$T^{\prime o}=\mu_{\alpha}(T^{o})$ and by $\alpha^{\prime}$ the new arc
obtained.

In this section we construct a partition bijection
$\varphi^{T^{o}}_{\alpha}:\mathcal{P}(G_{T^{o},\gamma})\to\mathcal{P}(G_{T^{\prime
o},\gamma})$. To this end, we first construct a partition bijection
$\phi^{T^{o}}_{\alpha}:\mathcal{CP}(T^{o},\gamma)\to\mathcal{CP}(T^{\prime
o},\gamma)$ between the complete $(T^{o},\gamma)$-paths.

We start with the polygon case.

###  7.1. Polygon case

In this subsection, let $\Sigma$ be the $n$-gon, i.e., the disk with $n$
marked points labeled $1,\cdots,n$ clockwise on the boundary.

####  7.1.1. Partition bijection $\phi^{T^{o}}_{\alpha}$

For any complete $(T^{o},\gamma)$-path $\overrightarrow{\xi}$ and $\zeta\in
T^{o}$, recall $m(\overrightarrow{\xi},\zeta)$ given in (2). In the case that
$\Sigma$ is a polygon, we have $m(\overrightarrow{\xi},\zeta)\in\\{1,0,-1\\}$.

Assume that $\alpha=(a,c)$ is the diagonal of the quadrilateral $(a,b,c,d)$ in
$T^{o}$, where $a,b,c,d$ are in the clockwise order. Then
$\alpha^{\prime}=(b,d)$.

Let
$\overrightarrow{\xi}=(\overrightarrow{\xi}_{1},\cdots,\overrightarrow{\xi}_{2c+1})$
be a complete $(T^{o},\gamma)$-path. We would like to construct a set of
complete $(T^{\prime o},\gamma)$-paths as follows.

Case I: $\gamma$ crosses with two parallel edges of the quadrilateral
$(a,b,c,d)$. We may assume that $\gamma$ crosses
$\cdots,(a,b),(a,c),(c,d),\cdots$ sequentially. Then
$\xi_{2j-2}=(a,b),\xi_{2j}=(a,c)$ and $\xi_{2j+2}=(c,d)$ for some
$j\in\\{2,\cdots,c-1\\}$.

(1) If $\overrightarrow{\xi}_{2j}=(a,c)$ then
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,a),(a,c),(c,d),(d,c))$.
We have $m(\overrightarrow{\xi},\alpha)=-1$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((a,b),(b,d),(d,b),(b,d),(d,c))$ in the sequence $\overrightarrow{\xi}$.

(2) If $\overrightarrow{\xi}_{2j}=(c,a)$ then there are four possibilities of
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$:
$((b,a),(a,c),(c,a),(a,c),(c,d))$, $((b,a),(a,c),(c,a),(a,d),(d,c))$,
$((a,b),(b,c),(c,a),(a,c),(c,d))$ and $((a,b),(b,c),(c,a),(a,d),(d,c))$.

(2.1) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((b,a),(a,c),(c,a),(a,c),(c,d))$,
we have $m(\overrightarrow{\xi},\alpha)=1$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((b,a),(a,d),(d,b),(b,c),(c,d))$ or $((b,a),(a,b),(b,d),(d,c),(c,d))$ in
the sequence $\overrightarrow{\xi}$.

(2.2) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((b,a),(a,c),(c,a),(a,d),(d,c))$,
we have $m(\overrightarrow{\xi},\alpha)=0$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((b,a),(a,d),(d,b),(b,d),(d,c))$ in the sequence $\overrightarrow{\xi}$.

(2.3) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,c),(c,a),(a,c),(c,d))$,
we have $m(\overrightarrow{\xi},\alpha)=0$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((a,b),(b,d),(d,b),(b,c),(c,d))$ in the sequence $\overrightarrow{\xi}$.

(2.4) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,c),(c,a),(a,d),(d,c))$,
we have $m(\overrightarrow{\xi},\alpha)=-1$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((a,b),(b,d),(d,b),(b,d),(d,c))$ in the sequence $\overrightarrow{\xi}$.

In each case, one can see that the new sequences
$\overrightarrow{\xi}^{\prime}$ obtained are complete $(T^{\prime
o},\gamma)$-paths and satisfy
$m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=-m(\overrightarrow{\xi},\alpha).$

Case II: $\gamma$ crosses with $(a,c)$ and two adjacent edges of the
quadrilateral $(a,b,c,d)$. We may assume that $\gamma$ crosses with
$\cdots,(a,b),(a,c),(a,d),\cdots$ sequentially. Then
$\xi_{2j-2}=(a,b),\xi_{2j}=(a,c)$ and $\xi_{2j+2}=(a,d)$ for some
$j\in\\{2,\cdots,c-1\\}$.

(1) If $\overrightarrow{\xi}_{2j}=(a,c)$ then
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,a),(a,c),(c,a),(a,d))$
or $((a,b),(b,a),(a,c),(c,d),(d,a))$.

(1.1) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,a),(a,c),(c,a),(a,d))$,
we have $m(\overrightarrow{\xi},\alpha)=0$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((a,b),(b,a),(a,d))$ in the sequence $\overrightarrow{\xi}$.

(1.2) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,a),(a,c),(c,d),(d,a))$,
we have $m(\overrightarrow{\xi},\alpha)=-1$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((a,b),(b,d),(d,a))$ in the sequence $\overrightarrow{\xi}$.

(2) If $\overrightarrow{\xi}_{2j}=(c,a)$ then
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,c),(c,a),(a,d),(d,a))$
or $((b,a),(a,c),(c,a),(a,d),(d,a))$.

(2.1) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((a,b),(b,c),(c,a),(a,d),(d,a))$,
we have $m(\overrightarrow{\xi},\alpha)=-1$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((a,b),(b,d),(d,a))$ in the sequence $\overrightarrow{\xi}$.

(2.2) In case
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})=((b,a),(a,c),(c,a),(a,d),(d,a))$,
we have $m(\overrightarrow{\xi},\alpha)=0$.

In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j},\overrightarrow{\xi}_{2j+1},\overrightarrow{\xi}_{2j+2})$
by $((b,a),(a,d),(d,a))$ in the sequence $\overrightarrow{\xi}$.

In each case, one can see that the new sequences
$\overrightarrow{\xi}^{\prime}$ obtained are complete $(T^{\prime
o},\gamma)$-paths and satisfy
$m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=-m(\overrightarrow{\xi},\alpha).$

Case III: $\gamma$ crosses two adjacent edges of the quadrilateral $(a,b,c,d)$
but not cross $(a,c)$. We may assume that $\gamma$ crosses with
$\cdots,(a,b),(b,c),\cdots$ sequentially. Then
$\xi_{2j-2}=(a,b),\xi_{2j}=(b,c)$ for some $j\in\\{2,\cdots,c\\}$. Then we
have
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})=((b,a),(a,b),(b,c))$
or $((a,b),(b,c),(c,b))$ or $((b,a),(a,c),(c,b))$.

(1) If
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})=((b,a),(a,b),(b,c))$,
we have $m(\overrightarrow{\xi},\alpha)=0$. In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})$
by $((b,a)$, $(a,b),(b,d),(d,b),(b,c))$ in the sequence
$\overrightarrow{\xi}$.

(2) If
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})=((a,b),(b,c),(c,b))$,
we have $m(\overrightarrow{\xi},\alpha)=0$. In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})$
by $((a,b)$, $(b,d),(d,b),(b,c),(c,b))$ in the sequence
$\overrightarrow{\xi}$.

(3) If
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})=((b,a),(a,c),(c,b))$,
we have $m(\overrightarrow{\xi},\alpha)=1$. In this case we replace
$(\overrightarrow{\xi}_{2j-2},\overrightarrow{\xi}_{2j-1},\overrightarrow{\xi}_{2j})$
by $((b,a)$, $(a,d),(d,b),(b,c),(c,b))$ or $((b,a),(a,b),(b,d),(d,c),(c,b))$
in the sequence $\overrightarrow{\xi}$.

In each case, one can see that the new sequences
$\overrightarrow{\xi}^{\prime}$ obtained are complete $(T^{\prime
o},\gamma)$-paths and satisfy
$m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=-m(\overrightarrow{\xi},\alpha).$

Case IV: $\gamma$ crosses with $(a,c)$ and exactly one edge of the
quadrilateral $(a,b,c,d)$. We may assume that $\gamma$ crosses with
$(a,c),(a,d),\cdots$ sequentially. Then $\xi_{2}=(a,c),\xi_{4}=(a,d)$. Thus we
have
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})=((b,a),(a,c),(c,a),(a,d))$
or $((b,a),(a,c),(c,d),(d,a))$ or $((b,c),(c,a),(a,d),(d,a))$.

(1) If
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})=((b,a),(a,c),(c,a),(a,d))$,
we have $m(\overrightarrow{\xi},\alpha)=0$. In this case we replace
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})$
by $((b,a)$, $(a,d))$ in the sequence $\overrightarrow{\xi}$.

(2) If
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})=((b,a),(a,c),(c,d),(d,a))$,
we have $m(\overrightarrow{\xi},\alpha)=-1$. In this case we replace
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})$
by $((b,d)$, $(d,a))$ in the sequence $\overrightarrow{\xi}$.

(3) If
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})=((b,c),(c,a),(a,d),(d,a))$,
we have $m(\overrightarrow{\xi},\alpha)=-1$. In this case we replace
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2},\overrightarrow{\xi}_{3},\overrightarrow{\xi}_{4})$
by $((b,d)$, $(d,a))$ in the sequence $\overrightarrow{\xi}$.

In each case, one can see that the new sequences
$\overrightarrow{\xi}^{\prime}$ obtained are complete $(T^{\prime
o},\gamma)$-paths and satisfy
$m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=-m(\overrightarrow{\xi},\alpha).$

Case V: $\gamma$ crosses with exactly one edge of the quadrilateral
$(a,b,c,d)$ but does not cross with $(a,c)$. We may assume that $\gamma$
crosses with $(b,c),\cdots$ sequentially. Then $\xi_{2}=(b,c)$. Thus we have
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2})=((a,b),(b,c))$ or
$((a,c),(c,b))$.

(1) If $(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2})=((a,b),(b,c))$, we
have $m(\overrightarrow{\xi},\alpha)=0$. In this case we replace
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2})$ by
$((a,b),(b,d),(d,b),(b,c))$ in the sequence $\overrightarrow{\xi}$.

(2) If $(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2})=((a,c),(c,b))$, we
have $m(\overrightarrow{\xi},\alpha)=1$. In this case we replace
$(\overrightarrow{\xi}_{1},\overrightarrow{\xi}_{2})$ by
$((a,b),(b,d),(d,c),(c,b))$ or $((a,d),(d,b),(b,c),(c,b))$ in the sequence
$\overrightarrow{\xi}$.

In each case, one can see that the new sequences
$\overrightarrow{\xi}^{\prime}$ obtained are complete $(T^{\prime
o},\gamma)$-paths and satisfy
$m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=-m(\overrightarrow{\xi},\alpha).$

From the above discussion, we have the following observation.

######  Lemma 7.1.

Let
$\overrightarrow{\xi}=(\overrightarrow{\xi}_{1},\cdots,\overrightarrow{\xi}_{2c+1})$
be a complete $(T^{o},\gamma)$-path. Assume that $\xi_{j}=\alpha=(a,c)$ for
some even $j$ such that $\xi_{j-1}\neq\alpha\neq\xi_{j+1}$.

  1. $(1)$

If $\overrightarrow{\xi}_{j}=(a,c)$ then
$(\overrightarrow{\xi}_{j-1},\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})=((b,a),(a,c),(c,d))$
or $((d,a),(a,c),(c,b))$.

  2. $(2)$

If $\overrightarrow{\xi}_{j}=(c,a)$ then
$(\overrightarrow{\xi}_{j-1},\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})=((b,c),(c,a),(a,d))$
or $((d,c),(c,a),(a,b))$.

Therefore, for any complete $(T^{o},\gamma)$-path $\overrightarrow{\xi}$, we
obtain a set of complete $(T^{\prime o},\gamma)$-paths, denoted by
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$, moreover, for any
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$,
we have
$m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=-m(\overrightarrow{\xi},\alpha).$

The construction of $\phi^{T^{o}}_{\alpha}$ can be summarized as follows:

Construction of $\phi^{T^{o}}_{\alpha}$: given any complete
$(T^{o},\gamma)$-path
$\overrightarrow{\xi}=(\overrightarrow{\xi}_{1},\cdots,\overrightarrow{\xi}_{2c+1})$.

If $m(\overrightarrow{\xi},\alpha)=1$, then the set
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$ contains all sequences of
oriented arcs $\overrightarrow{\xi}^{\prime}$, where
$\overrightarrow{\xi}^{\prime}$ is obtained via the following recipe:

  1. $(i)$

First remove a pair $(\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})$ in
case $\xi_{j}=\xi_{j+1}=\alpha$ and $\xi_{j-1}\neq\alpha$;

  2. $(ii)$

Next for any $\xi_{j}=\alpha$ with odd $j$,

    1. (a)

if $\overrightarrow{\xi}_{j}=(a,c)$, then replace $\overrightarrow{\xi}_{j}$
by $(a,b),(b,d),(d,c)$ or $(a,d),(d,b),(b,c)$;

    2. (b)

if $\overrightarrow{\xi}_{j}=(c,a)$, then replace $\overrightarrow{\xi}_{j}$
by $(c,b),(b,d),(d,a)$ or $(c,d),(d,b),(b,a)$;

If $m(\overrightarrow{\xi},\alpha)=0$, then the set
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$ contains all sequences of
oriented arcs $\overrightarrow{\xi}^{\prime}$, where
$\overrightarrow{\xi}^{\prime}$ can be obtained via the following recipe:

  1. $(i)$

First remove a pair $(\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})$ in
case $\xi_{j}=\xi_{j+1}=\alpha$ and $\xi_{j-1}\neq\alpha$;

  2. $(ii)$

Next, if $\gamma$ crosses with $\alpha^{\prime}=(b,d)$, then add $(b,d),(d,b)$
or $(d,b),(b,d)$ to make the sequences to complete $(T^{\prime
o},\gamma)$-paths;

If $m(\overrightarrow{\xi},\alpha)=-1$, then $\xi_{j}=\alpha$ for some even
$j$ such that $\xi_{j-1}\neq\alpha\neq\xi_{j+1}$. The set
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$ contains all sequences of
oriented arcs $\overrightarrow{\xi}^{\prime}$, where
$\overrightarrow{\xi}^{\prime}$ can be obtained via the following recipe:

  1. $(i)$

First,
$(\overrightarrow{\xi}_{j-1},\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})=((b,a),(a,c),(c,d))$;
$((b,c),(c,a),(a,d))$; $((d,a),(a,c),(c,b))$ or $((d,c),(c,a),(a,b))$ by Lemma
7.1.

    1. $(a)$

if
$(\overrightarrow{\xi}_{j-1},\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})=((b,a),(a,c),(c,d))$
or $((b,c),(c,a),(a,d))$, then replace it by $(b,d)$;

    2. $(b)$

if
$(\overrightarrow{\xi}_{j-1},\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})=((d,a),(a,c),(c,b))$
or $((d,c),(c,a),(a,b))$, then replace it by $(d,b)$;

  2. $(ii)$

Next, if $\gamma$ crosses with $\alpha^{\prime}=(b,d)$, then add $(b,d),(d,b)$
or $(d,b),(b,d)$ to make the sequences to complete $(T^{\prime
o},\gamma)$-paths.

######  Remark 7.2.

In particular, if $\gamma$ crosses neither $\alpha$ nor $\alpha^{\prime}$ then
$\mathcal{CP}(T^{o},\gamma)=\mathcal{CP}(T^{\prime o},\gamma)$ and
$\phi^{T^{o}}_{\alpha}$ is the identity map.

In summary, we obtain the following proposition.

######  Proposition 7.3.

With the foregoing notation. For any complete $(T^{o},\gamma)$-path
$\overrightarrow{\xi}$, we have

  1. $(1)$

$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})\subset\mathcal{CP}(T^{\prime
o},\gamma)$;

  2. $(2)$

$|\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})|=2^{[m(\overrightarrow{\xi},\alpha)]_{+}}$,
moreover, if $m(\overrightarrow{\xi},\alpha)=1$, assume that
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})=\\{\overrightarrow{\xi}^{\prime},\overrightarrow{\xi}^{\prime\prime}\\}$,
then $\overrightarrow{\xi}^{\prime}$ and $\overrightarrow{\xi}^{\prime\prime}$
are related by a twist at $\alpha^{\prime}$;

  3. $(3)$

$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})$
for some $\zeta\neq\xi$ if and only if $m(\overrightarrow{\xi},\alpha)=-1$,
and $\overrightarrow{\xi}$ and $\overrightarrow{\zeta}$ are related by a twist
on $\alpha$;

  4. $(4)$

for any
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$,
we have
$m(\overrightarrow{\xi},\alpha)=-m(\overrightarrow{\xi}^{\prime},\alpha^{\prime})$;

  5. $(5)$

for any $(a,b),(b,c),(c,d),(a,d),(a,c)\neq\zeta\in T^{o}$ and
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$,
we have
$m(\overrightarrow{\xi},\zeta)=m(\overrightarrow{\xi}^{\prime},\zeta)$.

Similarly, we can construct $\phi^{T^{o}}_{\alpha^{\prime}}$.

######  Theorem 7.4.

With the foregoing notation. $\phi^{T^{o}}_{\alpha}$ and $\phi^{T^{\prime
o}}_{\alpha^{\prime}}$ are partition bijections between
$\mathcal{CP}(T^{o},\gamma)$ and $\mathcal{CP}(T^{\prime o},\gamma)$,
moreover, $\phi^{T^{o}}_{\alpha}$ and $\phi^{T^{\prime o}}_{\alpha^{\prime}}$
are inverse to each other.

