonorientable surface of genus $2r - 2$, and it has $n$ boundary components. So, in this complement there can be at most a $4(r-1) + n - 4 = 4r + n -8$ dimensional simplex. Hence, there can be at most a $4r + n - 5$ dimensional simplex containing $a', x', y'$ on $N$. Since dim $ \lambda(\Delta) = 4r + n - 3 > 4r + n -8$ we get a contradiction.

\(b) Suppose $g = 2r + 1$ for some $r \in \mathbb{Z}$. Then the complement of $R$ (the projective plane with two boundary components $x, y$) in $N$ (see Figure \[Fig3\]) is a nonorientable surface with genus $2r $ and $n+2$ boundary components. By using Lemma \[dim\] we get the following: In this complement there is a $4r + n+ 2 - 4 = 4r + n -2$ dimensional simplex. Hence, there is a $4r + n +1$ dimensional simplex $\Delta$ containing $a, x, y$ on $N$. In this case the complement of the four holed sphere shown in Figure \[Fig3\] is a connected nonorientable surface of genus $2r - 1$, and it has $n$ boundary components. So, in this complement there can be at most a $4(r-1) + n - 2 = 4r + n -6$ dimensional simplex. Hence, there can be at most a $4r + n - 3$ dimensional simplex containing $a', x', y'$ on $N$. Since dim $\lambda(\Delta) = 4r + n + 1 > 4r + n - 3$, we get a contradiction.

=1.6in =1.7in

[**(ii)**]{} Now suppose $g < 4$. Since $g + n \geq 4$, we have $n \geq 1$. In this case, we complete $a$ to a curve configuration $a, x$ as shown in Figure \[Fig4\], using a boundary component $\partial_1$ of $N$. Let $T$ be the subsurface of $N$ having $x$ and $\partial_1$ as its boundary as shown in Figure \[Fig4\]. $T$ is a projective plane with two boundary components. The curves $a$ and $x$ can be completed to a pair of pants decomposition $P$ on $N$ which corresponds to a top dimensional maximal simplex in $\mathcal{C}(N)$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $x' \in \lambda([x]) \cap P'$. Since $a$ is adjacent to only $x$ w.r.t. $P$, we must have that $a'$ is adjacent to only $x'$ w.r.t. $P'$. There is a curve $z$ in $P$ different from $a$ and $x$ such that $x$ is adjacent to $z$ w.r.t. $P$. The only case that $a'$ is a separating curve where $a'$ is only adjacent to $x'$, and $x'$ is adjacent to at least two curves is that $a'$ is as shown in Figure \[Fig4\], i.e. there must be a four holed sphere, $R$, having $x'$ and three boundary components of $N$ (if exists) as its boundary components such that $a'$ divides $R$ into two pair of pants as shown in the figure. Hence, $a'$ cannot be a separating curve if $(g, n) = (2, 2)$.

=2.4in =2.2in

\(a) Suppose $g = 2, n = 3$. We complete $a$ to a pair of pants decomposition $P=\{a, b, c, d\}$ which corresponds to a top dimensional maximal simplex as shown in Figure \[Fig5\] (i). We see that $a$ is adjacent to only $b$ w.r.t. $P$. Let $a' \in \lambda([a])$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $b' \in \lambda([b]) \cap P'$, $c' \in \lambda([c]) \cap P'$, $d' \in \lambda([d]) \cap P'$. Since $a'$ is separating and $a'$ should be adjacent to only one curve w.r.t $P'$, there is a subsurface $T \subseteq N$ such that $T$ is a sphere with four boundary components where three of the boundary components of $T$ are the boundary components of $N$, $b'$ is a boundary component of $T$, and $a'$ divides $T$ into two pair of pants as shown in Figure \[Fig5\] (ii). Since $b$ is adjacent to $c$ and $d$, $b'$ is adjacent to $c'$ and $d'$. Since $c$ is adjacent to $d$, $c'$ is adjacent to $d'$.

Let $z, t$ be as shown in Figure \[Fig5\] (i). We see that $[z] \neq [t]$, each of $z$ and $t$ intersects $d$ essentially and each of them is disjoint from $b$ and $c$. Let $z' \in \lambda([z]), t' \in \lambda([t])$ such that each of $z'$ and $t'$ intersects each of $a', c', d'$ minimally. Since $\lambda$ is injective by Lemma \[inj\], $[z'] \neq [t']$. Since $z'$ is not in $P'$ and $z'$ is disjoint from $a', b'$ and $c'$, it has to intersect $d'$. Similarly $t'$ is not in $P'$ and $t'$ is disjoint from $a', b'$ and $c'$, and it has to intersect $d'$. So, $z'$ and $t'$ are curves with distinct isotopy classes which are both disjoint from $b', c'$ and intersect $d'$ nontrivially. By cutting $N$ along $b'$ and $c'$ we get a projective plane with two boundary components, $R$, containing $d'$. The curves $z'$ and $t'$ have to lie $R$. We get a contradiction as there are only two elements in the complex of curves in a projective plane with two boundary components (by Scharlemann’s result in [@Sc]), and $d'$ is one of them.

