int 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.

=2.2in =2.1in

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'