###### Proof.

We first show that $\phi^{T^{o}}_{\alpha}$ and $\phi^{T^{\prime
o}}_{\alpha^{\prime}}$ are partition bijections. We shall only consider
$\phi^{T^{o}}_{\alpha}$. By Proposition 7.3 (2), we have
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})\neq\emptyset$ for any
$\overrightarrow{\xi}\in\mathcal{CP}(T^{o},\gamma)$.

Suppose that
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})\cap\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})\neq\emptyset$
for some
$\overrightarrow{\xi},\overrightarrow{\zeta}\in\mathcal{CP}(T^{o},\gamma)$.
Assume that
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})\cap\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})$.
Then
$m(\overrightarrow{\xi},\alpha)=m(\overrightarrow{\zeta},\alpha)=-m(\overrightarrow{\xi^{\prime}},\alpha^{\prime})$
by Proposition 7.3 (4). If
$m(\overrightarrow{\xi},\alpha)=m(\overrightarrow{\zeta},\alpha)\leq 0$ then
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})=\\{\overrightarrow{\xi}^{\prime}\\}$
by Proposition 7.3 (2). If
$m(\overrightarrow{\xi},\alpha)=m(\overrightarrow{\zeta},\alpha)=1$, then
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})=\\{\overrightarrow{\xi}^{\prime},\overrightarrow{\xi}^{\prime\prime}\\}$
by Proposition 7.3 (2), where $\overrightarrow{\xi}^{\prime\prime}$ is the
complete $(T^{\prime o},\gamma)$-path related to
$\overrightarrow{\xi}^{\prime}$ by a twist on $\alpha^{\prime}$. Therefore we
have either
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})\cap\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})=\emptyset$
or
$\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha}(\overrightarrow{\zeta})$
for any
$\overrightarrow{\xi},\overrightarrow{\zeta}\in\mathcal{CP}(T^{o},\gamma)$.

According to the construction of $\phi^{T^{o}}_{\alpha}$ and $\phi^{T^{\prime
o}}_{\alpha^{\prime}}$, we see that
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})$
if and only if $\overrightarrow{\xi}\in\phi^{T^{\prime
o}}_{\alpha^{\prime}}(\overrightarrow{\xi}^{\prime})$ for any
$\overrightarrow{\xi}\in\mathcal{CP}(T^{o},\gamma)$ and
$\overrightarrow{\xi}^{\prime}\in\mathcal{CP}(T^{\prime o},\gamma)$. Thus
$\bigcup_{\overrightarrow{\xi}\in\mathcal{CP}(T^{o},\gamma)}\phi^{T^{o}}_{\alpha}(\overrightarrow{\xi})=\mathcal{CP}(T^{\prime
o},\gamma)$. Therefore, $\phi^{T^{o}}_{\alpha}$ is a partition bijection by
Remark 2.5. Similarly, $\phi^{T^{\prime o}}_{\alpha}$ is a partition
bijection. By Proposition 2.6, $\phi^{T^{o}}_{\alpha}$ and $\phi^{T^{\prime
o}}_{\alpha^{\prime}}$ are inverse to each other. $\square$

####  7.1.2. Partition bijection $\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$

Let $\alpha_{1},\cdots,\alpha_{k}$ be a set of arcs in $T^{o}$ such that
$\alpha_{i},\alpha_{j}$ are not two sides of any triangle in $T^{o}$ for any
$i\neq j$. Denote $T^{\prime
o}=\mu_{\alpha_{k}}\circ\cdots\circ\mu_{\alpha_{1}}(T^{o})$. Denote by
$\alpha^{\prime}_{i}$ the arc obtained from $T^{o}$ by flip at $\alpha_{i}$.
Assume that $\alpha_{i}=(a_{i},c_{i})$ and $\alpha_{i}$ is a diagonal of the
quadrilateral $(a_{i},b_{i},c_{i},d_{i})$ in $T^{o}$, where
$a_{i},b_{i},c_{i},d_{i}$ are in the clockwise order.

For any given complete $(T^{o},\gamma)$-path
$\overrightarrow{\xi}=(\overrightarrow{\xi}_{1},\cdots,\overrightarrow{\xi}_{2c+1})$,
we construct a subset
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$ of
complete $(T^{\prime o},\gamma)$-paths, which is obtained from
$\overrightarrow{\xi}$ via the following receipt:

  1. $(i)$

First, remove all the pairs
$(\overrightarrow{\xi}_{j},\overrightarrow{\xi}_{j+1})$ such that
$\xi_{j}=\xi_{j+1}=\alpha_{i}$ for some $i\in\\{1,\cdots,k\\}$ and
$\xi_{j-1}\neq\alpha_{i}$;

  2. $(ii)$

Next, for any $i\in\\{1,\cdots,k\\}$,

    1. $(a)$

if $m(\overrightarrow{\xi},\alpha_{i})=1$, we have $\xi_{j_{i}}=\alpha_{i}$
for some odd $j_{i}$, then replace $\overrightarrow{\xi}_{j_{i}}$ by
$(a_{i},b_{i}),(b_{i},d_{i}),(d_{i},c_{i})$ or
$(a_{i},d_{i}),(d_{i},b_{i}),(b_{i},c_{i})$ in case
$\overrightarrow{\xi}_{j_{i}}=(a,c)$ and replace
$\overrightarrow{\xi}_{j_{i}}$ by $(c_{i},b_{i}),(b_{i},d_{i}),(d_{i},a_{i})$
or $(c_{i},d_{i}),(d_{i},b_{i}),(b_{i},a_{i})$ in case
$\overrightarrow{\xi}_{j_{i}}=(c_{i},a_{i})$;

    2. $(b)$

if $m(\overrightarrow{\xi},\alpha_{i})=-1$, we have $\xi_{j_{i}}=\alpha_{i}$
for some even $j_{i}$, then replace
$(\overrightarrow{\xi}_{j_{i}-1},\overrightarrow{\xi}_{j_{i}},\overrightarrow{\xi}_{j_{i}+1})$
by $(b_{i},d_{i})$ in case
$(\overrightarrow{\xi}_{j_{i}-1},\overrightarrow{\xi}_{j_{i}},\overrightarrow{\xi}_{j_{i}+1})=((b_{i},a_{i}),(a_{i},c_{i}),(c_{i},d_{i}))$
or $((b_{i},c_{i}),(c_{i},a_{i}),(a_{i},d_{i}))$, and replace
$(\overrightarrow{\xi}_{j_{i}-1},\overrightarrow{\xi}_{j_{i}},\overrightarrow{\xi}_{j_{i}+1})$
by $(d_{i},b_{i})$ in case
$(\overrightarrow{\xi}_{j_{i}-1},\overrightarrow{\xi}_{j_{i}},\overrightarrow{\xi}_{j_{i}+1})=((d_{i},a_{i}),(a_{i},c_{i}),(c_{i},b_{i}))$
or $((d_{i},c_{i}),(c_{i},a_{i}),(a_{i},b_{i}))$;

  3. $(iii)$

Last, for all the $\alpha^{\prime}_{i}=(b_{i},d_{i})$ such that $\gamma$
crosses with $\alpha^{\prime}_{i}$, then add $(b_{i},d_{i}),(d_{i},b_{i})$ or
$(d_{i},b_{i}),(b_{i},d_{i})$ to make the sequences to complete $(T^{\prime
o},\gamma)$-paths.

As $\alpha_{i},\alpha_{j}$ are not two sides of any triangle in $T^{o}$ for
any $i\neq j$, the above construction is well-defined and we have

(20)
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})=\phi^{\mu_{\alpha_{k-1}}\circ\cdots\circ\mu_{\alpha_{1}}T^{o}}_{\alpha_{k}}\left(\cdots\phi^{\mu_{\alpha_{1}}T^{o}}_{\alpha_{2}}(\phi^{T^{o}}_{\alpha_{1}}(\overrightarrow{\xi}))\right),$

moreover, $\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$ does not depend on the
order of $\alpha_{1},\cdots,\alpha_{k}$.

We can also construct a subset $\phi^{T^{\prime
o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}(\overrightarrow{\xi}^{\prime})$
for any complete $(T^{\prime o},\gamma)$-path $\overrightarrow{\xi}^{\prime}$.

We have the following proposition which generalizes Proposition 7.3.

######  Proposition 7.5.

Let $\alpha_{1},\cdots,\alpha_{k}$ be a set of arcs in $T^{o}$ such that
$\alpha_{i}$ and $\alpha_{j}$ are not two sides of any triangle in $T^{o}$ for
any $i\neq j$. For any complete $(T^{o},\gamma)$-path $\overrightarrow{\xi}$,
we have

  1. $(1)$

$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})\subset\mathcal{CP}(T^{\prime
o},\gamma)$,

  2. $(2)$

$|\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})|=2^{\sum_{i=1}^{k}[m(\overrightarrow{\xi},\alpha_{i})]_{+}}$,
moreover, for any $\alpha_{i}$ with $m(\overrightarrow{\xi},\alpha_{i})=1$ and
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$,
there exists
$\overrightarrow{\xi}^{\prime\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$
such that $\overrightarrow{\xi}^{\prime}$ and
$\overrightarrow{\xi}^{\prime\prime}$ are related by a twist on
$\alpha^{\prime}_{i}$;

  3. $(3)$

if $m(\overrightarrow{\xi},\alpha_{i})=-1$ for some $\alpha_{i}$ then we have
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})$,
where $\overrightarrow{\zeta}$ is the complete
$(\prod_{i=1}^{k}\mu_{\alpha_{i}}(T^{o}),\gamma)$-path which is related to
$\overrightarrow{\xi}$ by a twist at $\alpha_{i}$;

  4. $(4)$

for any $i\in\\{1,\cdots,k\\}$ and
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$,
we have
$m(\overrightarrow{\xi},\alpha_{i})=-m(\overrightarrow{\xi}^{\prime},\alpha^{\prime}_{i}).$

  5. $(5)$

If $\zeta\in T^{o}$ and
$\zeta\neq(a_{i},b_{i}),(b_{i},c_{i}),(c_{i},d_{i}),(a_{i},d_{i}),(a_{i},c_{i})$
for all $i$, then for any
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$,
we have
$m(\overrightarrow{\xi},\zeta)=m(\overrightarrow{\xi}^{\prime},\zeta).$

The following theorem generalizes Theorem 7.4.

######  Theorem 7.6.

With the foregoing notation. $\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$ and
$\phi^{T^{\prime o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}$ are
partition bijections between $\mathcal{CP}(T^{o},\gamma)$ and
$\mathcal{CP}(T^{\prime o},\gamma)$ inverse to each other.

###### Proof.

It is clear that
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})\neq\emptyset$
for any $\overrightarrow{\xi}\in\mathcal{CP}(T^{o},\gamma)$.

Suppose that
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})\cap\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})\neq\emptyset$
for some $\overrightarrow{\xi}$ and $\overrightarrow{\zeta}$. Assume that
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})\cap\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})$.
Then
$m(\overrightarrow{\xi},\alpha_{i})=m(\overrightarrow{\zeta},\alpha_{i})=-m(\overrightarrow{\xi}^{\prime},\alpha^{\prime}_{i})$
for any $i$ by Proposition 7.5 (4). By Proposition 7.5 (2), we see that
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})$
contains all $\overrightarrow{\xi}^{\prime\prime}$ which are related to
$\overrightarrow{\xi}^{\prime}$ by a sequence of twists on some
$\alpha^{\prime}_{i}$ with $m(\overrightarrow{\xi},\alpha_{i})=1$. Therefore,
we have
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})$.

For any $\overrightarrow{\xi}\in\mathcal{CP}(T^{o},\gamma)$ and
$\overrightarrow{\xi}^{\prime}\in\mathcal{CP}(T^{\prime o},\gamma)$, we show
that
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$
if and only if $\overrightarrow{\xi}\in\phi^{T^{\prime
o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}(\overrightarrow{\xi}^{\prime})$
by induction on $k$. If $k=1$, it follows by Theorem 7.4. Suppose that it is
true for the cases less than $k$. If
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})$,
then
$\overrightarrow{\xi}^{\prime}\in\phi^{T^{o}}_{\alpha_{k}}(\overrightarrow{\eta})$
for some
$\overrightarrow{\eta}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k-1}}(\overrightarrow{\xi})$.
By induction hypothesis, we have
$\overrightarrow{\xi}\in\phi^{\prod_{i=1}^{k-1}\mu_{\alpha_{i}}T^{o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k-1}}(\overrightarrow{\eta})$
and
$\overrightarrow{\eta}\in\phi^{\prod_{i=1}^{k}\mu_{\alpha_{i}}T^{o}}_{\alpha^{\prime}_{k}}(\overrightarrow{\xi}^{\prime})$.
Thus,
$\overrightarrow{\xi}\in\phi^{\prod_{i=1}^{k-1}\mu_{\alpha_{i}}T^{o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k-1}}(\phi^{\prod_{i=1}^{k}\mu_{\alpha_{i}}T^{o}}_{\alpha^{\prime}_{k}}(\overrightarrow{\xi}^{\prime}))=\phi^{T^{\prime
o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}(\overrightarrow{\xi}^{\prime}).$
The statement is proved.

By Remark 2.5, $\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$ is a partition
bijection. Similarly, $\phi^{T^{\prime
o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}$ is a partition
bijection. By Proposition 2.6, $\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$
and $\phi^{T^{\prime o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}$ are
inverse to each other. $\square$

######  Remark 7.7.

$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})\cap\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})\neq\emptyset$
if and only if
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\xi})=\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(\overrightarrow{\zeta})$
if and only if $\overrightarrow{\xi}$ and $\overrightarrow{\zeta}$ are related
by a sequence of twists on some $\alpha_{i}$ such that
$m(\overrightarrow{\xi},\alpha_{i})=-1$.

####  7.1.3. Partition bijection $\varphi^{T^{o}}_{\alpha}$

Recall that there are natural bijective maps from $\mathcal{CP}(T^{o},\gamma)$
to $\mathcal{P}(G_{T^{o},\gamma})$ and from $\mathcal{CP}(T^{\prime
o},\gamma)$ to $\mathcal{P}(G_{T^{\prime o},\gamma})$, see Subsection 2.7.
Therefore, the partition bijection $\phi^{T^{o}}_{\alpha}$ from
$\mathcal{CP}(T^{o},\gamma)$ to $\mathcal{CP}(T^{\prime o},\gamma)$ induces a
partition bijection $\varphi^{T^{o}}_{\alpha}$ from
$\mathcal{P}(G_{T^{o},\gamma})$ to $\mathcal{P}(G_{T^{\prime o},\gamma})$,
that is, we have the following commutative diagram.

$\textstyle{\mathcal{CP}(T^{o},\gamma)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\phi^{T^{o}}_{\alpha}}$$\textstyle{\mathcal{CP}(T^{\prime
o},\gamma)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\mathcal{P}(G_{T^{o},\gamma})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi^{T^{o}}_{\alpha}}$$\textstyle{\mathcal{P}(G_{T^{\prime
o},\gamma}).}$

We also have the inverse partition bijection $\varphi^{T^{\prime
o}}_{\alpha^{\prime}}:\mathcal{P}(G_{T^{\prime
o},\gamma})\to\mathcal{P}(G_{T^{o},\gamma})$ of $\varphi^{T^{o}}_{\alpha}$.

Under the bijection between $\mathcal{CP}(T^{o},\gamma)$ and
$\mathcal{P}(G_{T^{o},\gamma})$, for any $P\in\mathcal{P}(G_{T^{o},\gamma})$,
denote by $\overrightarrow{\xi}(P)$ the perfect matching of $G_{T^{o},\gamma}$
corresponding to $P$. For any $\zeta\in T^{o}$, let

$\hat{m}(P,\zeta)=m(\overrightarrow{\xi}(P),\zeta),$

where $m(\overrightarrow{\xi}(P),\zeta)$ is given by (2).

In view of (2) and (9) (10), by the bijection between
$\mathcal{P}(G_{T^{o},\gamma})$ and $\mathcal{CP}(T^{o},\gamma)$, we have

$\hat{m}(P,\zeta)=m(P;\zeta)-n(G_{T^{o},\gamma};\zeta).$

Similarly we can define $\hat{m}(P^{\prime},\zeta)$ for any
$P^{\prime}\in\mathcal{P}(G_{T^{\prime o},\gamma})$ and $\zeta\in T^{\prime
o}$.

Assume that $\alpha=(a,c)$ and $\alpha$ is the diagonal of the quadrilateral
$(a,b,c,d)$ in $T^{o}$. Suppose that $a,b,c,d$ are in the clockwise order.

######  Proposition 7.8.

With the foregoing notation. Let $P\in\mathcal{P}(G_{T^{o},\gamma})$.

  1. $(1)$

We have $|\varphi_{\alpha}^{T^{o}}(P)|=2^{[\hat{m}(P,\alpha)]_{+}}$, moreover,

    1. (a)

if $\hat{m}(P,\alpha)=1$, then there is a tile $G^{\prime}$ of $G_{T^{\prime
o},\gamma}$ with diagonal labeled $\alpha^{\prime}$ such that any
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}$ can twist on $G^{\prime}$ and
$\varphi_{\alpha}^{T^{o}}$ is closed under the twist on $G^{\prime}$;

    2. (b)

if $\hat{m}(P,\alpha)=-1$, then there is a tile $G$ of $G_{T^{o},\gamma}$ with
diagonal labeled $\alpha$ such that $P$ can twist on $G$ and
$\varphi_{\alpha}^{T^{o}}(P)=\varphi_{\alpha}^{T^{o}}(\mu_{G}P)$;

  2. $(2)$

For any $P^{\prime}\in\varphi^{T^{o}}_{\alpha}(P)$, we have
$\hat{m}(P,\alpha)=-\hat{m}(P^{\prime},\alpha^{\prime}).$

  3. $(3)$

For any $\zeta\in T^{o}$ such that $\zeta\neq(a,b),(b,c),(c,d),(a,d),(a,c)$
and $P^{\prime}\in\phi^{T^{o}}_{\alpha}(P)$, we have

$\hat{m}(P,\zeta)=\hat{m}(P^{\prime},\zeta).$

###### Proof.

It follows immediately by Proposition 7.3. $\square$

######  Proposition 7.9.

Assume $P$ can twist on a tile $G_{l}$ with diagonal labeled $\tau_{i_{l}}$
and $P>\mu_{G_{l}}P$.