=3.65in =3.65in

=3.65in =3.65in

\(b) Suppose $g = 2, n > 3$. We complete $a$ to a pair of pants decomposition $P$, which corresponds to a top dimensional maximal simplex, as shown in Figure \[Fig7\] (i). We see that $a$ is adjacent to only $b$ w.r.t. $P$. Let $a' \in \lambda([a])$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $b' \in \lambda([b]) \cap P'$, $c' \in \lambda([c]) \cap P'$, $x_i' \in \lambda([x_i]) \cap P'$ for $i = 1, 2, \cdots, n-2$. Since $a'$ is separating and $a'$ should be adjacent to only one curve w.r.t $P'$, we see that there is a subsurface $T \subseteq N$ such that it is homeomorphic to sphere with four boundary components where three of the boundary components of $T$ are the boundary components of $N$, $b'$ is a boundary component of $T$, and $a'$ divides $T$ into two pair of pants as shown in Figure \[Fig7\] (ii). Since $b$ is adjacent to $c$ and $x_1$, $b'$ is adjacent to $c'$ and $x_1'$. Since $c$ is not adjacent to any other curve, $c'$ will not be adjacent to any other curve. This implies that $c'$ is either a 1-sided curve or a separating curve as shown in Figures \[Fig7\] (ii), \[Fig8\] (ii).

Suppose $c'$ is as shown in Figures \[Fig7\] (ii). By using the adjacency relations between $x_i$’s we get $P'$ as shown in Figure \[Fig7\] (ii).

Let $z, t$ be as shown in Figure \[Fig7\] (i). We see that $[z] \neq [t]$, each of $z$ and $t$ intersects $x_{n-2}$ essentially and each of them is disjoint from all the other curves in $P$. Let $z' \in \lambda([z]), t' \in \lambda([t])$ such that each of $z'$ and $t'$ intersects each curve in $P'$ minimally. Since $\lambda$ is injective by Lemma \[inj\], $[z'] \neq [t']$. We see that $z', t'$ are not in $P'$, and they both are disjoint from all the curves in $P'$ except for $x_{n-2}'$. We get a contradiction again similar to case (a), by using Scharlemann’s result about complex of curves of projective plane with two boundary components.

Similarly, if $c'$ is as shown in Figures \[Fig8\] (ii), by using the adjacency relations between $x_i$’s we get $P'$ as shown in \[Fig8\] (ii). By choosing curves $w, y$ as shown in Figures \[Fig8\] (i), we get a contradiction as in (a) and (b).

=2.5in

=2.3in =2.3in =2.3in

\(c) Now suppose $(g,n) = (3, 1)$. We complete $a$ to a pants decomposition $P$ as shown in Figure \[Fig4b\]. Let $P'$ be a set of pairwise disjoint elements in $\lambda([P]$ containing $a'$. Let $c' \in \lambda([c]) \cap P'$. If $a'$ is a separating simple closed curve on $N$, we get a contradiction by Lemma \[adjacent\], as $c'$ will be in one of the connected components in the complement of $a'$, and hence $c'$ cannot be adjacent to three curves w.r.t. $P'$ even though $c$ is adjacent to three curves w.r.t. $P$.

\(d) Suppose $g = 3, n = 2$. We complete $a$ to a pair of pants decomposition $P$, which corresponds to a top dimensional maximal simplex, as in Figure \[Fig9\] (i). Let $a, b, c, d, e$ be as shown in the figure. We see that $a$ is adjacent to $b$ and $e$ w.r.t. $P$. Let $a' \in \lambda([a])$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $b' \in \lambda([b]) \cap P'$, $c' \in \lambda([c]) \cap P'$, $d' \in \lambda([d]) \cap P'$, $e' \in \lambda([e]) \cap P'$. Since $a'$ is separating and $a'$ should be adjacent to $b'$ and $e'$ w.r.t $P'$, and $b'$ is adjacent to four curves $a', c', d', e'$, there is a subsurface $T \subseteq N$ such that $T$ is a sphere with four boundary components where two of the boundary components of $T$ are the boundary components of $N$, $b'$ and $c'$ are the other two boundary components of $T$, and $a'$ divides $T$ into two pair of pants as shown in Figure \[Fig9\] (iii). Since $e'$ is not adjacent to any other curve, $e'$ is a 1-sided curve as shown in the figure. Since $b$ is adjacent to $c$ and $d$, $b'$ is adjacent to $c'$ and $d'$, and we see that $c'$ and $d'$ are both 1-sided curves.

Let $z, t$ be as shown in Figure \[Fig9\] (ii). We have $[z] \neq [t]$, each of $z$ and $t$ intersects $e$ essentially and each of them is disjoint from $a, b, c$ and $d$. Let $z' \in \lambda([z]), t' \in \lambda([t])$ such that each of $z'$ and $t'$ intersects each of $a',b', c', d', e'$ minimally. Since $\lambda$ is injective by Lemma \[inj\], $[z'] \neq [t']$. Since $z'$ is not in $P'$ and $z'$ is disjoint from $a', b', c'$ and $d'$, it has to intersect $e'$. Similarly $t'$ is not in $P'$ and $t'$ is disjoint from $a', b', c'$ and $d'$, and it has to intersect $e'$. So, $z', t'$ are two curves with distinct isotopy classes which are both disjoint from $a'$ and $b'$, and intersect $e'$ nontrivially. This gives a contradiction as in (a), (b).

