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 $[a] \cup L_a$.

=2.5in =2.5in

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(