  1. $(1)$

if $\tau_{i_{l}}\neq\alpha,(a,b),(b,c),(c,d),(a,d)$ then there is a tile
$G^{\prime}_{[l]}$ of $G_{T^{\prime o},\gamma}$ with diagonal labeled
$\tau_{i_{l}}$ such that: any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can
twist on $G^{\prime}_{[l]}$, $P^{\prime}>\mu_{G^{\prime}_{[l]}}(P^{\prime})$,
and

$\varphi_{\alpha}^{T^{o}}(\mu_{G_{l}}P)=\mu_{G^{\prime}_{[l]}}(\varphi_{\alpha}^{T^{o}}(P)).$

  2. $(2)$

if $\tau_{i_{l}}=\alpha$, then $\hat{m}(P,\alpha)=-1$ and
$\\{P,\mu_{G_{l}}P\\}=\varphi_{\alpha^{\prime}}^{T^{\prime o}}(P^{\prime}),$
where $\varphi_{\alpha}^{T^{o}}(P)=\\{P^{\prime}\\}$.

  3. $(3)$

if $\tau_{i_{l}}\in\\{(a,b),(c,d)\\}$, then
$\hat{m}(P,\alpha)=\hat{m}(\mu_{G_{l}}P,\alpha)-1$. For
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$, there are tiles $G^{\prime}_{[l]}$
and $G^{\prime}_{[l^{\prime}]}$ of $G_{T^{\prime o},\gamma}$ with diagonal
labeled $\tau_{i_{l}}$ and $(b,d)$, respectively, such that

    1. (a)

$|[l]-[l^{\prime}]|=1$,

    2. (b)

$P^{\prime}$ can twist on $G^{\prime}_{[l]}$ and
$\mu_{G^{\prime}_{[l]}}P^{\prime}$ can twist on $G^{\prime}_{[l^{\prime}]}$,

    3. (c)

$P^{\prime}>\mu_{G^{\prime}_{[l]}}P^{\prime}>\mu_{G^{\prime}_{[l^{\prime}]}}\mu_{G^{\prime}_{[l]}}P^{\prime}$,

    4. (d)

$\mu_{G^{\prime}_{[l]}}P^{\prime}\in\varphi_{\alpha}^{T^{o}}(\mu_{G_{l}}P)$.

  4. $(4)$

if $\tau_{i_{l}}\in\\{(b,c),(a,d)\\}$, then
$\hat{m}(P,\alpha)=\hat{m}(\mu_{G_{l}}P,\alpha)+1$. For
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(\mu_{G_{l}}P)$, there are tiles
$G^{\prime}_{[l]}$ and $G^{\prime}_{[l^{\prime}]}$ of $G_{T^{\prime
o},\gamma}$ with diagonal labeled $\tau_{i_{l}}$ and $(b,d)$, respectively,
such that

    1. (a)

$|[l]-[l^{\prime}]|=1$,

    2. (b)

$P^{\prime}$ can twist on $G^{\prime}_{[l]}$ and
$\mu_{G^{\prime}_{[l]}}P^{\prime}$ can twist on $G^{\prime}_{[l^{\prime}]}$,

    3. (c)

$P^{\prime}<\mu_{G^{\prime}_{[l]}}P^{\prime}<\mu_{G^{\prime}_{[l^{\prime}]}}\mu_{G^{\prime}_{[l]}}P^{\prime}$,

    4. (d)

$\mu_{G^{\prime}_{[l]}}P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$.

###### Proof.

It follows by the construction of $\phi^{T^{o}}_{\alpha}$. $\square$

######  Example 7.10.

Let $\Sigma$ be the hexagon. If $T^{o}$ and $T^{\prime o}$ are triangulations
as shown in Figure 9,

Figure 9.

then the snake graphs $G_{T^{o},\gamma}$, $G_{T^{\prime o},\gamma}$ and the
partition bijections $\varphi_{\alpha}^{T^{o}}$,
$\varphi_{\alpha^{\prime}}^{T^{\prime o}}$ are shown in Figure 10.

Figure 10.

If $T^{o}$ and $T^{\prime o}$ are triangulations as shown in Figure 11, then
the snake graphs $G_{T^{o},\gamma}$, $G_{T^{\prime o},\gamma}$ and the
partition bijections $\varphi_{\alpha}^{T^{o}}$,
$\varphi_{\alpha^{\prime}}^{T^{\prime o}}$ are shown in Figure 11.

Figure 11.

According to Lemma 2.35, we have the following observation.

######  Lemma 7.11.

  1. $(1)$

If $\hat{m}(P_{-},\alpha)=1$, then one of the following cases happens: (1)
$\gamma$ crosses $(a,b),\alpha,(c,d)$ consecutively; (2) $\gamma$ starts from
$a$ then crosses $(c,d)$; (3) $\gamma$ starts from $c$ then crosses $(a,b)$.

  2. $(2)$

If $\hat{m}(P_{-},\alpha)=-1$, then one of the following cases happens: (1)
$\gamma$ crosses $(a,d),\alpha,(b,c)$ consecutively; (2) $\gamma$ starts from
$d$ then crosses $\alpha$, $(b,c)$ sequentially; (3) $\gamma$ starts from $b$
then crosses $\alpha,(a,d)$ sequentially.

####  7.1.4. Partition bijection
$\varphi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$

Let $\alpha_{1},\cdots,\alpha_{k}$ be a set of arcs in $T^{o}$ such that
$\alpha_{i},\alpha_{j}$ are not two sides of any triangle in $T^{o}$ for any
$i\neq j$. Denote by $\alpha^{\prime}_{i}$ the arc obtained from $T^{o}$ by
flip at $\alpha_{i}$. Assume that $\alpha_{i}=(a_{i},c_{i})$ and $\alpha_{i}$
is the diagonal of the quadrilateral $(a_{i},b_{i},c_{i},d_{i})$ in $T^{o}$,
where $a_{i},b_{i},c_{i},d_{i}$ are in the clockwise order. Denote $T^{\prime
o}=\mu_{\alpha_{k}}\circ\cdots\circ\mu_{\alpha_{1}}(T^{o})$.

The partition bijection
$\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}:\mathcal{CP}(T^{o},\gamma)\to\mathcal{CP}(T^{\prime
o},\gamma)$ induces a partition bijection
$\varphi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}:\mathcal{P}(G_{T^{o},\gamma})\to\mathcal{P}(G_{T^{\prime
o},\gamma})$, i.e., we have the following commutative diagram.

$\textstyle{\mathcal{CP}(T^{o},\gamma)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}}$$\textstyle{\mathcal{CP}(T^{\prime
o},\gamma)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\mathcal{P}(G_{T^{o},\gamma})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}}$$\textstyle{\mathcal{P}(G_{T^{\prime
o},\gamma}).}$

We have the inverse partition bijection $\varphi^{T^{\prime
o}}_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}\mathcal{P}(G_{T^{\prime
o},\gamma})\to\mathcal{P}(G_{T^{o},\gamma})$ of
$\varphi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}$.

The following proposition generalizes Proposition 7.8, follows by Proposition
7.5.

######  Proposition 7.12.

With the foregoing notation. Let $P\in\mathcal{P}(G_{T^{o},\gamma})$.

  1. $(1)$

We have
$|\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(P)|=2^{\sum_{i=1}^{k}[\hat{m}(P,\alpha_{i})]_{+}}$,
moreover,

    1. (a)

if $\hat{m}(P,\alpha_{i})=1$ for some $i$, then there is a tile $G^{\prime}$
of $G_{T^{\prime o},\gamma}$ with diagonal labeled $\alpha^{\prime}_{i}$ such
that any $P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}$ can
twist on $G^{\prime}$ and $\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}$ is
closed under the twist on $G^{\prime}$;

    2. (b)

if $\hat{m}(P,\alpha_{i})=-1$ for some $i$, then there is a tile $G$ of
$G_{T^{o},\gamma}$ with diagonal labeled $\alpha_{i}$ such that $P$ can twist
on $G$ and
$\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(P)=\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(\mu_{G}P)$;

  2. $(2)$

For any $P^{\prime}\in\varphi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(P)$ and
$i\in\\{1,\cdots,k\\}$, we have
$\hat{m}(P,\alpha_{i})=-\hat{m}(P^{\prime},\alpha^{\prime}_{i}).$

  3. $(3)$

For any $P^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(P)$ and
$\zeta\in T^{o}$ with
$\zeta\neq(a_{i},b_{i}),(b_{i},c_{i}),(c_{i},d_{i}),(a_{i},d_{i}),(a_{i},c_{i})$
for all $i$, we have $\hat{m}(P,\zeta)=\hat{m}(P^{\prime},\zeta).$

  4. $(4)$

For any $P^{\prime}\in\phi^{T^{o}}_{\alpha_{1},\cdots,\alpha_{k}}(P)$ and
$\zeta\in\\{(a_{i},b_{i}),(b_{i},c_{i}),(c_{i},d_{i}),(a_{i},d_{i})\\}$, the
number of non-$\alpha_{i}$-mutable edges labeled $\zeta$ in $P$ equals the
number of non-$\alpha^{\prime}_{i}$-mutable edges labeled $\zeta$ in
$P^{\prime}$.

The following proposition follows by Proposition 7.9.

######  Proposition 7.13.

Assume $P$ can twist on a tile $G_{l}$ with diagonal labeled $\tau_{i_{l}}$
and $P>\mu_{G_{l}}P$.

  1. $(1)$

If
$\tau_{i_{l}}\neq(a_{i},b_{i}),(b_{i},c_{i}),(c_{i},d_{i}),(a_{i},d_{i}),\alpha_{i}$
for any $i$, then there is a tile $G^{\prime}_{[l]}$ of $G_{T^{\prime
o},\gamma}$ with diagonal labeled $\tau_{i_{l}}$ such that: for any
$P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(P)$, $P^{\prime}$
can twist on $G^{\prime}_{[l]}$,
$P^{\prime}>\mu_{G^{\prime}_{[l]}}(P^{\prime})$, and
$\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(\mu_{G_{l}}P)=\mu_{G^{\prime}_{[l]}}(\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(P)).$

  2. $(2)$

If $\tau_{i_{l}}=\alpha_{i}$ for some $i$, then
$\mu_{G_{l}}P\in\varphi_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}^{T^{\prime
o}}(P^{\prime})$ for any
$P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(P)$.

  3. $(3)$

If $\tau_{i_{l}}\in\\{(a_{i},b_{i}),(c_{i},d_{i})\\}$ for some $i$, then
$\hat{m}(P,\alpha_{i})=\hat{m}(\mu_{G_{l}}P,\alpha_{i})-1$. Moreover, there is
a tile $G^{\prime}_{[l]}$ of $G_{T^{\prime o},\gamma}$ with diagonal labeled
$\tau_{i_{l}}$ such that: any
$P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(P)$ can twist on
$G^{\prime}_{[l]}$ and
$P^{\prime}>\mu_{G^{\prime}_{[l]}}P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(\mu_{G_{l}}P)$.

  4. $(4)$

If $\tau_{i_{l}}\in\\{(b_{i},c_{i}),(a_{i},d_{i})\\}$ for some $i$, then
$\hat{m}(P,\alpha_{i})=\hat{m}(\mu_{G_{l}}P,\alpha_{i})+1$. Moreover, there is
a tile $G^{\prime}_{[l]}$ of $G_{T^{\prime o},\gamma}$ with diagonal labeled
$(a_{i},b_{i})$ such that: any
$P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(\mu_{G_{l}}P)$
can twist on $G^{\prime}_{[l]}$ and
$P^{\prime}<\mu_{G^{\prime}_{[l]}}P^{\prime}\in\varphi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o}}(\mu_{G_{l}}P)$.

###  7.2. General case

In this subsection, we assume that $\Sigma$ is an arbitrary orbifold without
orbifold points of weight $2$. Let $T^{o}$ be an ideal triangulation and
$\alpha\in T^{o}$ be a non-self-folded arc. Denote $T^{\prime
o}=\mu_{\alpha}(T^{o})$ and by $\alpha^{\prime}$ the new edge obtained. Let
$\gamma$ be an arc in $\Sigma$. Let $P_{n}$ be the canonical polygon of
$\gamma$ in $T^{o}$ with morphism $f:P_{n}\to\Sigma$ and triangle
$\hat{T}^{o}$. Assume that $f(\hat{\gamma})=\gamma$ for some diagonal
$\hat{\gamma}$ in $P_{n}$.

Assume that $\alpha$ is a diagonal of the quadrilateral
$(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})$ in $T^{o}$ and
$\alpha_{1},\alpha_{3}$ are in the clockwise direction of $\alpha$.

Denote by $\eta_{\alpha}$ the cardinality of $f^{-1}(\alpha)$. We may label
$f^{-1}(\alpha)$ as $\hat{\alpha}_{1},\cdots,\hat{\alpha}_{\eta_{\alpha}}$ in
order according to the orientation of $\hat{\gamma}$. We may further assume
that $\hat{\alpha}_{i}$ is not a boundary edge in $P_{n}$ for all
$i\in\\{1,\cdots,\eta_{\alpha}\\}$, otherwise we may extend $P_{n}$ to a
larger polygon. As $\alpha$ is not a folded arc, $\hat{\alpha}_{i}$ and
$\hat{\alpha}_{j}$ are not in a same triangle in $\hat{T}^{o}$ for $i\neq j$.
Denote $\hat{T}^{\prime
o}=\mu_{\hat{\alpha}_{\eta_{\alpha}}}\cdots\mu_{\hat{\alpha}_{1}}\hat{T}^{o}$.

####  7.2.1. Partition bijection $\phi^{T^{o}}_{\alpha}$

Using Theorem 7.6, as $\hat{\alpha}_{i}$ and $\hat{\alpha}_{j}$ are not in a
same triangle in $\hat{T}^{o}$ for $i\neq j$, we have a partition bijection

(21)
$\phi^{\hat{T}^{o}}_{\hat{\alpha}_{1},\cdots,\hat{\alpha}_{\eta_{\alpha}}}:\mathcal{CP}(\hat{T}^{o},\hat{\gamma})\to\mathcal{CP}(\hat{T}^{\prime
o},\hat{\gamma}).$

Since $f$ induces bijections between $\mathcal{CP}(\hat{T}^{o},\hat{\gamma})$
and $\mathcal{CP}(T^{o},\gamma)$, and between $\mathcal{CP}(\hat{T}^{\prime
o},\hat{\gamma})$ and $\mathcal{CP}(T^{\prime o},\gamma)$, we have
$\phi^{\hat{T}^{o}}_{\hat{\alpha}_{1},\cdots,\hat{\alpha}_{\eta_{\alpha}}}$
induces a partition bijection

(22)
$\phi^{T^{o}}_{\alpha}:\mathcal{CP}(T^{o},\gamma)\to\mathcal{CP}(T^{\prime
o},\gamma).$

That is, we have the following commutative diagram.

$\textstyle{\mathcal{CP}(\hat{T}^{o},\hat{\gamma})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\phi^{\hat{T}^{o}}_{\hat{\alpha}_{1},\cdots,\hat{\alpha}_{\eta_{\alpha}}}}$$\textstyle{\mathcal{CP}(\hat{T}^{\prime
o},\hat{\gamma})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\mathcal{CP}(T^{o},\gamma)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi^{T^{o}}_{\alpha}}$$\textstyle{\mathcal{CP}(T^{\prime
o},\gamma).}$

For any complete $(T^{o},\gamma)$-path
$\overrightarrow{\xi}=(\overrightarrow{\xi}_{1},\cdots,\overrightarrow{\xi}_{2c+1})$,
denote by
$\overrightarrow{\hat{\xi}}=(\overrightarrow{\hat{\xi}}_{1},\cdots,\overrightarrow{\hat{\xi}}_{2c+1})$
the corresponding complete $(\hat{T}^{o},\hat{\gamma})$-path under $f$. We may
associate $\overrightarrow{\xi}$ with an integer vector

(23) ${\bf
m}(\overrightarrow{\xi},\alpha)=(m(\overrightarrow{\hat{\xi}},\hat{\alpha}_{1}),\cdots,m(\overrightarrow{\hat{\xi}},\hat{\alpha}_{\eta_{\alpha}})),$

where each $m(\overrightarrow{\hat{\xi}},\hat{\alpha}_{i})$ is given by (2).

For any $i\in\\{1,\cdots,\eta_{\alpha}\\}$, denote by
$\hat{\alpha}^{\prime}_{i}$ the new arc in $P_{n}$ obtained from $\hat{T}^{o}$
by flip at $\hat{\alpha}_{i}$. For any complete $(T^{\prime o},\gamma)$-path
$\overrightarrow{\xi}^{\prime}=(\overrightarrow{\xi}^{\prime}_{1},\cdots,\overrightarrow{\xi}^{\prime}_{2c^{\prime}+1})$,
denote by
$\overrightarrow{\hat{\xi}^{\prime}}=(\overrightarrow{\hat{\xi}^{\prime}}_{1},\cdots,\overrightarrow{\hat{\xi}^{\prime}}_{2c^{\prime}+1})$
the corresponding complete
$(\mu_{\hat{\alpha}_{\eta_{\alpha}}}\cdots\mu_{\hat{\alpha}_{1}}\hat{T}^{o},\hat{\beta})$-path
under $f$. Similarly, we may associate $\overrightarrow{\xi}^{\prime}$ with an
integer vector

${\bf
m}(\overrightarrow{\xi}^{\prime},\alpha^{\prime})=(m(\overrightarrow{\hat{\xi}^{\prime}},\hat{\alpha}^{\prime}_{1}),\cdots,m(\overrightarrow{\hat{\xi}^{\prime}},\hat{\alpha}^{\prime}_{\eta_{\alpha}})).$

Similarly, we define the partition bijection

(24) $\phi^{T^{\prime o}}_{\alpha^{\prime}}:\mathcal{CP}(T^{\prime
o},\gamma)\to\mathcal{CP}(T^{o},\gamma),$

which is induced by the partition bijection $\phi^{\hat{T}^{\prime
o}}_{\hat{\alpha}^{\prime}_{1},\cdots,\hat{\alpha}^{\prime}_{\eta_{\alpha}}}$.