=3.5in =3.5in

\(e) Suppose $g = 3, n > 2$. We complete $a$ to a pair of pants decomposition $P= \{a, b, c, d, x_1, \cdots, x_{n-1}\}$ which corresponds to a top dimensional maximal simplex as shown in Figure \[Fig10\] (i). We see that $a$ is adjacent to $b$ and $x_1$ w.r.t. $P$. Let $a' \in \lambda([a])$. Let $P'$ be a set of pairwise disjoint elements of $\lambda([P])$ containing $a'$. Let $b' \in \lambda([b]) \cap P'$, $c' \in \lambda([c]) \cap P'$, $d' \in \lambda([d]) \cap P'$, $x_i' \in \lambda([x_i]) \cap P'$, for $i= 1, \cdots, n-1$. Since $a'$ is separating and $a'$ should be adjacent to $b'$ and $x_1'$ w.r.t. $P'$, there is a subsurface $T \subseteq N$ such that it is a sphere with four boundary components where two of the boundary components of $T$ are the boundary components of $N$, and $b'$ and $x_1'$ are the other two boundary components of $T$, and $a'$ divides $T$ into two pair of pants as shown in Figure \[Fig10\] (ii). Since $b$ is adjacent to $a, c, d$ and $x_1$, $b'$ is adjacent to $a', c', d'$ and $x_1'$. So, there is a pair of pants in $P'$ having $b', c', d'$ on its boundary. Since $x_1$ is adjacent to $x_2$ w.r.t. $P$, there is a pair of pants in $P'$ having $x_1'$ and $x_2'$ on its boundary. Since $x_1'$ is not adjacent to any other curve the third boundary component of this pair of pants must be a boundary component of $N$ as shown in the figure. Similarly since $x_2$ is adjacent to $x_3$ w.r.t. $P$, there is a pair of pants in $P'$ having $x_2'$ and $x_3'$ on its boundary. Since $x_2'$ is not adjacent to any other curve the third boundary component of this pair of pants must be a boundary component of $N$ as shown in the figure. Continuing like this we get a pair of pants in $P'$ such that it has $x_{n-2}'$ and $x_{n-1}'$ on its boundary. Since $g=3$, $x_{n-1}'$ is only adjacent to $x_{n-2}'$, the curves $c', d'$ are only adjacent to $b'$, and we already have $n-1$ boundary components, we see that $b', c', x_{n-1}'$ should be all be 1 sided curves as shown in the figure.

Let $z, t$ be as shown in Figure \[Fig10\] (i). We have $[z] \neq [t]$, each of $z$ and $t$ intersects $x_{n-1}$ essentially, and each of them is disjoint from all the other curves in $P$. Let $z' \in \lambda([z]), t' \in \lambda([t])$ such that each of $z'$ and $t'$ intersects each curve in $P'$ minimally. Since $\lambda$ is injective by Lemma \[inj\], $[z'] \neq [t']$. We see that $z', t'$ are not in $P'$, and they both are disjoint from all the curves in $P'$ except for $x_{n-1}'$. This means that $z', t'$ both are in the projective plane with two boundary components, where one of the boundary components is $x_{n-1'}$ and the other is a boundary component of $N$ as shown in the figure. The curve $x_{n-2}'$ is a one sided curve in this subsurface. The existence of such $z', t'$ gives a contradiction as in cases (a), (b), (c).

We have seen that $a'$ cannot be a 1-sided simple closed curve with orientable complement, it cannot be a 2-sided nonseparating simple closed curve and it cannot be a separating simple closed curve. Hence, $a'$ is a 1-sided simple closed curve whose complement is nonorientable.

Proof of the Main Result
========================

In this section we will prove our main result when $(g, n) = (3, 0)$ or $g + n \geq 5$. Together with Theorem \[A\] this will complete our proof of the main theorem for $g + n \leq 3$ or $g + n \geq 5$.\
First we consider two graphs on $N$ as given in [@AK]: If $g=1$, let $\mathcal{A}$ be the set of isotopy classes of all 1-sided simple closed curves on $N$. If $g \geq 2$ let $\mathcal{A}$ be the set of isotopy classes of all 1-sided simple closed curves which have nonorientable complements on $N$. Let $X(N)$ be the graph with vertex set $\mathcal{A}$ such that two distinct vertices in $X(N)$ are connected by an edge if and only if they have representatives intersecting transversely at one point.

Let $\widetilde{X}(N)$ be a subgraph of $X(N)$ with the vertex set $\mathcal{A}$. Two distinct vertices $\alpha$ and $\beta$ are connected by an edge in $\widetilde{X}(N)$ if $\alpha$ and $\beta$ have representatives a and b intersecting transversely at one point such that

\(i) either $g \geq 4$ and the surface $N_{a \cup b}$ obtained by cutting $N$ along $a$ and $b$ is connected (Since $N_a$ and $N_b$ are nonorientable, it is easy to see that $N_{a \cup b}$ is also nonorientable in this case.),

\(ii) or $1 \leq g \leq 3$ and the Euler characteristic of one of the connected components of $N_{a \cup b}$ is at most $-2$.

We will use some connectivity results about these graphs given by Atalan-Korkmaz in [@AK].\
If $a$ is a nontrivial simple closed curve on $N$, [*the link*]{}, $L_a$ of $a$, is defined as the full subcomplex spanned by all the vertices of $\mathcal{C}(N)$ which have representatives disjoint from $a$. [*The star*]{} $St_a$ of $a$ is defined as the subcomplex of $\mathcal{C}(N)$ consisting of all simplices in $\mathcal{C}(N)$ containing $[a]$ and the faces of all such simplices. So, the link $L_a$ of $a$ is the subcomplex of $\mathcal{C}(N)$ whose simplices are those simplices of $St_a$ which do not contain $[a]$. A nontrivial simple closed curve $b$ is called [*dual*]{} to $a$ if they intersect transversely only once. Let $D_a$ be the set of isotopy classes of nontrivial 1-sided simple closed curves that are dual to $a$ on $N$.

