]$. 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]).

=3.7in =3.7in

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 $N$.

=2.5in

\[1-sided-cn\] Let $g \geq 2$. Suppose that $(g, n) = (3, 0)$ or $g+n \geq 4$. Let $\lambda : \mathcal{C}(N) \rightarrow