####  7.2.2. Partition bijection $\varphi^{T^{o}}_{\alpha}$

Since $\mathcal{CP}(T^{o},\gamma)\cong\mathcal{P}(G_{T^{o},\gamma})$ and
$\mathcal{CP}(T^{\prime o},\gamma)\cong\mathcal{P}(G_{T^{\prime o},\gamma})$,
we have the partition bijection
$\phi^{T^{o}}_{\alpha}:\mathcal{CP}(T^{o},\gamma)\to\mathcal{CP}(T^{\prime
o},\gamma)$ in (22) induces partition bijections
$\varphi^{T^{o}}_{\alpha}:\mathcal{P}(G_{T^{o},\gamma})\to\mathcal{P}(G_{T^{\prime
o},\gamma})$. Similarly, we have the inverse partition bijection
$\varphi^{T^{\prime o}}_{\alpha^{\prime}}:\mathcal{P}(G_{T^{\prime
o},\gamma})\to\mathcal{P}(G_{T^{o},\gamma})$.

In this subsection, we study the properties of $\varphi^{T^{o}}_{\alpha}$ and
$\varphi^{T^{\prime o}}_{\alpha^{\prime}}$.

For any $P\in\mathcal{P}(G_{T^{o},\gamma})$, denote by
$\overrightarrow{\xi}(P)$ the corresponding complete $(T^{o},\gamma)$-path and
$\hat{P}$ the corresponding perfect matching of
$G_{\hat{T}^{o},\hat{\gamma}}$, let

${\bf m}(P,\alpha)={\bf m}(\overrightarrow{\xi}(P),\alpha),$

where ${\bf m}(\overrightarrow{\xi}(P),\alpha)$ is given by (23).

Assume that ${\bf m}(P,\alpha)=(m_{1}(P),\cdots,m_{\eta_{\alpha}}(P))$. In
view of (9) and (10), we have

$\sum_{i=1}^{\eta(\alpha)}m_{i}(P)=m(P;\alpha)-n(G_{T^{o},\gamma};\alpha).$

Similarly we can define ${\bf m}(P^{\prime};\alpha^{\prime})$ for any
$P^{\prime}\in\mathcal{P}(G_{T^{\prime o},\alpha^{\prime}})$.

Among
$f^{-1}(\alpha)=\\{\hat{\alpha}_{1},\cdots,\hat{\alpha}_{\eta_{\alpha}}\\}$,
assume that $\hat{\gamma}$ crosses
$\hat{\alpha}_{i_{1}},\cdots,\hat{\alpha}_{i_{l}}$ sequentially, among
$\\{\hat{\alpha}^{\prime}_{1},\cdots,\hat{\alpha}^{\prime}_{\eta_{\alpha}}\\}$,
assume that $\hat{\gamma}$ crosses
$\hat{\alpha}^{\prime}_{j_{1}},\cdots,\hat{\alpha}^{\prime}_{j_{l^{\prime}}}$
sequentially.

The following two Propositions follows by Proposition 7.12 (1) (2).

######  Proposition 7.14.

Assume that ${\bf m}(P,\alpha)=(m_{1}(P),\cdots,m_{\eta_{\alpha}}(P))$.

  1. $(1)$

If $m_{k}(P)=-1$ for some $k\in\\{1,\cdots,\eta_{\alpha}\\}$, then there
exists a tile $G(k)$ of $G_{T^{o},\gamma}$ with diagonal labeled $\alpha$ such
that $P$ can twist on the tile $G(k)$.

  2. $(2)$

If $m_{k}(P)=1$ for some $k\in\\{1,\cdots,\eta_{\alpha}\\}$, then there exists
a tile $G^{\prime}(k)$ of $G_{T^{\prime o},\gamma}$ with diagonal labeled
$\alpha^{\prime}$ such that any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can
twist on $G^{\prime}(k)$.

######  Proposition 7.15.

For any $P\in\mathcal{P}(G_{T^{o},\gamma})$ with ${\bf
m}(P,\alpha)=(m_{1}(P),\cdots,m_{\eta_{\alpha}}(P))$, we have

$|\varphi^{T^{o}}_{\alpha}(P)|=2^{\sum_{i=1}^{\eta_{\alpha}}[m_{i}(P)]_{+}},$

more precisely, for any $P^{\prime}\in\varphi^{T^{o}}_{\alpha}(P)$,

  1. $(1)$

we have ${\bf m}(P^{\prime},\alpha^{\prime})=-{\bf m}(P,\alpha)$,

  2. $(2)$

for any $k$ such that $m_{k}(P)=1$, $P^{\prime}$ can twist on the tiles
$G^{\prime}(k)$ and
$\mu_{G^{\prime}(k)}P^{\prime}\in\varphi^{T^{o}}_{\alpha}(P)$, where
$G^{\prime}(k)$ are the tiles given in Proposition 7.14.

######  Lemma 7.16.

For any $P\in\mathcal{\mathcal{missing}}P(G_{T^{o},\gamma})$ and
$P^{\prime}\in\varphi^{T^{o}}_{\alpha}(P)$, for any $(\alpha\neq)\tau\in
T^{o}$,

  1. $(1)$

if $\tau\neq\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, then we have
$\hat{m}(P;\tau)=\hat{m}(P^{\prime};\tau)$;

  2. $(2)$

if $\tau\in\\{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\\}$, then the number
of edges in $P$ labeled $\tau$ which are not $\alpha$-mutable equals the
number of edges in $P^{\prime}$ labeled $\tau$ which are not
$\alpha^{\prime}$-mutable.

###### Proof.

It follows by Proposition 7.12 (3) (4). $\square$

The following Proposition follows by Propositions 7.12 and 7.13.

######  Proposition 7.17.

Assume that $P$ can twist on a tile $G_{l}$ with diagonal labeled
$\tau_{i_{l}}$ and $P>\mu_{G_{l}}P$. Assume that ${\bf
m}(P,\alpha)=(m_{1}(P),\cdots,m_{\eta_{\alpha}}(P))$,

  1. $(1)$

if $\tau_{i_{l}}\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, then
there is a tile $G^{\prime}_{[l]}$ of $G_{T^{\prime o},\gamma}$ with diagonal
labeled $\tau_{i_{l}}$ such that any
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can twist on $G^{\prime}_{[l]}$,
$P^{\prime}>\mu_{G^{\prime}_{[l]}}(P^{\prime})$, and
$\varphi_{\alpha}^{T^{o}}(\mu_{G_{l}}P)=\mu_{G^{\prime}_{[l]}}(\varphi_{\alpha}^{T^{o}}(P)),$
moreover, the following equalities hold:
(a)
$m^{\pm}(P,G_{l};\tau_{i_{l}})=m^{\pm}(P^{\prime},G^{\prime}_{[l]};\tau_{i_{l}})$;
(b) $n(G_{l}^{\pm};\tau_{i_{l}})=n(G^{\prime\pm}_{[l]};\tau_{i_{l}})$.
In fact $G^{\prime}_{[l]}$ corresponds to the same crossing point of $\gamma$
and $\tau_{i_{l}}$ as $G_{l}$.

  2. $(2)$

if $\tau_{i_{l}}=\alpha$, then $m_{k}(P)=-1$ and $G_{l}=G(k)$ for some $k$,
where $G(k)$ is the tile given in Proposition 7.14, moreover,
$\mu_{G_{l}}P\in\varphi_{\alpha^{\prime}}^{T^{\prime o}}(P^{\prime})$ for any
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$.

  3. $(3)$

if $\tau_{i_{l}}\in\\{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\\}$ and
$\tau_{i_{l}}$ is not the radius of any self-folded triangle in $T^{o}$ or
$T^{\prime o}$, without loss of generality, we may assume
$\tau_{i_{l}}=\alpha_{1}$, then there exists a tile $G^{\prime}_{[l]}$ of
$G_{T^{\prime o},\gamma}$ with diagonal labeled $\alpha_{1}$ such that any
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(\mu_{G_{l}}P)$ can twist on
$G^{\prime}_{[l]}$ and
$P^{\prime}>\mu_{G^{\prime}_{[l]}}P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$.
Moreover, if the $\alpha^{\prime}$-mutable edges in $P^{\prime}$ are labeled
$\alpha_{2},\alpha_{4}$ then
(a)
$m^{\pm}(P,G_{l};\alpha_{1})=m^{\pm}(P^{\prime},G^{\prime}_{[l]};\alpha_{1})$,
(b) $n(G_{l}^{\pm};\alpha_{1})=n(G^{\prime\pm}_{[l]};\alpha_{1})$.

######  Proposition 7.18.

Assume that $\alpha_{1}=\alpha_{2}$ and the diagonal of $G_{l}$ is labeled
$\alpha_{1}$. We further assume that $rel(G_{l},T^{o})=1$ and
$N(G_{l}),W(G_{l})$ are labeled $\alpha$. If $P$ can twist on $G_{l}$ and
$P>Q:=\mu_{G_{l}}P$, denote ${\bf
m}(P,\alpha)=(m_{1}(P),\cdots,m_{\eta_{\alpha}}(P)),{\bf
m}(Q,\alpha)=(m_{1}(Q),\cdots,m_{\eta_{\alpha}}(Q))$, then
$m_{k}(P)=m_{k}(Q)-1,m_{k+1}(P)=m_{k+1}(Q)+1$ for some $k$ and
$m_{i}(P)=m_{i}(Q)$ for any $i\neq k,k+1$. Consequently, we have
$(m_{k}(P),m_{k+1}(P))=(-1,1)$, $(0,0)$ or $(0,1)$.

  1. $(1)$

In case $(m_{k}(P),m_{k+1}(P))=(-1,1)$, we have
$(m_{k}(\mu_{G_{l}}P),m_{k+1}(\mu_{G_{l}}P))=(0,0)$. Assume the tile left to
$G^{\prime}(k+1)$ is $G^{\prime}$. Then the diagonal of $G^{\prime}$ is
labeled $\alpha_{1}$ and any $Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(Q)$ can
twist on $G^{\prime}$, see Figure 12. Moreover,
(a)
$\mu_{G^{\prime}}Q^{\prime},\mu_{G^{\prime}(k+1)}\mu_{G^{\prime}}Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$,
(b)
$Q^{\prime}<\mu_{G^{\prime}}Q^{\prime}<\mu_{G^{\prime}(k+1)}\mu_{G^{\prime}}Q^{\prime}$,
(c),
(d) $\Omega(Q^{\prime};G^{\prime})=\Omega(Q;G_{l})+\Omega(P;G_{l+1})$.
In particular,
$\varphi_{\alpha}^{T^{o}}(P)=\\{\mu_{G^{\prime}}Q^{\prime},\mu_{G^{\prime}(k+1)}\mu_{G^{\prime}}Q^{\prime}\mid
Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(Q)\\}$.

Figure 12.

  2. $(2)$

In case $(m_{k}(P),m_{k+1}(P))=(0,0)$, we have
$(m_{k}(Q),m_{k+1}(\mu_{G_{l}}P))=(1,-1)$. Assume the tile right to
$G^{\prime}(k)$ is $G^{\prime}$. Then the diagonal of $G^{\prime}$ is labeled
$\alpha_{1}$ and any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can twist on
$G^{\prime}$, see Figure 13. Moreover,
(a)
$\mu_{G^{\prime}}P^{\prime},\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime}\in\varphi_{\alpha}^{T^{o}}(Q)$,
(b)
$P^{\prime}>\mu_{G^{\prime}}P^{\prime}>\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime}$,
(c)
$\Omega(Q;G_{l})=\Omega(\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime};G^{\prime}(k))+\Omega(\mu_{G^{\prime}}P^{\prime};G^{\prime})$,
In particular,
$\varphi_{\alpha}^{T^{o}}(Q)=\\{\mu_{G^{\prime}}P^{\prime},\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime}\mid
P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)\\}$.

Figure 13.

  3. $(3)$

In case $(m_{k}(P),m_{k+1}(P))=(0,1)$, we have $(m_{k}(Q),m_{k+1}(Q))=(1,0)$.
There is a unique tile $G^{\prime}$ between $G^{\prime}(k)$ and
$G^{\prime}(k+1)$ and its diagonal is labeled $\alpha_{1}$. Assume that
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ with
$P^{\prime}<\mu_{G^{\prime}(k+1)}P^{\prime}$. Then $P^{\prime}$ can twist on
$G^{\prime}$, see Figure 14. Moreover,
(a)
$\mu_{G^{\prime}}P^{\prime},\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime}\in\varphi_{\alpha}^{T^{o}}(Q)$,
(b)
$\mu_{G^{\prime}(k+1)}P^{\prime}>P^{\prime}>\mu_{G^{\prime}}P^{\prime}>\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime}$,
(c)
$\Omega(Q;G_{l})=\Omega(\mu_{G^{\prime}(k)}\mu_{G^{\prime}}P^{\prime};G^{\prime}(k))+\Omega(\mu_{G^{\prime}}P^{\prime};G^{\prime})$.

Figure 14.

  4. $(4)$

In all cases, for any $Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(Q)$ we have

$m(Q;\alpha_{1})-n(G_{T^{o},\widetilde{\beta}};\alpha_{1})=m(Q^{\prime};\alpha_{1})-n(G_{T^{\prime
o},\widetilde{\beta}};\alpha_{1})+m(Q^{\prime};\alpha^{\prime})-n(G_{T^{\prime
o},\widetilde{\beta}};\alpha^{\prime}).$

###### Proof.

Since $P>Q$, we have $N(G_{l})\in P,W(G_{l})\in Q$. Thus
$m_{k}(P)=m_{k}(Q)-1,m_{k+1}(P)=m_{k+1}(Q)+1$ for some $k$ and
$m_{i}(P)=m_{i}(Q)$ for any $i\neq k,k+1$. It follows that
$m_{k}(P)\in\\{-1,0\\}$ and $m_{k+1}(P)\in\\{0,1\\}$. Since
$\widetilde{\beta}$ can not cross $\alpha,\alpha_{1},\alpha$ consecutively, we
have $(m_{k}(P),m_{k+1}(P))\neq(-1,0)$. Therefore,
$(m_{k}(P),m_{k+1}(P))=(-1,1)$, $(0,0)$ or $(0,1)$.

We only consider the case that $(m_{k}(P),m_{k+1}(P))=(-1,1)$, as the other
cases can be proved similarly. We have the diagonal of $G_{l-1}$ is labeled
$\alpha$ and $S(G_{l-1})\in P\cap Q$. We may assume
$rel(G^{\prime}(k+1),T^{\prime o})=1$. As $E(G_{l})$ is labeled $\alpha_{4}$,
from the construction of snake groups, we have $E(G^{\prime}(k+1))$ is labeled
$\alpha_{4}$ and the diagonal of $G^{\prime}$ is labeled $\alpha_{1}$. For any
$Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(Q)$, since $m_{k}(Q)=m_{k}(P)+1=0$, we
have $m_{k}(Q^{\prime})=0$ by Proposition 7.15. Thus we have $N(G^{\prime})\in
Q^{\prime}$. As $S(G_{l-1})$ is labeled $\alpha_{1}$ and $S(G_{l-1})\in Q$, we
have $S(G^{\prime})\in Q^{\prime}$. Thus $Q^{\prime}$ can twist on
$G^{\prime}$. For any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$, since
$m_{k}(P)=-1$, we have $m_{k}(P^{\prime})=1$ by Proposition 7.15. Thus we have
$W(G^{\prime})\in P^{\prime}$. Therefore,
$\mu_{G^{\prime}}Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$. By Proposition
7.14 (2), $\mu_{G^{\prime}}Q^{\prime}$ can twist on $G^{\prime}(k+1)$ and
$\mu_{G^{\prime}(k+1)}\mu_{G^{\prime}}Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$.
$Q^{\prime}<\mu_{G^{\prime}}Q^{\prime}<\mu_{G^{\prime}(k+1)}\mu_{G^{\prime}}Q^{\prime}$
is immediate. By Proposition 7.12 (3) (4), we have

$m^{\pm}(Q,G_{l};\alpha_{1})=m^{\pm}(Q^{\prime},G^{\prime};\alpha_{1})+m^{\pm}(\mu_{G^{\prime}}Q^{\prime},G^{\prime}(k+1);\alpha^{\prime})-n((G^{\prime}(k+1))^{\pm};\alpha^{\prime}),$
$m^{+}(Q^{\prime},G^{\prime};\alpha_{1})=m^{+}(Q,G_{l};\alpha_{1})+m^{+}(P,G_{l-1};\alpha)-n(G_{l-1}^{+};\alpha)-1,$
$m^{-}(Q^{\prime},G^{\prime};\alpha_{1})=m^{-}(Q,G_{l};\alpha_{1})+m^{-}(P,G_{l-1};\alpha)-n(G_{l-1}^{-};\alpha)-1.$

We have $n(G_{l}^{\pm};\alpha_{1})=n(G^{\prime\pm};\alpha_{1})$. Therefore, we
have

$\Omega(Q;G_{l})=\Omega(Q^{\prime};G^{\prime})+\Omega(\mu_{G^{\prime}}Q^{\prime};G^{\prime}(k+1)),\;\;\;\Omega(Q^{\prime};G^{\prime})=\Omega(Q;G_{l})+\Omega(P;G_{l+1})$

and

$m(Q;\alpha_{1})-n(G_{T^{o},\widetilde{\beta}};\alpha_{1})=m(Q^{\prime};\alpha_{1})-n(G_{T^{\prime
o},\widetilde{\beta}};\alpha_{1})+m(Q^{\prime};\alpha^{\prime})-n(G_{T^{\prime
o},\widetilde{\beta}};\alpha^{\prime}).$

$\square$

######  Lemma 7.19.

We have $P^{\prime}_{-}\in\varphi^{T^{o}}_{\alpha}(P_{-})$.