The following theorem is given by Atalan-Korkmaz in [@AK].

\[id\] Let $N$ be a connected, nonorientable surface of genus $g \geq 1$ with $n$ holes, and $g + n \geq 5$. Let $a$ and $b$ be two 1-sided simple closed curves that are dual to each other such that $[a]$ and $[b]$ are two vertices in $\widetilde{X}(N)$ which are connected by an edge in $\widetilde{X}(N)$. Let $h$ be a mapping class. If $h(\gamma) = \gamma$ for every vertex $\gamma$ in the set $(St_a \cup D_a) \cap (St_b \cup D_b)$, then $h$ is the identity.

=2.5in =2.5in

We prove a similar result below. Suppose $(g, n) = (3, 0)$. When we remove the identifications on $N$ that are shown in the first part of Figure \[Fig(3,0)\], we get a sphere with three boundary components, $z, t, w$, as shown in the second part. The reflection of this sphere about the $xy$-plane which changes the orientation on each curve in $\{z, t, w, x, y\}$ shown in the figure induces a homeomorphism $R$ on $N$. We call $R$ a reflection homeomorphism on $N$. Let $r = [R]$.

\[id2\] Suppose $(g, n)= (3, 0)$. Let $a$, $b$ be two 1-sided simple closed curves with nonorientable complements such that they are dual to each other. Let $h$ be a mapping class. If $h(\gamma) = \gamma$ for every vertex $\gamma$ in the set $(St_a \cup D_a) \cap (St_b \cup D_b)$, then $h$ is the identity or $r$.

Let $a$ and $b$ be two 1-sided simple closed curves with nonorientable complements such that they are dual to each other. Let $T$ be a regular neighborhood of $a \cup b$. $T$ is a genus one surface with two boundary components, say $x$ and $y$. Since $(g, n) = (3, 0)$, $a$ and $b$ are dual to each other, and $N_a$ and $N_b$ are nonorientable, we can see that each of $x$ and $y$ bounds a genus one surface with one boundary component, mobius band, as shown in the Figure \[Fig(3,0)\]. Let $h$ be a mapping class such that $h(\gamma) = \gamma$ for every vertex $\gamma$ in the set $(St_a \cup D_a) \cap (St_b \cup D_b)$. Let $H$ be a representative of $h$ such that $H$ fixes the curves $x, y, a$ and $b$. If $H$ fixes the orientation of $x$ then it also fixes the orientation of $y$ by Corollary 4.6 in [@K2] as it also fixes $a$ and $b$ up to isotopy. Then $H$ is isotopic to identity on $T$, and also $H$ is isotopic to identity on the two mobius bands on each side of $T$. Since the Dehn twists about $x$ and $y$ are trivial, we see that $H$ is isotopic to identity on $N$. If $H$ changes the orientation of $x$ then it also changes the orientation of $y$ by Corollary 4.6 in [@K2] as it also fixes $a$ and $b$ up to isotopy. In this case the composition $H \circ R$ fixes the orientation of both of $x$ and $y$ where $R$ is the reflection homeomorphism defined above. By the above argument $H \circ R$ is isotopic to identity. Hence, $H$ is isotopic to $R$.

\
We prove our main result with the following theorem:

Let $N$ be a compact, connected, nonorientable surface of genus $g$ with $n$ boundary components. Suppose that $(g, n) =(3, 0)$ or $g + n \geq 5$. If $\lambda : \mathcal{C}(N) \rightarrow \mathcal{C}(N)$ is a superinjective simplicial map, then $\lambda$ is induced by a homeomorphism $h : N \rightarrow N$ (i.e $\lambda([a]) = [h(a)]$ for every vertex $[a]$ in $\mathcal{C}(N)$).

Assume that $g + n \geq 5$. We will first prove the result using induction on $g$.

[**Case i:**]{} Suppose that $g=1$. Let $a$ be a 1-sided simple closed curve. Let $a' \in \lambda([a])$. By Lemma \[1-sided-co\] we know that $a'$ is a 1-sided simple closed curve. Since $g=1$, both of $N_a$ and $N_{a'}$ are orientable. So, there is a homeomorphism $f: N \rightarrow N$ such that $f(a) = a'$. Let $f_{\#}$ be the simplicial automorphism induced by $f$ on $\mathcal{C}(N)$. Then $f_{\#}^{-1} \circ \lambda $ fixes $[a]$. By replacing $f_{\#}^{-1} \circ \lambda $ by $\lambda$ we can assume that $\lambda([a]) = [a]$. The simplicial map $\lambda$ restricts to $\lambda_a : L_a \rightarrow L_a$, where $L_a$ is the link of $[a]$. It is easy to see that $\lambda_a$ is a superinjective simplicial map. Since $L_a \cong \mathcal{C}(N_a)$, we get a superinjective simplicial map $\lambda_a :  \mathcal{C}(N_a)\rightarrow \mathcal{C}(N_a)$. By following the proof of Lemma \[inj\], we see that $\lambda_a$ is injective. Since $N_a$ is a sphere with at least five boundary components, by Shackleton’s result in [@Sh], there is a homeomorphism $G_a: N_a \rightarrow N_a$ such that $\lambda_a$ is induced by $(G_a)_{\#}$.