###### Proof.

Assume $P^{\prime}$ is the minimum element in
$\varphi^{T^{o}}_{\alpha}(P_{-})$. For any tile $G^{\prime}$ that $P^{\prime}$
can twist on, as $P_{-}$ is the minimum perfect matching, we have
$P^{\prime}<\mu_{G^{\prime}}P^{\prime}$ by Proposition 7.17 and Proposition
7.18. It follows that $P^{\prime}=P^{\prime}_{-}$. $\square$

The following result follows by Lemma 7.11.

######  Lemma 7.20.

For $P_{-}$, we have $m_{i}(P_{-})m_{j}(P_{-})\geq 0$ for any
$i,j\in\\{1,\cdots,\eta_{\alpha}\\}$.

###### Proof.

Otherwise, $m_{i}(P_{-})m_{j}(P_{-})<0$ for some
$i,j\in\\{1,\cdots,\eta_{\alpha}\\}$. We see that $\gamma$ intersects itself
by Lemma 7.11, a contradiction. $\square$

##  8\. Partition bijection between triangles

With the notation in Section 4.2. Let $q$ be a puncture. Let $\Delta(T^{o},q)$
(resp. $\Delta(T^{\prime o},q)$) be the set of triangles incident to $q$ in
$T^{o}$ (resp. $T^{\prime o}=\mu_{\alpha}(T^{o})$). In this section, we give a
partition bijection $\phi_{\alpha}^{T^{o}}:\Delta(T^{o},q)\to\Delta(T^{\prime
o},q)$.

###  8.1. Once-puncture polygon case

We first consider the case that $\Sigma$ is the once-punctured polygon
$P_{n,1}$, the disk with $n$ marked points labeled $1,\cdots,n$ clockwise and
one puncture labeled $0$. Let $T^{o}$ be a triangulation of $P_{n,1}$. Assume
that $\Delta(T^{o},0)=\\{\Delta_{1},\cdots,\Delta_{t}\\}$ and for any
$i\in\\{1,\cdots,t\\}$ the three edges of $\Delta_{i}$ are labeled
$\tau_{i^{-}},\tau_{i},\tau_{[i]}$, where we denote

$\begin{array}[]{ccl}i^{+}&=&\left\\{\begin{array}[]{ll}i+1,&\mbox{if $1\leq
i\leq t-1$},\\\ 1,&\mbox{if
$i=t$},\end{array}\right.\end{array},\begin{array}[]{ccl}i^{-}&=&\left\\{\begin{array}[]{ll}i-1,&\mbox{if
$2\leq i\leq t$},\\\ t,&\mbox{if $i=1$}.\end{array}\right.\end{array}$

Fix an arc $\alpha\in T^{o}$, we have one of the following cases happens: (1)
$\alpha=\tau_{i}$ for some $i$; (2) $\alpha=\tau_{[i]}$ for some $i$; (3) the
rest cases.

For any $\Delta_{j}\in\Delta(T^{o},0)$, we associate with a subset
$\phi_{\alpha}^{T^{o},q}(\Delta_{j})\subset\Delta(T^{\prime o},0)$ as follows.

Case I: $\alpha=\tau_{i}$ for some $i$. Then $\Delta(T^{\prime
o},0)=\\{(\tau_{i^{-}},\tau_{i^{+}},\alpha^{\prime}),\Delta_{j}\mid j\neq
i,i^{+}\\}$. In this case, let

$\begin{array}[]{ccl}\phi_{\alpha}^{T^{o},q}(\Delta_{j})&=&\left\\{\begin{array}[]{ll}(\tau_{i^{-}},\tau_{i^{+}},\alpha^{\prime}),&\mbox{if
$j=i$ or $i^{+}$},\\\ \Delta_{j},&\mbox{if $j\neq
i,i^{+}$},\end{array}\right.\end{array}$

Case II: $\alpha=\tau_{[i]}$ for some $i$. Then $\tau_{i-1},\alpha^{\prime}$
are the two sides of some triangle $\hat{\Delta}_{1}$ in $T^{\prime o}$ and
$\tau_{i},\alpha^{\prime}$ are the two sides of some triangle
$\hat{\Delta}_{2}$ in $T^{\prime o}$. We have $\Delta(T^{\prime
o},0)=\\{\hat{\Delta}_{1},\hat{\Delta}_{2},\Delta_{j}\mid j\neq i\\}$.

In this case, let

$\begin{array}[]{ccl}\phi_{\alpha}^{T^{o},q}(\Delta_{j})&=&\left\\{\begin{array}[]{ll}\\{\hat{\Delta}_{1},\hat{\Delta}_{2}\\},&\mbox{if
$j=i$},\\\ \Delta_{j},&\mbox{if $j\neq i$},\end{array}\right.\end{array}$

Case III: the rest cases. In this case we have $\Delta(T^{\prime
o},0)=\Delta(T^{o},0)$. In this case let
$\phi_{\alpha}^{T^{o},q}(\Delta_{j})=\Delta_{j}$ for all
$j\in\\{1,\cdots,s\\}$.

Similarly we can associate a subset $\phi_{\alpha}^{T^{\prime
o},q}(\Delta^{\prime})$ of $\Delta(T^{o},0)$ for any
$\Delta^{\prime}\in\Delta(T^{\prime o},0)$.

Assume that $\Delta(T^{\prime
o},0)=\\{\Delta^{\prime}_{1},\cdots,\Delta^{\prime}_{s^{\prime}}\\}$ and for
any $i\in\\{1,\cdots,s^{\prime}\\}$ the three edges of $\Delta^{\prime}_{i}$
are labeled $\tau^{\prime}_{i^{-}},\tau^{\prime}_{i},\tau^{\prime}_{[i]}$.

######  Proposition 8.1.

With the foregoing notation. $\phi_{\alpha}^{T^{o},q}$ is a partition
bijection from $\Delta(T^{o},0)$ to $\Delta(T^{\prime o},0)$ with inverse
$\phi_{\alpha}^{T^{\prime o},q}$. For any $\Delta_{i}\in\Delta(T^{o},0)$ we
have

$|\phi_{\alpha}^{T^{o},q}(\Delta_{i})|=2^{[m(\Delta_{i};\alpha)]_{+}},$

where $m(\Delta_{i};\alpha)$ is given by (11), moreover, if
$m(\Delta_{i};\alpha)=1$, then $\alpha=\tau_{[i]}$ and
$\phi_{\alpha}^{T^{o},q}(\Delta_{i})=\\{\Delta^{\prime}_{j},\Delta^{\prime}_{j+1}\\}$
for some $j$ such that $\tau^{\prime}_{j}=\alpha^{\prime}$.

###### Proof.

The result follows easily by checking all the above three cases. $\square$

The following result is clear.

######  Lemma 8.2.

For any $\Delta_{i}\in\Delta(T^{o},0)$ and
$\Delta^{\prime}\in\phi_{\alpha}^{T^{o},q}(\Delta_{i})$, we have
$m(\Delta_{i};\alpha)=-m(\Delta^{\prime};\alpha^{\prime})$.

Next, we generalize the above construction for a sequence of non-self-folded
arcs $\alpha_{1},\cdots,\alpha_{k}\in T^{o}$. Let
$\alpha_{1},\cdots,\alpha_{k}\in T^{o}$ be a sequence of arcs such that
$\alpha_{i},\alpha_{j}$ are not two sides of any triangulation in $T^{o}$ for
any $i\neq j$. Denote by $\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}$ the
new arcs obtained from $T^{o}$ by flip at $\alpha_{1},\cdots,\alpha_{k}$,
respectively. For any triangle $\Delta_{i}\in\Delta(T^{o},0)$, recall that the
three sides of $\Delta_{i}$ are $\tau_{i-1},\tau_{i}$ and $\tau_{[i]}$, one
and exactly one of the following cases happens:

  1. (i)

Case I: $\tau_{i-1}=\alpha_{a}$ for some $a\in\\{1,\cdots,k\\}$;

  2. (ii)

Case II: $\tau_{i}=\alpha_{a}$ for some $a\in\\{1,\cdots,k\\}$;

  3. (iii)

Case III: $\tau_{[i]}=\alpha_{a}$ for some $a\in\\{1,\cdots,k\\}$;

  4. (iv)

Case IV: $\tau_{i-1},\tau_{i},\tau_{[i]}\neq\alpha_{a}$ for any
$a\in\\{1,\cdots,k\\}$.

We associate a subset
$\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})\subset\Delta(\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o},0)$
as follows:

  1. (i)

In case I, we have $\tau_{i-2},\tau_{i},\alpha^{\prime}_{a}$ form a triangle
in $\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o}$, let

$\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})=(\tau_{i-2},\tau_{i},\alpha^{\prime}_{a});$

  2. (ii)

In case II, we have $\tau_{i-1},\tau_{i+1},\alpha^{\prime}_{a}$ form a
triangle in $\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o}$, let

$\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})=(\tau_{i-1},\tau_{i+1},\alpha^{\prime}_{a});$

  3. (iii)

In case III, we have $\tau_{i-1},\alpha^{\prime}$ are the two sides of some
triangle $\hat{\Delta}_{1}$ in $\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o}$
and $\tau_{i},\alpha^{\prime}$ are the two sides of some triangle
$\hat{\Delta}_{2}$ in $\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o}$, let

$\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})=\\{\hat{\Delta}_{1},\hat{\Delta}_{2}\\};$

  4. (iv)

In case IV, let

$\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})=\Delta_{i}.$

Similarly, we can associate a subset
$\phi_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}^{\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o},q}(\Delta^{\prime}_{i})\subset\Delta(T^{o},0)$
for any
$\Delta^{\prime}_{i}\in\Delta(\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o},0)$.

######  Proposition 8.3.

With the foregoing notation. We have
$\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}$ and
$\phi_{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{k}}^{\mu_{\alpha_{k}}\cdots\mu_{\alpha_{1}}T^{o},q}$
are partition bijections inverse to each other, moreover, for any
$\Delta_{i}\in\Delta(0)$ we have

$|\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})|=2^{\sum_{j=1}^{k}[m(\Delta_{i};\alpha_{j})]_{+}},$

where $m(\Delta_{i};\alpha_{j})$ is given by (11), moreover, if
$m(\Delta_{i};\alpha_{j})=1$ for some $j$, then $\alpha_{j}=\tau_{[i]}$ and
$\phi_{\alpha}^{T^{o},q}(\Delta_{i})=\\{\Delta^{\prime}_{j^{\prime}},\Delta^{\prime}_{j^{\prime}+1}\\}$
for some $j^{\prime}$ such that
$\tau^{\prime}_{j^{\prime}}=\alpha^{\prime}_{i}$.

###### Proof.

The result follows easily by checking all the above three cases. $\square$

Note that $\sum_{j=1}^{k}[m(\Delta_{i}(q);\alpha_{j})]_{+}=1,0$ or $-1$.

The following result follows immediately by Lemma 8.2.

######  Lemma 8.4.

For any $\Delta_{i}\in\Delta(T^{o},0)$ and
$\Delta^{\prime}\in\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i})$,
we have $m(\Delta_{i};\alpha_{j})=-m(\Delta^{\prime};\alpha^{\prime}_{j})$ for
any $j\in\\{1,\cdots,k\\}$.

###  8.2. General case

The following is a generalization of “canonical polygon” in [2] to “canonical
once - polygon”.

######  Theorem 8.5.

Let $T^{o}$ be an ideal triangulation of $\Sigma$. For any puncture $q$ of
$\Delta$, assume that
$\Delta(T^{o},q)=\\{\Delta_{1}(q),\cdots,\Delta_{s}(q)\\}$ with the three
sides of $\Delta_{i}(q)$ are $\tau_{i-1},\tau_{i}$ and $\tau_{[i]}$. Then
there is a morphism $\pi$ from $P_{s,1}$ to $\Sigma$ such that
$\pi(0i)=\tau_{i}$ for all $i=1,\cdots,s$ and $\pi(i,i^{+})=\tau_{[i^{+}]}$.

We call $P_{s,1}$ together with the morphism $\pi$ the _canonical-once
punctured polygon_ for $q$ concerning $T^{o}$. It is clear that
$\Delta_{i}(q)$ corresponds to a unique triangle $\Delta_{i}$ incident to $0$
in $P_{s,1}$ for any $i$.

######  Example 8.6.

In the twice punctured digon with triangulation $T^{o}$, as shown in the right
below figure. We have $\alpha_{2}=\alpha_{4}$, $\alpha_{[2]}=\alpha_{5}$,
$\alpha_{[3]}=\alpha_{3}=\alpha_{[4]}$ and $\alpha_{[5]}=\alpha_{1}$. Then the
canonical once-punctured polygon for $0$ is shown in the left below figure.

Figure 15. Canonical once-punctured polygon for puncture $0$

For any non-self-folded arc $\alpha\in T^{o}$, let $T^{\prime
o}=\mu_{\alpha}(T^{o})$, we now construct partition bijection
$\phi_{\alpha}^{T^{o},q}$ from $\Delta(T^{o},q)$ to $\Delta(T^{\prime o},0)$.

In case $\alpha\notin\\{\alpha_{i},\alpha_{[i]}\mid i=1,\cdots,s\\}$, we have
$\Delta(T^{\prime o},0)=\Delta(T^{o},0)$. Then let
$\phi_{\alpha}^{T^{o},q}(\Delta_{i})=\Delta_{i}$ for any $i=1,\cdots,s$.

In case $\alpha\in\\{\alpha_{i},\alpha_{[i]}\mid i=1,\cdots,s\\}$, we can
extend the canonical once punctured polygon $(P_{s,1},\pi)$ to a bigger once
punctured polygon $P_{n,1}$ for some $n>s$ with an ideal triangulation
$\hat{T}^{o}$ and a morphism $\pi^{\prime}:P_{n,1}\to\Sigma$ such that (1)
$\pi^{\prime}|_{P_{s,1}}=\pi$, (2) the preimages of $\alpha$ are inner arcs,
and (3) $\pi^{\prime}(\Delta)$ is a triangle in $T^{o}$ for any triangle
$\Delta$ in $\hat{T}^{o}$.

Assume that
$\pi^{-1}(\alpha)=\\{\alpha_{1},\cdots,\alpha_{k}\\}\subset\hat{T}^{o}$. It is
easy to see that $\alpha_{i},\alpha_{j}$ are not two sides of any
triangulation in $\hat{T}^{o}$ for all $i\neq j$ from the construction of the
canonical once punctured polygon.

For any $\Delta_{i}(q)\in\Delta(T^{o},q)$, let
$\phi_{\alpha}^{T^{o},q}(\Delta_{i}(q))=\pi^{\prime}(\phi_{\alpha_{1},\cdots,\alpha_{k}}^{T^{o},q}(\Delta_{i}))$,
where $\Delta_{i}$ is the triangle incident to $0$ in the canonical once
punctured polygon. Similarly, we can define $\phi_{\alpha}^{T^{\prime o},q}$.

######  Proposition 8.7.

With the foregoing notation. $\phi_{\alpha}^{T^{o},q}$ is a partition
bijection from $\Delta(T^{o},q)$ to $\Delta(T^{\prime o},q)$ with inverse
$\phi_{\alpha}^{T^{\prime o},q}$, moreover, for any
$\Delta_{i}(q)\in\Delta(T^{o},q)$ we have

$|\phi_{\alpha}^{T^{o},q}(\Delta_{i}(q))|=2^{[m(\Delta_{i}(q);\alpha)]_{+}},$

where $m(\Delta_{i}(q);\alpha)$ is given by (11).

###### Proof.

It follows immediately by Proposition 8.3. $\square$

The following lemma is clear.

######  Lemma 8.8.

We have $\Delta_{1}^{\prime}(q)\in\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))$,
where $\Delta^{\prime}_{1}(q)$ is the first triangle incident to $q$ in
$T^{\prime o}$.

By Lemma 8.4, we have

######  Lemma 8.9.

For any $\Delta\in\Delta(T^{o},q)$ and
$\Delta^{\prime}\in\phi_{\alpha}^{T^{o},q}(\Delta)$, we have
$m(\Delta;\alpha)=-m(\Delta^{\prime};\alpha^{\prime})$.

######  Lemma 8.10.

For any $\Delta\in\Delta(T^{o},q)$ and
$\Delta^{\prime}\in\phi_{\alpha}^{T^{o},q}(\Delta)$, for any $\tau\in T^{o}$,

  1. $(1)$

if $\tau\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, then
$m(\Delta;\tau)=m(\Delta^{\prime};\tau)$;

  2. $(2)$

if $\tau\in\\{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\\}$ and
$m(\Delta;\alpha)=0$, then $m(\Delta;\tau)=m(\Delta^{\prime};\tau)$.

## Notation

We fix some notation throughout the rest of this paper. Fix an arc
$\widetilde{\beta}$, an ideal triangulation $T^{o}$ and a non-self-folded
$\alpha\in T^{o}$, let $T^{\prime o}=\mu_{\alpha}T^{o}$ and $\alpha^{\prime}$
be the new arc obtained. Suppose that $\alpha$ is a diagonal of the
quadrilateral $(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})$ in $T^{o}$ and
$\alpha_{1},\alpha_{3}$ are in the clockwise direction of $\alpha$. As
$\alpha$ is not folded in $T^{o}$,
$\\{\alpha_{1},\alpha_{3}\\}\neq\\{\alpha_{2},\alpha_{4}\\}$.

Denote by $G_{T^{o},\widetilde{\beta}}$ (resp. $G_{T^{\prime
o},\widetilde{\beta}}$) the associated snake graph with tiles
$G_{1},\cdots,G_{c}$ (resp. $G^{\prime}_{1},\cdots,G^{\prime}_{c^{\prime}}$)
in order and $\mathcal{P}(G_{T^{o},\widetilde{\beta}})$ (resp.
$\mathcal{P}(G_{T^{\prime o},\widetilde{\beta}})$) the set of all perfect
matching. Denote by $P_{\pm}$ (resp. $P^{\prime}_{\pm}$) the maximal/minimal
perfect matching. Denote by $\tau_{i_{l}}$ (resp. $\tau^{\prime}_{i_{l}}$) the
label of the diagonal of $G_{l}$ (resp. $G^{\prime}_{l}$).

If $t(\widetilde{\beta})=q$ is a puncture, denote by
$\Delta_{1}(q),\Delta_{2}(q),\cdots,\Delta_{t}(q)$ (resp.
$\Delta^{\prime}_{1}(q),\Delta^{\prime}_{2}(q),\cdots,\Delta^{\prime}_{t^{\prime}}(q)$)
the triangles incident to $q$ in $T^{o}$ (resp. $T^{\prime o}$) in clockwise
order such that $\widetilde{\beta}$ crosses $\Delta_{1}(q)$ (resp.
$\Delta^{\prime}_{1}(q)$) or $\widetilde{\beta}$ is the common side of
$\Delta_{1}(q)$ and $\Delta_{t}(q)$ (resp. $\Delta^{\prime}_{1}(q)$ and
$\Delta^{\prime}_{t^{\prime}}(q)$). Denote the common side of $\Delta_{i}(q)$
and $\Delta_{i+1}(q)$ (resp. $\Delta^{\prime}_{i}(q)$ and
$\Delta^{\prime}_{i+1}(q)$) by $\tau_{i}(q)$ (resp. $\tau^{\prime}_{i}(q)$).
Denote the third side of $\Delta_{i}(q)$ (resp. $\Delta^{\prime}_{i}(q)$) by
$\tau_{[i]}(q)$ (resp. $\tau^{\prime}_{[i]}(q)$).

If $s(\widetilde{\beta})=p$ is a puncture, denote by
$\Delta_{1}(p),\Delta_{2}(p),\cdots,\Delta_{s}(p)$ (resp.
$\Delta^{\prime}_{1}(p),\Delta^{\prime}_{2}(p),\cdots,\Delta^{\prime}_{s^{\prime}}(p)$)
the triangles incident to $p$ in $T^{o}$ (resp. $T^{\prime o}$) in clockwise
order such that $\widetilde{\beta}$ crosses $\Delta_{1}(p)$ (resp.
$\Delta^{\prime}_{1}(p)$) or $\widetilde{\beta}$ is the common side of
$\Delta_{1}(p)$ and $\Delta_{s}(p)$ (resp. $\Delta^{\prime}_{1}(p)$ and
$\Delta^{\prime}_{s^{\prime}}(p)$). Denote the common side of $\Delta_{i}(p)$
and $\Delta_{i+1}(p)$ (resp. $\Delta^{\prime}_{i}(p)$ and
$\Delta^{\prime}_{i+1}(p)$) by $\tau_{i}(p)$ (resp. $\tau^{\prime}_{i}(p)$).
Denote the third side of $\Delta_{i}(p)$ (resp. $\Delta^{\prime}_{i}(p)$) by
$\tau_{[i]}(p)$ (resp. $\tau^{\prime}_{[i]}(p)$).

If $\widetilde{\beta}\notin T^{o}$ and $t(\widetilde{\beta})=q$, denote

$E_{1}(q)=\begin{cases}E(G_{c}),&\mbox{if
$rel(G_{c},T^{o})=1$},\vspace{1mm}\\\ N(G_{c}),&\mbox{if
$rel(G_{c},T^{o})=-1$,}\end{cases}\;\;\;E_{2}(q)=\begin{cases}N(G_{c}),&\mbox{if
$rel(G_{c},T^{o})=1$},\vspace{1mm}\\\ E(G_{c}),&\mbox{if
$rel(G_{c},T^{o})=-1$.}\end{cases}$

If $\widetilde{\beta}\notin T^{o}$ and $s(\widetilde{\beta})=p$, denote

$E_{1}(p)=\begin{cases}W(G_{1}),&\mbox{if
$rel(G_{1},T^{o})=1$},\vspace{1mm}\\\ S(G_{1}),&\mbox{if
$rel(G_{1},T^{o})=-1$,}\end{cases}\;\;\;E_{2}(p)=\begin{cases}S(G_{1}),&\mbox{if
$rel(G_{1},T^{o})=1$},\vspace{1mm}\\\ W(G_{1}),&\mbox{if
$rel(G_{1},T^{o})=-1$.}\end{cases}$

If $\widetilde{\beta}\notin T^{\prime o}$ and $t(\widetilde{\beta})=q$, denote

$E^{\prime}_{1}(q)=\begin{cases}E(G^{\prime}_{c^{\prime}}),&\mbox{if
$rel(G^{\prime}_{c^{\prime}},T^{\prime o})=1$},\vspace{1mm}\\\
N(G^{\prime}_{c^{\prime}}),&\mbox{if $rel(G^{\prime}_{c^{\prime}},T^{\prime
o})=-1$,}\end{cases}\;\;\;E^{\prime}_{2}(q)=\begin{cases}N(G^{\prime}_{c^{\prime}}),&\mbox{if
$rel(G^{\prime}_{c^{\prime}},T^{\prime o})=1$},\vspace{1mm}\\\
E(G^{\prime}_{c^{\prime}}),&\mbox{if $rel(G^{\prime}_{c^{\prime}},T^{\prime
o})=-1$.}\end{cases}$

If $\widetilde{\beta}\notin T^{o}$ and $s(\widetilde{\beta})=p$, denote

$E^{\prime}_{1}(p)=\begin{cases}W(G^{\prime}_{1}),&\mbox{if
$rel(G^{\prime}_{1},T^{\prime o})=1$},\vspace{1mm}\\\
S(G^{\prime}_{1}),&\mbox{if $rel(G^{\prime}_{1},T^{\prime
o})=-1$,}\end{cases}\;\;\;E^{\prime}_{2}(p)=\begin{cases}S(G^{\prime}_{1}),&\mbox{if
$rel(G^{\prime}_{1},T^{\prime o})=1$},\vspace{1mm}\\\
W(G^{\prime}_{1}),&\mbox{if $rel(G^{\prime}_{1},T^{\prime
o})=-1$.}\end{cases}$

Let
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}),\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$
or $\mathcal{L}(T^{o},\widetilde{\beta}^{(p,q)})$. Let
$\mathcal{L}^{\prime}=\mathcal{L}(T^{\prime
o},\widetilde{\beta}),\mathcal{L}(T^{\prime o},\widetilde{\beta}^{(q)})$ or
$\mathcal{L}(T^{\prime o},\widetilde{\beta}^{(p,q)})$ correspondingly. Denote
the minimal elements in $\mathcal{L}$ and $\mathcal{L}^{\prime}$ by ${\bf
P}_{-}$ and ${\bf P}^{\prime}_{-}$, respectively.

##  9\. Partition bijection between lattices $\mathcal{L}$ and
$\mathcal{L}^{\prime}$

For any pair ${\bf P},{\bf Q}\in\mathcal{L}$, assume that ${\bf P}$ covers
${\bf Q}$, then we have that ${\bf P}$ and ${\bf Q}$ are either related by a
twist on a tile $G_{l}$ or related by two adjacent triangles incident to $p$
or $q$, assume $\tau$ is the label of the diagonal or the common side of the
triangles. In both cases, we say that ${\bf P}$ covers ${\bf Q}$ and _related
by $\tau$_. We denote by

$\Omega({\bf P},{\bf Q})=w({\bf P})-w({\bf Q}).$

We say that ${\bf P},{\bf Q}\in\mathcal{L}$ are _related by a twist at $\tau$_
if either ${\bf P}$ covers ${\bf Q}$ and related by $\tau$ or ${\bf Q}$ covers
${\bf P}$ and related by $\tau$.

In this section, we construct a partition bijection from $\mathcal{L}$ to
$\mathcal{L}^{\prime}$ via the partition bijections
$\varphi^{T^{o}}_{\alpha}:\mathcal{P}(G_{T^{o},\widetilde{\beta}})\to\mathcal{P}(G_{T^{\prime
o},\widetilde{\beta}})$ and
$\phi_{\alpha}^{T^{o},q}:\Delta(T^{o},q)\to\Delta(T^{\prime o},q)$, which are
given in Section 7.

#### Partition bijection
$\pi:\mathcal{L}(T^{o},\widetilde{\beta})\to\mathcal{L}(T^{\prime
o},\widetilde{\beta})$

Herein, we assume that $\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta})$ and
$\mathcal{L}^{\prime}=\mathcal{L}(T^{\prime o},\widetilde{\beta})$.

Recall that for any $P\in\mathcal{P}(G_{T^{o},\widetilde{\beta}})$ we have
${\bf
m}(P,\alpha)=(m_{1}(P),\cdots,m_{\eta_{\alpha}}(P))\in\\{-1,0,1\\}^{\eta_{\alpha}}$.

######  Definition 9.1.

[22, 23] For any $P\in\mathcal{P}(G_{T^{o},\widetilde{\beta}})$, we choose
pairs of indices in $\\{1,\cdots,\eta_{\alpha}\\}$ via the following
algorithm:

  1. $(i)$

If $m_{i}(P)\geq 0$ for all $i=1,\cdots,\eta_{\alpha}$ or $m_{i}(P)\leq 0$ for
all $i=1,\cdots,\eta_{\alpha}$, then nothing is chosen;

  2. $(ii)$

If $m_{i}(P)m_{j}(P)=-1$ for some $i,j$, choose the pair of indices $(a,b)$
with $a<b$ such that (1) $m_{a}(P)m_{b}(P)=-1$, (2) $m_{\ell}(P)=0$ for all
$a<\ell<b$, (3) $m_{a}(P)m_{\ell}(P)\geq 0$ for all $\ell<a$.

  3. $(iii)$

Delete $m_{a}(P)$ and $m_{b}(P)$ from ${\bf m}(P,\alpha)$. Then return to step
(i).

We call any above chosen pair $(a,b)$ of indices an _${\bf m}(P,\alpha)$
-pair_.

######  Example 9.2.

If ${\bf m}(P,\alpha)=(1,1,0,-1,0,1,-1,1)$. Then all of the ${\bf
m}(P,\alpha)$-pairs are $(2,4)$ and $(6,7)$. If ${\bf
m}(P,\alpha)=(1,1,-1,0-1,1)$. Then all the the ${\bf m}(P,\alpha)$-pairs are
$(2,3)$ and $(1,5)$.

The following is immediate.

######  Lemma 9.3.

For ${\bf m}(P,\alpha)$, we have

  1. $(1)$

${\bf m}(P,\alpha)$-pairs and $-{\bf m}(P,\alpha)$-pairs coincide.

  2. $(2)$

For any ${\bf m}(P,\alpha)$-pair $(a,b)$ with $a<b$ we have
$\sum_{a<\ell<b}m_{\ell}(P)=0.$

  3. $(3)$

If $\sum m_{i}(P)\geq 0$ then for any $a$ with $m_{a}(P)<0$ we have $a$ is in
some ${\bf m}(P,\alpha)$-pair.

  4. $(4)$

If $\sum m_{i}(P)\leq 0$ then for any $a$ with $m_{a}(P)>0$ we have $a$ is in
some ${\bf m}(P,\alpha)$-pair.

Let $(a,b)$ be an ${\bf m}(P,\alpha)$-pair. Then either
$m_{a}(P)=1,m_{b}(P)=-1$ or $m_{a}(P)=-1,m_{b}(P)=1$. By Proposition 7.14, in
case $m_{a}(P)=1,m_{b}(P)=-1$, we have $P$ can twist on the tile $G(b)$ and
any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can twist on $G^{\prime}(a)$;
in case $m_{a}(P)=-1,m_{b}(P)=1$, we have $P$ can twist on the tile $G(a)$ and
any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can twist on $G^{\prime}(b)$,
where $G(a),G(b)$ (resp. $G^{\prime}(a),G^{\prime}(b)$) are the tiles with
diagonal labeled $\alpha$ (resp. $\alpha^{\prime}$) given in Proposition 7.14.

_Construction of $\pi$._ For any $P\in\mathcal{L}$, $\pi(P)$ contains all
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ satisfy the following condition:
for any ${\bf m}(P,\alpha)$-pair $(a,b)$,

$\bullet$ if $m_{a}(P)=1,m_{b}(P)=-1$ then the labels of $P^{\prime}\cap
edge(G^{\prime}(a))$ and $P\cap edge(G(b))$ coincide;

$\bullet$ if $m_{a}(P)=-1,m_{b}(P)=1$ then the labels of $P^{\prime}\cap
edge(G^{\prime}(b))$ and $P\cap edge(G(a))$ coincide.

Similarly, we can construction a set
$\pi^{\prime}(P^{\prime})\subset\mathcal{L}$ for any
$P^{\prime}\in\mathcal{L}^{\prime}=\mathcal{P}(G_{T^{\prime
o},\widetilde{\beta}})$.

######  Proposition 9.4.

For any $P\in\mathcal{L}=\mathcal{P}(G_{T^{o},\widetilde{\beta}})$ we have

$|\pi(P)|=2^{[\sum m_{i}(P)]_{+}},$

to be precise, for any $P^{\prime}\in\pi(P)$, we have

  1. $(1)$

${\bf m}(P^{\prime},\alpha^{\prime})=-{\bf m}(P,\alpha)$,

  2. $(2)$

if $m_{k}(P)=1$ and $k$ is not in any ${\bf m}(P,\alpha)$-pair then
$P^{\prime}$ can twist on the tiles $G^{\prime}(k)$ and
$\mu_{G^{\prime}(k)}P^{\prime}\in\pi(P)$.

where $G^{\prime}(k)$ are the tiles given in Proposition 7.14.

###### Proof.

It follows by Proposition 7.15 and the construction of $\pi(P)$. $\square$

######  Remark 9.5.

Let $P\in\mathcal{L}$.

  1. $(1)$

Denote $r=[\sum m_{i}(P)]_{+}$. Denote by
$\\{k_{1},\cdots,k_{r}\\}\subset\\{1,\cdots,\eta_{\alpha}\\}$ the set of
indices that are not in any ${\bf m}(P,\alpha)$-pair and $m_{k_{j}}=1$. Assume
$k_{1}<k_{2}<\cdots<$ $k_{r}$. By Proposition 9.4, we may write $\pi(P)$ as
$\\{P(\vec{c})\mid\vec{c}=(c_{1},\cdots,c_{r})\in\\{0,1\\}^{r}\\}.$ To be
precise, for any $j\in\\{1,\cdots,r\\}$ the edges labeled
$\alpha_{1},\alpha_{3}$ of $G^{\prime}(k_{j})$ are in $P(\vec{c})$ if
$c_{j}=1$ and the the edges labeled $\alpha_{2},\alpha_{4}$ of
$G^{\prime}(k_{j})$ are in $P(\vec{c})$ if $c_{j}=0$.

  2. $(2)$

Under the order of $\mathcal{L}^{\prime}=\mathcal{P}(G_{T^{\prime
o},\widetilde{\beta}})$ in Section 4, we have $P(\vec{c})\leq P(\vec{c}\hskip
2.0pt^{\prime})$ if and only if $c_{j}\leq c^{\prime}_{j}$ for all
$j\in\\{1,\cdots,r\\}$.

#### Partition bijection
$\pi:\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})\to\mathcal{L}(T^{\prime
o},\widetilde{\beta}^{(q)})$.

Herein, we assume that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$ and
$\mathcal{L}^{\prime}=\mathcal{L}(T^{\prime o},\widetilde{\beta}^{(q)})$.
Recall in (11) that

$\begin{array}[]{rcl}m(\Delta_{j}(q);\alpha)&=&\text{number of edges labeled
}\alpha\text{ in }\\{\tau_{[j]}(q)\\}\vspace{2mm}\\\ &-&\text{number of edges
labeled }\alpha\text{ in }\\{\tau_{j-1}(q),\tau_{j}(q)\\}.\end{array}$

Note that if $m(\Delta_{j}(q);\alpha)=1$ then $\alpha=\tau_{[j]}(q)$ and thus
either $\tau_{j-1}(q)=\alpha_{1},\tau_{j}(q)=\alpha_{4}$ or
$\tau_{j-1}(q)=\alpha_{3},\tau_{j}(q)=\alpha_{2}$.

For ${\bf P}=(P,\Delta_{j}(q))\in\mathcal{L}$, denote ${\bf m}({\bf
P};\alpha)=({\bf m}(P,\alpha),m(\Delta_{j}(q);\alpha))$. As $\alpha$ is not a
self-folded arc in $T^{o}$, we have $m(\Delta_{j}(q);\alpha)\in\\{-1,0,1\\}$
and thus ${\bf m}({\bf P};\alpha)\in\\{-1,0,1\\}^{\eta_{\alpha}+1}$.

######  Definition 9.6.

Assume that ${\bf m}({\bf P};\alpha)=(m_{1}({\bf
P}),\cdots,m_{\eta_{\alpha}}({\bf P}),m_{\eta_{\alpha}+1}({\bf P}))$. We
choose pairs of indices $(a,b)$ via the following algorithm:

  1. $(i)$

If $m_{\eta_{\alpha}}({\bf P})m_{\eta_{\alpha}+1}({\bf P})=-1$, then first
choose $(\eta_{\alpha},\eta_{\alpha}+1)$ and all of the $(m_{1}({\bf
P}),\cdots,m_{\eta_{\alpha}-1}({\bf P}))$-pairs as in Definition 9.1;

  2. $(ii)$

If $m_{\eta_{\alpha}}({\bf P})m_{\eta_{\alpha}+1}({\bf P})\neq-1$, then we
choose all of the $(m_{1}({\bf P}),\cdots,m_{\eta_{\alpha}+1}({\bf P})$-pairs
as in Definition 9.1.

The chosen pairs are called _${\bf m}({\bf P};\alpha)$ -pairs_.

The following is immediate.