Let $\partial_a$ be the boundary component of $N_a$ which came by cutting $N$ along $a$. We can see that $G_a (\partial_a) = \partial_a$ as follows: Since $g=1$ there are at least four boundary components of $N$. Let $b, c$ be the curves as shown in the Figure \[bdcor\] (i) ($b$ separates two of the boundary components of $N$). Complete $\{a, b, c\}$ to a top dimensional pair of pants decomposition $P$ on $N$. We assumed that $\lambda([a])=[a]$. Since $a, b, c$ are pairwise disjoint, there exist $b', c'$ some representatives of $\lambda(b)$ and $\lambda(c)$ respectively, such that $a, b', c'$ are pairwise disjoint. Let $P'$ be the set of pairwise disjoint representatives of $\lambda([P])$ containing $a, b', c'$. Since $a$ is adjacent to $b$ and $c$ w.r.t. $P$, $a$ is adjacent to $b'$ and $c'$ w.r.t. $P'$ by Lemma \[adjacent\]. Since $a$ is 1-sided there is a genus one subsurface with two boundary components, say $T$, such that $a$ is in $T$ and the boundary components of $T$ are $b', c'$. Since $b$ is adjacent to only $a$ and $c$ w.r.t. $P$, and nonadjacency is preserved by $\lambda$ by Lemma \[nonadjacent\], we see that $b'$ must be adjacent to only $a$ and $c'$. We know that $b'$ cannot be a 1-sided curve as $g=1$. This shows that $a, b', c'$ are as shown in the Figure \[bdcor\] (ii) (i.e. $b'$ separates two of the boundary components of $N$). Since $\lambda_a$ is induced by $(G_a)_{\#}$, they agree on $[b]$ and $[c]$. This shows that $G_a (\partial_a) = \partial_a$.

By composing $G_a$ with a homeomorphism isotopic to identity, we can assume that $G_a$ maps antipodal points on the boundary $\partial_a$ to antipodal points. So, $G_a$ induces a homeomorphism $g_a: N \rightarrow N$ such that $g_a (a)= a$ and $(g_a)_{\#}$ agrees with $\lambda$ on every vertex of $[a] \cup L_a$. So, $(g_a)_{\#}^{-1} \circ \lambda$ fixes every vertex in $[a] \cup L_a$. Let $D_a$ be the set of isotopy classes of 1-sided simple closed curves that are dual to $a$.

Claim 1: $(g_a)_{\#}$ agrees with $\lambda$ on every vertex of $\{[a]\} \cup L_a \cup D_a$.

Proof of Claim 1: We already know that $(g_a)_{\#}$ agrees with $\lambda$ on every vertex of $\{[a]\} \cup L_a$. Let $d$ be a 1-sided simple closed curve that is dual to $a$. Let $T$ be a regular neighborhood of $a \cup d$. We see that $T$ is a real projective plane with two boundary components, say $x, y$. Since $(g_a)_{\#}^{-1} \circ \lambda$ is identity on ${[a]} \cup L_a$, $(g_a)_{\#}^{-1} \circ \lambda$ is identity on the complement of $T$. Since $(g_a)_{\#}^{-1} \circ \lambda$ is a superinjective simplicial map, there exists $d' \in \lambda([d])$ such that $d'$ is disjoint from the complement of $T$, so $d'$ is in $T$. Since $d$ is the only nontrivial simple closed curve in $T$ which is not isotopic to $a$ by Scharlemann’s Theorem in [@Sc], we see that $d'$ is isotopic to $d$. So, $((g_a)_{\#}^{-1} \circ \lambda) ([d]) = [d]$. Hence, $(g_a)_{\#}^{-1} \circ \lambda$ is identity on ${[a]} \cup L_a \cup D_a$.

=2.2in =2.2in

=2.2in =2.2in

=2.2in

Claim 2: Let $v$ be any 1-sided simple closed curve on $N$. Then, $(g_v)_{\#} = (g_a)_{\#}$ on $\mathcal{C}(N)$.

Proof of Claim 2: Since $g=1$, $\widetilde{X}(N)$ is connected by Theorem 3.10 in [@AK]. So, we can find a sequence $a \rightarrow a_1 \rightarrow a_2 \rightarrow  \cdots \rightarrow  a_n=v$ of 1-sided simple closed curves connecting $a$ to $v$ such that each consecutive pair is connected by an edge in $\widetilde{X}(N)$. By Claim 1, $(g_a)_{\#}$ agrees with $\lambda$ on $\{[a]\} \cup L_a \cup D_a$, and $(g_{a_1})_{\#}$ agrees with $\lambda$ on $\{[a_1]\} \cup L_{a_1} \cup D_{a_1}$. So, we see that $(g_a)_{\#} ^{-1} (g_{a_1})_{\#}$ fixes every vertex in $([a] \cup L_a \cup D_a) \cap ([a_1] \cup L_{a_1} \cup D_{a_1})$. We note that this is the same thing as fixing every vertex in $(St_a \cup D_a) \cap (St_{a_1} \cup D_{a_1})$ since the vertex sets of them are equal. By Theorem \[id\], we get $(g_a)_{\#} = (g_{a_1})_{\#}$ on $\mathcal{C}(N)$. By using the sequence, we get $(g_a)_{\#} = (g_{v})_{\#}$ on $\mathcal{C}(N)$.

Since genus is 1 there is no nonseparating 2-sided curve on $N$. Since every separating curve is in the link, $L_r$, of some 1-sided curve $r$, we see that $(g_a)_{\#}$ agrees with $\lambda$ on $\mathcal{C}(N)$. This proves the theorem for $g=1$.