######  Lemma 9.7.

For ${\bf m}(P,\alpha)$, we have

  1. $(1)$

${\bf m}(P,\alpha)$-pairs and $-{\bf m}(P,\alpha)$-pairs coincide.

  2. $(2)$

For any ${\bf m}(P,\alpha)$-pair $(a,b)$ with $a<b$ we have
$\sum_{a<\ell<b}m_{\ell}(P)=0.$

  3. $(3)$

If $\sum m_{i}(P)\geq 0$ then for any $a$ with $m_{a}(P)<0$ we have $a$ is in
some ${\bf m}(P,\alpha)$-pair.

  4. $(4)$

If $\sum m_{i}(P)\leq 0$ then for any $a$ with $m_{a}(P)>0$ we have $a$ is in
some ${\bf m}(P,\alpha)$-pair.

_Construction of $\pi$._ For ${\bf P}=(P,\Delta_{j}(q))\in\mathcal{L}$,
$\pi({\bf P})$ contains all
$(P^{\prime},\Delta^{\prime})\in\varphi_{\alpha}^{T^{o}}(P)\times\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))$
satisfy the following condition: for any ${\bf
m}(P,\Delta_{j}(q);\alpha)$-pair111We use ${\bf
m}(P,\Delta_{j}(q);\alpha)$-pair as in Definition 9.15 to ensure Proposition
9.12 holds $(a,b)$,

$\bullet$ if $m_{a}({\bf P})=1,m_{b}({\bf P})=-1$ with $b\neq\eta_{\alpha}+1$
then the labels of $P^{\prime}\cap edge(G^{\prime}(a))$ and $P\cap edge(G(b))$
coincide;

$\bullet$ if $m_{a}({\bf P})=-1,m_{b}({\bf P})=1$ with $b\neq\eta_{\alpha}+1$
then the labels of $P^{\prime}\cap edge(G^{\prime}(b))$ and $P\cap edge(G(a))$
coincide;

$\bullet$ if $m_{a}({\bf P})=1,m_{b}({\bf P})=-1$ with $b=\eta_{\alpha}+1$
then the edges labeled $\tau_{j-2}(q)$ and $\tau_{[j]}(q)$ of
$edge(G^{\prime}(a))$ are in $P^{\prime}$ in case $\tau_{j-1}(q)=\alpha$, or
the edges labeled $\tau_{j+1}(q)$ and $\tau_{[j]}(q)$ of $edge(G^{\prime}(a))$
are in $P^{\prime}$ in case $\tau_{j}(q)=\alpha$;

$\bullet$ if $m_{a}({\bf P})=-1,m_{b}({\bf P})=1$ with $b=\eta_{\alpha}+1$,

  1. $(i)$

in case $\tau_{j-1}(q)=\alpha_{1},\tau_{j}(q)=\alpha_{4}$, then
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{3},\alpha_{4}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{1}$ and $\alpha_{3}$, or
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{1},\alpha_{2}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{2}$ and $\alpha_{4}$.

  2. $(ii)$

in case $\tau_{j-1}(q)=\alpha_{3},\tau_{j}(q)=\alpha_{2}$, then
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{1},\alpha_{2}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{1}$ and $\alpha_{3}$, or
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{3},\alpha_{4}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{2}$ and $\alpha_{4}$.

Similarly, we can construction a set $\pi^{\prime}({\bf
P}^{\prime})\subset\mathcal{L}$ for any ${\bf
P}^{\prime}\in\mathcal{L}^{\prime}=\mathcal{P}(G_{T^{\prime
o},\widetilde{\beta}^{(q)}})$.

By Propositions 7.15, 8.7, from the construction, the following proposition
follows.

######  Proposition 9.8.

For any ${\bf P}=(P,\Delta_{j}(q))\in\mathcal{L}$ we have

$|\pi({\bf P})|=2^{[\sum m_{i}({\bf P})]_{+}},$

to be precise, for any ${\bf
P}^{\prime}=(P^{\prime},\Delta^{\prime})\in\pi({\bf P})$, we have

  1. $(1)$

${\bf m}({\bf P}^{\prime};\alpha^{\prime})=-{\bf m}({\bf P};\alpha)$;

  2. $(2)$

if $k\neq\eta_{\alpha}+1$ is not in any ${\bf m}({\bf P};\alpha)$-pair and
$m_{k}({\bf P})=1$, then $P^{\prime}$ can twist on $G^{\prime}(k)$ and
$(\mu_{G^{\prime}(k)}P^{\prime},\Delta^{\prime})\in\pi({\bf P})$, where
$G^{\prime}(k)$ is the tile given in Proposition 7.14;

  3. $(3)$

if $\eta(\alpha)+1$ is not in any ${\bf m}({\bf P};\alpha)$-pair and
$m_{\eta_{\alpha}+1}({\bf P})=1$, then
$(P^{\prime},\Delta^{\prime\prime})\in\pi({\bf P})$, where
$\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))=\\{\Delta^{\prime},\Delta^{\prime\prime}\\}$.

######  Lemma 9.9.

For any ${\bf P}=(P,\Delta_{j}(q))\in\mathcal{L}$, if
$m(\Delta_{j}(q);\alpha)=1$, then
$\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))=\\{\Delta^{\prime}_{1},\Delta^{\prime}_{2}\\}$
satisfies that
$(P^{\prime},\Delta^{\prime}_{1})<(P^{\prime},\Delta^{\prime}_{2})$ for any
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$.

###### Proof.

Otherwise, by Lemma 4.5, we have
$\Delta^{\prime}_{1}(q)\in\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))$ and there
exist $P^{\prime},Q^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ such that
$E^{\prime}_{1}(q)\in P^{\prime},E^{\prime}_{2}(q)\in Q^{\prime}$. Thus by
Proposition 7.15, the diagonal of $G^{\prime}_{c^{\prime}}$ is labeled
$\alpha^{\prime}$. Thus $m(\Delta^{\prime}_{1}(q),\alpha^{\prime})=1$. It
contradicts to $m(\Delta_{j}(q);\alpha)=1$ and Lemma 8.9. $\square$

######  Remark 9.10.

Denote $r=[\sum m_{i}({\bf P})]_{+}$. Denote by
$\\{k_{1},\cdots,k_{r}\\}\subset\\{1,\cdots,\eta_{\alpha}+1\\}$ the set of
indices that are not in any ${\bf m}({\bf P};\alpha)$-pair and
$m_{k_{\ell}}({\bf P})=1$ for $\ell=1,\cdots,r$. Suppose that
$k_{1}<k_{2}<\cdots<k_{r}$. By Proposition 9.8, we may write $\pi({\bf P})$ as
$\\{{\bf P}(\vec{c})\mid\vec{c}\in\\{0,1\\}^{r}\\},$ to be precise, assume
that ${\bf P}(\vec{c})=(P^{\prime},\Delta^{\prime}),$ for any
$\ell\in\\{1,\cdots,r\\}$,

  1. $(i)$

if $k_{\ell}\neq\eta_{\alpha}+1$ then the edges labeled
$\alpha_{1},\alpha_{3}$ of $G^{\prime}(k_{\ell})$ are in $P$ in case
$c_{\ell}=1$ and the the edges labeled $\alpha_{2},\alpha_{4}$ of
$G^{\prime}(k_{\ell})$ are in $P$ in case $c_{\ell}=0$;

  2. $(ii)$

if $k_{\ell}=\eta_{\alpha}+1$, then $\Delta^{\prime}=\Delta^{\prime}_{1}$ in
case $c_{\ell}=0$ and $\Delta^{\prime}=\Delta^{\prime}_{2}$ in case
$c_{\ell}=1$, where $\Delta^{\prime}_{1}$ and $\Delta^{\prime}_{2}$ are given
by Lemma 9.9.

######  Lemma 9.11.

Suppose that $\widetilde{\beta}\notin T^{\prime o}$. For any ${\bf
P}=(P,\Delta_{j}(q))\in\mathcal{L}$,

  1. $(1)$

if
$\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))=\\{\Delta^{\prime}_{1}(q),\Delta^{\prime}_{2}(q)\\}$
and $E^{\prime}_{2}(q)\in P^{\prime}$ for some
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$, then $m_{\eta_{\alpha}}({\bf
P})=-1,m_{\eta_{\alpha}+1}({\bf P})=1$;

  2. $(2)$

If
$\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))=\\{\Delta^{\prime}_{t^{\prime}}(q),\Delta^{\prime}_{1}(q)\\}$
and $E^{\prime}_{1}(q)\in P^{\prime}$ for some
$P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$, then $m_{\eta_{\alpha}}({\bf
P})=-1,m_{\eta_{\alpha}+1}({\bf P})=1$.

Consequently, we have $(\eta_{\alpha},\eta_{\alpha}+1)$ is an ${\bf m}({\bf
P};\alpha)$-pair.

###### Proof.

We shall only prove the first statement, as the second one can be proved
similarly. As
$\phi_{\alpha}^{T^{o},q}(\Delta)=\\{\Delta^{\prime}_{1}(q),\Delta^{\prime}_{2}(q)\\}$,
we see that $\tau^{\prime}_{1}(q)=\alpha$ and the diagonal of
$G^{\prime}_{c^{\prime}}$ is labeled $\alpha_{1}$ or $\alpha_{3}$. It follows
that $m(\Delta^{\prime}_{1}(q))=-1$ and $E^{\prime}_{2}(q)$ is labeled
$\alpha^{\prime}$. Thus $m(\Delta_{j}(q))=1$ and
$m_{\eta_{\alpha}}(P^{\prime})=1$. It follows that $m_{\eta_{\alpha}}({\bf
P})=-m_{\eta_{\alpha}}(P^{\prime})=-1$ and $m_{\eta_{\alpha}+1}({\bf
P})=m(\Delta_{j}(q))=1$. $\square$

######  Proposition 9.12.

Suppose that $r=\sum m_{i}({\bf P})\geq 0$. Given any $k$, for
$\vec{c},\vec{c}\hskip 2.27621pt^{\prime}\in\\{0,1\\}^{r}$ such that
$c_{k}=1,c^{\prime}_{k}=0$ and $c_{a}=c^{\prime}_{a}$ for $a\neq k$, then
${\bf P}(\vec{c})$ covers ${\bf P}(\vec{c}\hskip 2.27621pt^{\prime})$.

###### Proof.

Assume that ${\bf P}(\vec{c}\hskip
2.27621pt^{\prime})=(P^{\prime}_{1},\Delta^{\prime}_{1})$ and ${\bf
P}(\vec{c})=(P^{\prime}_{2},\Delta^{\prime}_{2})$. By Remark 9.10, we have
either $P^{\prime}_{1}=P^{\prime}_{2}$ or
$\Delta^{\prime}_{1}=\Delta^{\prime}_{2}$.

(Case I) $P^{\prime}_{1}=P^{\prime}_{2}$. We have
$(P^{\prime}_{2},\Delta^{\prime}_{2})=(P^{\prime}_{1},\Delta^{\prime}_{2})>(P^{\prime}_{1},\Delta^{\prime}_{1})$
by Remark 9.10 (2). By Lemma 4.5, $(P^{\prime}_{1},\Delta^{\prime}_{2})$ does
not cover $(P^{\prime}_{1},\Delta^{\prime}_{1})$ if and only if either
$E^{\prime}_{2}(q)\in
P^{\prime}_{1},\Delta^{\prime}_{2}=\Delta^{\prime}_{1}(q),\Delta^{\prime}_{1}=\Delta^{\prime}_{2}(q)$
or $E^{\prime}_{1}(q)\in
P^{\prime}_{1},\Delta^{\prime}_{2}=\Delta^{\prime}_{t^{\prime}}(q),\Delta^{\prime}_{1}=\Delta^{\prime}_{1}(q)$.
We have $(\eta_{\alpha},\eta_{\alpha}+1)$ is an ${\bf m}({\bf P};\alpha)$-pair
by Lemma 9.11. From the construction of $\pi$, we see that
$(P^{\prime}_{1},\Delta^{\prime}_{1})$ and
$(P^{\prime}_{1},\Delta^{\prime}_{2})$ can not belong to $\pi({\bf P})$, a
contradiction.

(Case II) $\Delta^{\prime}_{1}=\Delta^{\prime}_{2}$. We have
$P^{\prime}_{1}<P^{\prime}_{2}=\mu_{G^{\prime}_{l}}P^{\prime}_{1}$ for some
tile $G^{\prime}_{l}$ with diagonal labeled $\alpha^{\prime}$ by Remark 9.10
(1). If
$(P^{\prime}_{2},\Delta^{\prime}_{2})=(\mu_{G^{\prime}_{l}}P^{\prime}_{1},\Delta^{\prime}_{1})$
does not cover $(P^{\prime}_{1},\Delta^{\prime}_{1})$, by Lemma 4.4, we have
$\Delta^{\prime}_{1}=\Delta^{\prime}_{1}(q)$ and
$G^{\prime}_{l}=G^{\prime}_{c^{\prime}}$ is the last tile of $G_{T^{\prime
o},\widetilde{\beta}}$. It follows that
$m_{\eta_{\alpha}+1}(P^{\prime}_{1},\Delta^{\prime}_{1})=1$ and
$m_{\eta_{\alpha}}(P^{\prime}_{1},\Delta^{\prime}_{1})=-1$. Thus
$m_{\eta_{\alpha}+1}({\bf P})=-1$ and $m_{\eta_{\alpha}}({\bf P})=1$. It
follows that $(\eta_{\alpha},\eta_{\alpha}+1)$ is an ${\bf m}({\bf
P};\alpha)$-pair. From the construction of $\pi$, we see that
$(P^{\prime}_{1},\Delta^{\prime}_{1})$ and
$(P^{\prime}_{2},\Delta^{\prime}_{1})$ can not belong to $\pi({\bf P})$, a
contradiction. $\square$

#### Partition bijection
$\pi_{\alpha}^{T^{o},p,q}:\mathcal{L}(T^{o},\widetilde{\beta}^{(p,q)})\to\mathcal{L}(T^{\prime
o},\widetilde{\beta}^{(p,q)})$

Herein we assume that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(p,q)})$ and
$\mathcal{L}^{\prime}=\mathcal{L}(T^{\prime o},\widetilde{\beta}^{(p,q)})$.

Recall in (12) that

$\begin{array}[]{rcl}m(\Delta_{i}(p);\alpha)&=&\text{number of edges labeled
}\alpha\text{ in }\\{\tau_{[i]}(p)\\}\vspace{2mm}\\\ &-&\text{number of edges
labeled }\alpha\text{ in }\\{\tau_{i-1}(p),\tau_{i}(p)\\}.\end{array}$

######  Lemma 9.13.

If $m(\Delta_{i}(p);\alpha)=1$ and $m(\Delta_{j}(q);\alpha)=-1$, assume that
$\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))=\\{\Delta^{\prime\prime}\\}$, then
there is a unique $\Delta^{\prime}\in\phi_{\alpha}^{T^{o},p}(\Delta_{i}(p))$
such that

$x^{T}(\Delta_{i}(p))x^{T}(\Delta_{j}(q))=x^{T^{\prime}}(\Delta^{\prime})x^{T^{\prime}}(\Delta^{\prime\prime}).$

###### Proof.

Since $m(\Delta_{j}(q);\alpha)=-1$, we have the following cases:
$\Delta_{j}(q)=\\{\alpha_{1},\alpha_{4},\alpha\\}$,
$\tau_{[j]}(q)=\alpha_{1}$;
$\Delta_{j}(q)=\\{\alpha_{1},\alpha_{4},\alpha\\}$,
$\tau_{[j]}(q)=\alpha_{4}$;
$\Delta_{j}(q)=\\{\alpha_{2},\alpha_{3},\alpha\\}$,
$\tau_{[j]}(q)=\alpha_{2}$;
$\Delta_{j}(q)=\\{\alpha_{2},\alpha_{3},\alpha\\}$,
$\tau_{[j]}(q)=\alpha_{3}$. We shall only consider the case that
$\Delta_{j}(q)=\\{\alpha_{1},\alpha_{4},\alpha\\}$,
$\tau_{[j]}(q)=\alpha_{1}$, as the remaining cases can be proved similarly.
Then $x^{T}(\Delta_{j}(q))=x_{\alpha_{1}}x^{-1}_{\alpha_{4}}x^{-1}_{\alpha}$,
$\Delta^{\prime\prime}=\\{\alpha_{3},\alpha_{4},\alpha^{\prime}\\}$ with
$m(\Delta^{\prime\prime};\alpha^{\prime})=1$ and thus
$x^{T^{\prime}}(\Delta^{\prime\prime})=x_{\alpha^{\prime}}x^{-1}_{\alpha_{3}}x^{-1}_{\alpha_{4}}$

Since $m(\Delta_{i}(p);\alpha)=1$, we have $\tau_{[i]}(p)=\alpha$ and
$\Delta_{i}(p)=\\{\alpha_{1},\alpha_{4},\alpha\\}$ or
$\\{\alpha_{2},\alpha_{3},\alpha\\}$.

In case $\Delta_{i}(p)=\\{\alpha_{1},\alpha_{4},\alpha\\}$, then
$x^{T}(\Delta_{i}(p))=x_{\alpha}x^{-1}_{\alpha_{1}}x^{-1}_{\alpha_{4}}$. Thus
$\Delta^{\prime}=\\{\alpha_{3},\alpha_{4},\alpha^{\prime}\\}$ with
$m(\Delta^{\prime};\alpha_{3})=1$. Therefore
$x^{T}(\Delta_{i}(p))x^{T}(\Delta_{j}(q))=\frac{1}{x^{2}_{\alpha_{4}}}=x^{T^{\prime}}(\Delta^{\prime})x^{T^{\prime}}(\Delta^{\prime\prime}).$
The uniqueness of $\Delta^{\prime}$ follows as $\alpha_{2}=\alpha_{3}$ and
$\alpha_{1}=\alpha_{4}$ can not hold simultaneous.