[**Case ii:**]{} Suppose that $g=2$. Since $g +n \geq 5$, $N$ has at least three boundary components. Let $a, b$ be as in Figure \[Fig11\]. Let $P(a)$, $P(b)$ be the connected components of $\widetilde{X}(N)$ that contains $a$, $b$ respectively. By Theorem 3.10 in [@AK], $\widetilde{X}(N)$ has two connected components $P(a)$ and $P(b)$. Since $\lambda$ is superinjective and $a, b$ are pairwise disjoint, there exist $a' \in \lambda([a])$, $b' \in \lambda([b])$ such that $a', b'$ are pairwise disjoint. By Lemma \[1-sided-cn\] we know that $a'$ and $b'$ are 1-sided simple closed curves with nonorientable complements since $a$ and $b$ are. So, there is a homeomorphism $f: N \rightarrow N$ such that $f(a) = a'$. Let $f_{\#}$ be the simplicial automorphism induced by $f$ on $\mathcal{C}(N)$. We see that $f_{\#}^{-1} \circ \lambda $ fixes $[a]$ as in case (i). By replacing $f_{\#}^{-1} \circ \lambda $ by $\lambda$ we can assume that $\lambda([a]) = [a]$. The simplicial map $\lambda$ restricts to a superinjective map $\lambda_{a} : L_{a} \rightarrow L_{a}$. Since $L_{a} \cong \mathcal{C}(N_{a})$, we get a superinjective simplicial map $\lambda_{a} :  \mathcal{C}(N_{a})\rightarrow \mathcal{C}(N_{a})$. Since $N_a$ is a genus one nonorientable surface with at least four boundary components, by case (i) there is a homeomorphism $G_{a}: N_{a} \rightarrow N_{a}$ such that $\lambda_{a}$ is induced by $(G_{a})_{\#}$.

Let $\partial_a$ be the boundary component of $N_a$ which came by cutting $N$ along $a$. We can see that $G_a (\partial_a) = \partial_a$ as follows: Since $g=2$ there at least three boundary components of $N$. Let $b, c$ be the curves as shown in the Figure \[bdcor\] (iii). We see that $b$ is a 1-sided curve and $c$ separates a genus two subsurface containing $a, b$. Complete $\{a, b, c\}$ to a top dimensional pair of pants decomposition $P$ on $N$. We assumed that $\lambda([a])=[a]$. Since $a, b, c$ are pairwise disjoint, there exist $b', c'$ some representatives of $\lambda(b)$ and $(\lambda(c)$ respectively, such that $a, b', c'$ are pairwise disjoint. Let $P'$ be the set of pairwise disjoint representatives of $\lambda([P])$ containing $a, b', c'$. Since $a$ is adjacent to $b$ and $c$ w.r.t. $P$, $a$ is adjacent to $b'$ and $c'$ w.r.t. $P'$ by Lemmma \[adjacent\]. Since $a$ is 1-sided there is a genus one subsurface with two boundary components, say $T$, such that $a$ is in $T$ and the boundary components of $T$ are $b', c'$. Since $b$ is adjacent to only $a, c$ w.r.t. $P$, and nonadjacency is preserved by $\lambda$ by Lemma \[nonadjacent\], we see that $b'$ must be only adjacent to and $a, c'$. We also know that $b'$ must be a 1-sided curve whose complement is nonorientable, as $b$ is such a curve (see Lemma \[1-sided-cn\]). This shows that $a, b', c'$ are as shown in the Figure \[bdcor\] (iv). Since $\lambda_a$ is induced by $(G_a)_{\#}$, they agree on $[b]$ and $[c]$. This shows that $G_a (\partial_a) = \partial_a$. We note that when $g \geq 3$, we can use the curves shown in Figure \[bdcor\] (iv) to get similar results.

Now we continue as follows: As in case (i), by composing $G_a$ with a homeomorphism isotopic to identity, we can assume that $G_a$ maps antipodal points on the boundary $\partial_a$ to antipodal points. So, $G_a$ induces a homeomorphism $g_a: N \rightarrow N$ such that $g_a (a)= a$ and $(g_a)_{\#}$ agrees with $\lambda$ on every vertex of $[a] \cup L_a$. So, $(g_a)_{\#}^{-1} \circ \lambda$ fixes every vertex in $[a] \cup L_a$.

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As in the proof of case i, we get $(g_{a})_{\#}$ agrees with $\lambda$ on $\{[a]\} \cup L_{a} \cup D_{a}$. Also as in that proof, we can see that if $v$ is any 1-sided simple closed curve $v$ such that $[v] \in P(a)$ (where $P(a)$ is the connected component of $\widetilde{X}(N)$ that contains $a$), then $(g_a)_{\#} = (g_{v})_{\#}$ on $\mathcal{C}(N)$. Hence, we can see that there exists a homeomorphism $h_1$ such that our original superinjective map $\lambda$ agrees with $(h_1)_{\#}$, on every vertex of $\{[a]\} \cup L_{a} \cup D_{a} \cup \{[v]\} \cup L_{v} \cup D_{v}$ for any 1-sided simple closed curve $v$ such that $[v] \in P(a)$. Similarly, there exists a homeomorphism, $h_2$, such that $\lambda$ agrees with $(h_2)_{\#}$ on every vertex of $\{[b]\} \cup L_{b} \cup D_{b} \cup \{[w]\} \cup L_{w} \cup D_{w}$ for any 1-sided simple closed curve $w$ such that $[w] \in P(b)$. So, $(h_1)_{\#}^{-1} (h_2)_{\#}$ fixes everything in the intersection of these two sets, for every such $v, w$.

Claim: $(h_1)_{\#} = (h_2)_{\#}$.

Proof of Claim: Let $c, d, e$ be as in Figure \[Fig11\]. Let $T$ be the genus 2 subsurface containing $a$ and $b$ and bounded by $c$ as shown in the figure. Since $(h_1)_{\#}^{-1} (h_2)_{\#}$ fixes $[c]$, (by composing with a map isotopic to identity if necessary) we may assume that ${h_1}^{-1} h_2$ fixes $c$. By the above argument we know that ${h_1}^{-1} h_2$ fixes every nontrivial simple closed curve up to isotopy in the complement of $T$. Since the complement of $T$ together with $c$ is a sphere with at least four boundary components, ${h_1}^{-1} h_2$ fixes $c$ and every nontrivial simple closed curve up to isotopy in the complement of $T$, ${h_1}^{-1} h_2$ is isotopic to identity in the complement of $T$ by Lemma 7.1 in [@AK]. So, ${h_1}^{-1} h_2$ is isotopic to identity on $c$. Since ${h_1}^{-1} h_2$ fixes $a$, $b$ up to isotopy, it fixes the curve $e$ shown in the figure up to isotopy, see the proof of Theorem \[A\] and Figure \[figure2\], and recall that the vertex set of the curve complex in $T$ is $\{[e], [a], t_e^m ([a]) : m \in \mathbb{Z} \}$. By looking at the complex we see that the map ${h_1}^{-1} h_2$ actually fixes every nontrivial simple closed curve on $T$ up to isotopy. Since it is also isotopic to identity on $c$, ${h_1}^{-1} h_2$ is isotopic to identity on $T$. This implies that $(h_1)_{\#}^{-1} (h_2)_{\#} = t_c^m$ for some $m \in \mathbb{Z}$. Since $(h_1)_{\#}^{-1} (h_2)_{\#}$ also fixes $[d]$, we have $m=0$, so $(h_1)_{\#}^{-1} (h_2)_{\#} = [id]$.

Let $h=h_1$. If $x$ is a 1-sided simple closed curve with nonorientable complement (i.e. $[x]$ is a vertex in $\widetilde{X}(N)$), then $[x]$ is in one of $P(a)$ or $P(b)$ by Theorem 3.10 in [@AK]. So, by the above arguments $\lambda$ agrees with $h_{\#}$ on $\{[x]\} \cup L_{x} \cup D_{x}$. Since any nontrivial simple closed curve is in the dual or link of a 1-sided simple closed curve whose complement is nonorientable, $\lambda$ agrees with $h_{\#}$ on $\mathcal{C}(N)$.

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[**Case iii:**]{} Suppose that $g=3$. $N$ has at least two boundary components. Let $a, b, c$ be as in Figure \[Fig12\]. Let $P(a)$, $P(b)$ and $P(c)$ be the connected components of $\widetilde{X}(N)$ that contains $a$, $b$ and $c$ respectively. By Theorem 3.10 in [@AK], $\widetilde{X}(N)$ has three connected components $P(a)$, $P(b)$ and $P(c)$. As in the proof of case (ii), there exist homeomorphisms $h_1, h_2, h_3$ such that $\lambda$ agrees with $(h_1)_\#$ on every vertex of $\{[a]\} \cup L_{a} \cup D_{a} \cup \{[v]\} \cup L_{v} \cup D_{v}$ for any 1-sided simple closed curve $v$ such that $[v] \in P(a)$, $\lambda$ agrees with $(h_2)_\#$ on every vertex of $\{[b]\} \cup L_{b} \cup D_{b} \cup \{[w]\} \cup L_{w} \cup D_{w}$ for any 1-sided simple closed curve $w$ such that $[w] \in P(b)$, and $\lambda$ agrees with $(h_3)_\#$ on every vertex of $\{[c]\} \cup L_{c} \cup D_{c} \cup \{[z]\} \cup L_{z} \cup D_{z}$ for any 1-sided simple closed curve $z$ such that $[z] \in P(c)$. Let $d, e, f, g$ be as in Figure \[Fig12\]. By following the proof of case (ii) and using the curves $d, e$ we can see that $(h_1)_\# = (h_2)_\#$. (We note that in this case the complement of $T$ will be a genus one surface with at least 3 boundary components, and the difference map will be isotopic to identity on the complement of $T$, see Lemma 7.1 in [@AK]. The rest of the proof will follow as in case (ii).) Similarly, by using the curves $f, g$ we can see that $(h_2)_\# = (h_3)_\#$. Hence, we have $(h_1)_\# = (h_2)_\# = (h_3)_\#$.

Let $h=h_1$. If $x$ is a 1-sided simple closed curve with nonorientable complement (i.e. $[x]$ is a vertex in $\widetilde{X}(N)$), then $[x]$ is in one of $P(a)$ or $P(b)$ or $P(c)$ by Theorem 3.10 in [@AK]. By the above arguments and the proof of case (ii), we see that $\lambda$ agrees with $h_{\#}$ on $\{[x]\} \cup L_{x} \cup D_{x}$. Since any nontrivial simple closed curve is in the dual or link of a 1-sided simple closed curve whose complement is nonorientable, $\lambda$ agrees with $h_{\#}$ on $\mathcal{C}(N)$.

[**Case iv:**]{} We will prove the remaining cases when $g + n \geq 5$ by using induction on $g$. Assume that the theorem is true for some genus $g-1$ where $g-1 \geq 3$. We will prove that it is true for genus $g$. Let $a$ be a 1-sided simple closed curve on $N$ such that $N_a$ is nonorientable. Let $a' \in \lambda([a])$. By Lemma \[1-sided-cn\], $a'$ is a 1-sided simple closed curve with nonorientable complement. There is a homeomorphism $f: N \rightarrow N$ such that $f(a) = a'$. Let $f_{\#}$ be the simplicial automorphism induced by $f$ on $\mathcal{C}(N)$. Then $f_{\#}^{-1} \circ \lambda$ fixes $[a]$. By replacing $f_{\#}^{-1} \circ \lambda $ by $\lambda$ we can assume that $\lambda([a]) = [a]$. The simplicial map $\lambda$ restricts to a superinjective simplicial map $\lambda_a : L_a \rightarrow L_a$ as before. Since $L_a \cong \mathcal{C}(N_a)$, we get a superinjective simplicial map $\lambda_a :  \mathcal{C}(N_a)\rightarrow \mathcal{C}(N_a)$. By the induction assumption, there is a homeomorphism $G_a: N_a \rightarrow N_a$ such that $\lambda_a$ is induced by $(G_a)_{\#}$.

By following the proof of case (i), we see the following: There is a homeomorphism $g_a$ such that $\lambda$ agrees with $(g_a)_\#$ on $\{[a]\} \cup L_a \cup D_a$, and if $v$ is a 1-sided simple closed curve with nonorientable complement such that $[v]$ is connected to $[a]$ by a path in $\widetilde{X}(N)$, then $(g_v)_{\#} = (g_a)_{\#}$ on $\mathcal{C}(N)$. We also see that $\lambda$ agrees with $(g_v)_\#$ on $\{[v]\} \cup L_v \cup D_v$. Let $w$ be any 1-sided simple closed curve with nonorientable complement. Between $[a]$ and $[w]$ there is a path in $\widetilde{X}(N)$ since $\widetilde{X}(N)$ is connected by Theorem 3.10 in [@AK]. So, we have $(g_a)_{\#} = (g_w)_{\#}$ on $\mathcal{C}(N)$. Since the isotopy class of every nontrivial simple closed curve is in the link or dual of some 1-sided simple closed curve with nonorientable complement, we see that $\lambda$ agrees with $(g_a)_\#$ on $\mathcal{C}(N)$. By induction, we get our result for $g + n \geq 5$.\

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Now we prove the remaining case for the theorem. Suppose $(g, n) = (3, 0)$. Consider the curves $a, b, c$ as shown in Figure \[Figlast\]. Let $Q(a), Q(b), Q(c)$ be the connected components of $a, b, c$ in $X(N)$ respectively. By using Lemma \[1-sided-cn\], Theorem \[A\], Theorem \[id2\], and following the proof of case iii, we see that there exist homeomorphisms $h_1, h_2, h_3$ such that $\lambda$ agrees with $(h_1)_\#$ on $\{[a]\} \cup L_{a} \cup D_{a} \cup \{[v]\} \cup L_{v} \cup D_{v}$ for any 1-sided simple closed curve $v$ such that $[v] \in Q(a)$, $\lambda$ agrees with $(h_2)_\#$ on $\{[b]\} \cup L_{b} \cup D_{b} \cup \{[w]\} \cup L_{w} \cup D_{w}$ for any 1-sided simple closed curve $w$ such that $[w] \in Q(b)$, and $\lambda$ agrees with $(h_3)_\#$ on $\{[c]\} \cup L_{c} \cup D_{c} \cup \{[z]\} \cup L_{z} \cup D_{z}$ for any 1-sided simple closed curve $z$ such that $[z] \in Q(c)$. Let $d, e$ be as in Figure \[Figlast\]. By following the proof of case iii, and using the curves $d, e$ we can see that $(h_1)_\# = (h_2)_\#$ or $(h_1)_\# = (h_2)_\# \circ R_\#$ where $R$ is the reflection homeomorphism that we described at the beginning of this section. Similarly, we can see that $(h_2)_\# = (h_3)_\#$ or $(h_2)_\# = (h_3)_\# \circ R_\#$. Since $R_\#$ fixes the isotopy class of every nontrivial simple closed curve, the action of all these maps $(h_1)_\#$, $(h_2)_\#$ and $(h_3)_\#$ are the same on $\mathcal{C}(N)$.

Let $h=h_1$. If $x$ is a 1-sided simple closed curve with nonorientable complement, then $[x]$ is in one of $Q(a)$ or $Q(b)$ or $Q(c)$ by Theorem 3.9 in [@AK]. By the above arguments and the proof of case (i), we see that $\lambda$ agrees with $h_{\#}$ on $\{[x]\} \cup L_{x} \cup D_{x}$. Since any nontrivial simple closed curve is in the dual or link of a 1-sided simple closed curve whose complement is nonorientable, $\lambda$ agrees with $h_{\#}$ on $\mathcal{C}(N)$. This completes the proof.

[**Acknowledgments**]{}

We thank Peter Scott for some discussions and his comments about this paper. We thank Mustafa Korkmaz for some discussions. We also thank Ferihe Atalan for some discussions.

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