In case $\Delta_{i}(p)=\\{\alpha_{2},\alpha_{3},\alpha\\}$, then
$x^{T}(\Delta_{i}(p))=x_{\alpha}x^{-1}_{\alpha_{2}}x^{-1}_{\alpha_{3}}$. Thus
$\Delta^{\prime}=\\{\alpha_{1},\alpha_{2},\alpha^{\prime}\\}$ with
$m(\Delta^{\prime};\alpha_{1})=1$. Therefore
$x^{T}(\Delta_{i}(p))x^{T}(\Delta_{j}(q))=\frac{x_{\alpha_{1}}}{x_{\alpha_{2}}x_{\alpha_{3}}x_{\alpha_{4}}}=x^{T^{\prime}}(\Delta^{\prime})x^{T^{\prime}}(\Delta^{\prime\prime}).$
The uniqueness of $\Delta^{\prime}$ follows as $\alpha_{2}=\alpha_{3}$ and
$\alpha_{1}=\alpha_{4}$ can not hold simultaneous. $\square$

Similarly, we have the following result.

######  Lemma 9.14.

If $m(\Delta_{i}(p);\alpha)=-1$ and $m(\Delta_{j}(q);\alpha)=1$, assume that
$\phi_{\alpha}^{T^{o},p}(\Delta_{i}(p))=\\{\Delta^{\prime}\\}$, then there is
a unique $\Delta^{\prime\prime}\in\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))$ such
that

$x^{T}(\Delta_{i}(p))x^{T}(\Delta_{j}(q))=x^{T^{\prime}}(\Delta^{\prime})x^{T^{\prime}}(\Delta^{\prime\prime}).$

For ${\bf P}=(\Delta_{i}(p),P,\Delta_{j}(q))\in\mathcal{L}$, denote ${\bf
m}({\bf P};\alpha)=(m(\Delta_{i}(p);\alpha),{\bf
m}(P,\alpha),m(\Delta_{j}(q);\alpha))$. As $\alpha$ is not a self-folded arc
in $T^{o}$, we have $m(\Delta_{j}(q);\alpha)\in\\{-1,0,1\\}$ and thus ${\bf
m}({\bf P};\alpha)\in\\{-1,0,1\\}^{\eta_{\alpha}+1}$. Denote ${\bf m}({\bf
P};\alpha)=(m_{0}({\bf P}),m_{1}({\bf P}),\cdots,m_{\eta_{\alpha}}({\bf
P}),m_{\eta_{\alpha}+1}({\bf P}))$.

######  Definition 9.15.

For ${\bf m}({\bf P};\alpha)$, we choose pairs of indices $(a,b)$ via the
following algorithm:

  1. $(i)$

If $m_{0}({\bf P})m_{1}({\bf P})=-1$, then choose $(0,1)$ and let ${\bf
m}^{\prime}=(m_{2}({\bf P}),\cdots,m_{\eta_{\alpha}+1}({\bf P}))$, otherwise
let ${\bf m}^{\prime}={\bf m}({\bf P};\alpha)$;

  2. $(ii)$

In ${\bf m}^{\prime}$, if $m_{\eta_{\alpha}}({\bf P})m_{\eta_{\alpha}+1}({\bf
P})=-1$, then choose $(\eta_{\alpha},\eta_{\alpha}+1)$ and let ${\bf
m}^{\prime\prime}$ be obtained by delete $m_{\eta_{\alpha}}({\bf
P}),m_{\eta_{\alpha}+1}({\bf P})$ from ${\bf m}^{\prime}$, otherwise let ${\bf
m}^{\prime\prime}={\bf m}^{\prime}$;

  3. $(iii)$

Choose all of the ${\bf m}^{\prime\prime}$-pairs as in Definition 9.1.

The chosen pairs are called _${\bf m}({\bf P};\alpha)$ -pairs_.

The following is immediate.

######  Lemma 9.16.

For ${\bf m}(P,\alpha)$, we have

  1. $(1)$

${\bf m}(P,\alpha)$-pairs and $-{\bf m}(P,\alpha)$-pairs coincide.

  2. $(2)$

For any ${\bf m}(P,\alpha)$-pair $(a,b)$ with $a<b$ we have
$\sum_{a<\ell<b}m_{\ell}(P)=0.$

  3. $(3)$

If $\sum m_{i}(P)\geq 0$ then for any $a$ with $m_{a}(P)<0$ we have $a$ is in
some ${\bf m}(P,\alpha)$-pair.

  4. $(4)$

If $\sum m_{i}(P)\leq 0$ then for any $a$ with $m_{a}(P)>0$ we have $a$ is in
some ${\bf m}(P,\alpha)$-pair.

Note that if $m_{0}({\bf P})=1$ then $\alpha=\tau_{[i]}(p)$ and thus either
either $\tau_{j-1}(p)=\alpha_{1},\tau_{j}(p)=\alpha_{4}$ or
$\tau_{j-1}(p)=\alpha_{3},\tau_{j}(p)=\alpha_{2}$. Similarly, if
$m_{\eta_{\alpha}+1}({\bf P})=1$ then $\alpha=\tau_{[j]}(q)$ and thus either
$\tau_{j-1}(q)=\alpha_{1},\tau_{j}(q)=\alpha_{4}$ or
$\tau_{j-1}(q)=\alpha_{3},\tau_{j}(q)=\alpha_{2}$.

_Construction of $\pi$._ For ${\bf
P}=(\Delta_{i}(p),P,\Delta_{j}(q))\in\mathcal{L}$, $\pi({\bf P})$ contains all
$(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime})\in\phi_{\alpha}^{T^{o},p}(\Delta_{i}(p))\times\varphi_{\alpha}^{T^{o}}(P)\times\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))$
satisfy the following condition: for any ${\bf m}({\bf P};\alpha)$-pair
$(a,b)$,

$\bullet$ if $m_{a}({\bf P})=1,m_{b}({\bf P})=-1$ with $a\neq
0,b\neq\eta_{\alpha}+1$ then the labels of $P^{\prime}\cap
edge(G^{\prime}(a))$ and $P\cap edge(G(b))$ coincide;

$\bullet$ if $m_{a}({\bf P})=-1,m_{b}({\bf P})=1$ with $a\neq
0,b\neq\eta_{\alpha}+1$ then the labels of $P^{\prime}\cap
edge(G^{\prime}(b))$ and $P\cap edge(G(a))$ coincide;

$\bullet$ if $m_{a}({\bf P})=1,m_{b}({\bf P})=-1$ with
$a=0,b\neq\eta_{\alpha}+1$,

  1. $(i)$

in case $\tau_{j-1}(p)=\alpha_{1},\tau_{j}(p)=\alpha_{4}$, then
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{3},\alpha_{4}\\}$ when the edges
in $P\cap edge(G(b))$ labeled $\alpha_{1}$ and $\alpha_{3}$, or
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{1},\alpha_{2}\\}$ when the edges
in $P\cap edge(G(b))$ labeled $\alpha_{2}$ and $\alpha_{4}$.

  2. $(ii)$

in case $\tau_{j-1}(p)=\alpha_{3},\tau_{j}(p)=\alpha_{2}$, then
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{1},\alpha_{2}\\}$ when the edges
in $P\cap edge(G(b))$ labeled $\alpha_{1}$ and $\alpha_{3}$, or
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{3},\alpha_{4}\\}$ when the edges
in $P\cap edge(G(b))$ labeled $\alpha_{2}$ and $\alpha_{4}$.

$\bullet$ if $m_{a}({\bf P})=-1,m_{b}({\bf P})=1$ with
$a=0,b\neq\eta_{\alpha}+1$, then the edges labeled $\tau_{i-2}(p)$ and
$\tau_{[i]}(p)$ of $edge(G^{\prime}(b))$ are in $P^{\prime}$ in case
$\tau_{i-1}(p)=\alpha$, or the edges labeled $\tau_{i+1}(p)$ and
$\tau_{[i]}(p)$ of $edge(G^{\prime}(a))$ are in $P^{\prime}$ in case
$\tau_{i}(p)=\alpha$;

$\bullet$ if $m_{a}({\bf P})=1,m_{b}({\bf P})=-1$ with $a\neq
0,b=\eta_{\alpha}+1$ then the edges labeled $\tau_{j-2}(q)$ and
$\tau_{[j]}(q)$ of $edge(G^{\prime}(a))$ are in $P^{\prime}$ in case
$\tau_{j-1}(q)=\alpha$, or the edges labeled $\tau_{j+1}(q)$ and
$\tau_{[j]}(q)$ of $edge(G^{\prime}(a))$ are in $P^{\prime}$ in case
$\tau_{j-1}(q)=\alpha$;

$\bullet$ if $m_{a}({\bf P})=-1,m_{b}({\bf P})=1$ with $a\neq
0,b=\eta_{\alpha}+1$,

  1. $(i)$

in case $\tau_{j-1}(q)=\alpha_{1},\tau_{j}(q)=\alpha_{4}$, then
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{3},\alpha_{4}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{1}$ and $\alpha_{3}$, or
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{1},\alpha_{2}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{2}$ and $\alpha_{4}$.

  2. $(ii)$

in case $\tau_{j-1}(q)=\alpha_{3},\tau_{j}(q)=\alpha_{2}$, then
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{1},\alpha_{2}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{1}$ and $\alpha_{3}$, or
$\Delta^{\prime}=\\{\alpha^{\prime},\alpha_{3},\alpha_{4}\\}$ when the edges
in $P\cap edge(G(a))$ labeled $\alpha_{2}$ and $\alpha_{4}$.

$\bullet$ if $m_{a}({\bf P})=1,m_{b}({\bf P})=-1$ with
$a=0,b=\eta_{\alpha}+1$, then $\Delta^{\prime}$ is given by Lemma 9.13.

$\bullet$ if $m_{a}({\bf P})=-1,m_{b}({\bf P})=1$ with
$a=0,b=\eta_{\alpha}+1$, then $\Delta^{\prime\prime}$ is given by Lemma 9.14.

Similarly, we can construction a set $\pi^{\prime}({\bf
P}^{\prime})\subset\mathcal{L}$ for any ${\bf
P}^{\prime}\in\mathcal{L}^{\prime}=\mathcal{P}(G_{T^{\prime
o},\widetilde{\beta}^{(p,q)}})$.

By Propositions 7.15, 8.7, from the construction, the following proposition
follows.

######  Proposition 9.17.

For any ${\bf P}=(\Delta_{i}(p),P,\Delta_{j}(q))\in\mathcal{L}$ we have

$|\pi({\bf P})|=2^{[\sum m_{a}({\bf P})]_{+}},$

more precisely, for any ${\bf
P}^{\prime}=(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime})\in\pi({\bf
P})$,

  1. $(1)$

${\bf m}({\bf P}^{\prime})=-{\bf m}({\bf P})$;

  2. $(2)$

if $k(\neq 0,\eta_{\alpha}+1)$ is not in any ${\bf m}({\bf P};\alpha)$-pair
and $m_{k}({\bf P})=1$, then $P^{\prime}$ can twist on $G^{\prime}(k)$ and
$(\Delta^{\prime},\mu_{G^{\prime}(k)}P^{\prime},\Delta^{\prime\prime})\in\pi({\bf
P})$;

  3. $(3)$

if $0$ is not in any ${\bf m}({\bf P};\alpha)$-pair and $m_{0}({\bf P})=1$,
then there is a triangle $\widetilde{\Delta}^{\prime}\in\Delta_{p}(T^{\prime
o})$ such that $\widetilde{\Delta}^{\prime}$ and $\Delta^{\prime}$ share the
common side $\alpha^{\prime}$, and
$(\widetilde{\Delta}^{\prime},P^{\prime},\Delta^{\prime\prime})\in\pi({\bf
P})$.

  4. $(4)$

if $\eta(\alpha)+1$ is not in any ${\bf m}({\bf P};\alpha)$-pair and
$m_{\eta_{\alpha}+1}({\bf P})=1$, then there is a triangle
$\widetilde{\Delta}^{\prime\prime}\in\Delta_{q}(T^{\prime o})$ such that
$\widetilde{\Delta}^{\prime\prime}$ and $\Delta^{\prime\prime}$ share the
common side $\alpha^{\prime}$, and
$(\Delta^{\prime},P^{\prime},\widetilde{\Delta}^{\prime\prime})\in\pi({\bf
P})$.

where $G^{\prime}(k)$ are the tiles given in Proposition 7.14.

The following lemmas are similar to Lemma 9.9, the proofs are also similar, we
omit them.

######  Lemma 9.18.

For any ${\bf P}=(\Delta_{i}(p),P,\Delta_{j}(q))\in\mathcal{L}$, if
$m(\Delta_{j}(q);\alpha)=1$, then
$\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))=\\{\Delta^{\prime\prime}_{1},\Delta^{\prime\prime}_{2}\\}$
satisfies that
$(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime}_{1})<(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime}_{2})$
for any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ and
$\Delta^{\prime}\in\phi_{\alpha}^{T^{o},p}(\Delta_{i}(p))$.

######  Lemma 9.19.

For any ${\bf P}=(\Delta_{i}(p),P,\Delta_{j}(q))\in\mathcal{L}$, if
$m(\Delta_{i}(p);\alpha)=1$, then
$\phi_{\alpha}^{T^{o},p}(\Delta_{i}(p))=\\{\Delta^{\prime}_{1},\Delta^{\prime}_{2}\\}$
satisfies that
$(\Delta^{\prime}_{1},P^{\prime},\Delta^{\prime})<(\Delta^{\prime}_{2},P^{\prime},\Delta^{\prime})$
for any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ and
$\Delta^{\prime}\in\phi_{\alpha}^{T^{o},q}(\Delta_{j}(q))$.

######  Remark 9.20.

Denote $r=[\sum m_{i}({\bf P})]_{+}$. Denote by
$\\{k_{1},\cdots,k_{r}\\}\subset\\{1,\cdots,\eta_{\alpha}+1\\}$ the set of
indices that are not in any ${\bf m}({\bf P};\alpha)$-pair and
$m_{k_{\ell}}({\bf P})=1$ for $\ell=1,\cdots,r$. Suppose that
$k_{1}<k_{2}<\cdots<k_{r}$. By Proposition 9.17, we may write $\pi({\bf P})$
as $\\{{\bf P}(\vec{c})\mid\vec{c}\in\\{0,1\\}^{r}\\},$ to be precise, assume
that ${\bf P}(\vec{c})=(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime}),$
for any $\ell\in\\{1,\cdots,r\\}$,

  1. $(i)$

if $k_{\ell}\neq 0,\eta_{\alpha}+1$, then the edges labeled
$\alpha_{1},\alpha_{3}$ of $G^{\prime}(k_{j})$ are in $P^{\prime}$ in case
$c_{j}=1$ and the the edges labeled $\alpha_{2},\alpha_{4}$ of
$G^{\prime}(k_{j})$ are in $P^{\prime}$ in case $c_{j}=0$;

  2. $(ii)$

if $k_{\ell}=0$, then $\Delta^{\prime}=\Delta^{\prime}_{1}$ in case
$c_{\ell}=0$ and $\Delta^{\prime}=\Delta^{\prime}_{2}$ in case $c_{\ell}=1$,
where $\Delta^{\prime}_{1}$ and $\Delta^{\prime}_{2}$ are given by Lemma 9.19;

  3. $(iii)$

if $k_{\ell}=\eta_{\alpha}+1$, then
$\Delta^{\prime\prime}=\Delta^{\prime\prime}_{1}$ in case $c_{\ell}=0$ and
$\Delta^{\prime\prime}=\Delta^{\prime\prime}_{2}$ in case $c_{\ell}=1$, where
$\Delta^{\prime\prime}_{1}$ and $\Delta^{\prime\prime}_{2}$ are given by Lemma
9.18.

######  Proposition 9.21.

Suppose that $r=\sum m_{a}({\bf P})\geq 0$. Given any $k$, assume that
$\vec{c},\vec{c}\hskip 2.27621pt^{\prime}$ in $\\{0,1\\}^{r}$ satisfy
$c_{k}=1,c^{\prime}_{k}=0$ and $c_{a}=c^{\prime}_{a}$ for $a\neq k$, then
${\bf P}(\vec{c})$ covers ${\bf P}(\vec{c}\hskip 2.27621pt^{\prime})$.

###### Proof.

Assume that ${\bf P}(\vec{c}\hskip
2.27621pt^{\prime})=(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime\prime}_{1})$
and ${\bf
P}(\vec{c})=(\Delta^{\prime}_{2},P^{\prime}_{2},\Delta^{\prime\prime}_{2})$.
By Remark 9.20, one of the following three cases happens: (1)
$P^{\prime}_{1}=P^{\prime}_{2}$ and
$\Delta^{\prime\prime}_{1}=\Delta^{\prime\prime}_{2}$, (2)
$\Delta^{\prime}_{1}=\Delta^{\prime}_{2}$ and
$\Delta^{\prime\prime}_{1}=\Delta^{\prime\prime}_{2}$, (3)
$\Delta^{\prime}_{1}=\Delta^{\prime}_{2}$ and $P^{\prime}_{1}=P^{\prime}_{2}$.

In case $P^{\prime}_{1}=P^{\prime}_{2}$ and
$\Delta^{\prime\prime}_{1}=\Delta^{\prime\prime}_{2}$, we have
$(\Delta^{\prime}_{2},P^{\prime\prime}_{2},\Delta^{\prime\prime}_{2})=(\Delta^{\prime}_{2},P^{\prime\prime}_{1},\Delta^{\prime\prime}_{2})>(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime}_{1})$
by Remark 9.20 (2). By Lemma 4.12,
$(\Delta^{\prime}_{2},P^{\prime\prime}_{2},\Delta^{\prime\prime}_{2})=(\Delta^{\prime}_{2},P^{\prime\prime}_{1},\Delta^{\prime\prime}_{2})$
covers $(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime}_{1})$.

Similarly,
$(\Delta^{\prime}_{2},P^{\prime\prime}_{2},\Delta^{\prime\prime}_{2})$ covers
$(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime}_{1})$ in case
$\Delta^{\prime}_{1}=\Delta^{\prime}_{2}$ and $P^{\prime}_{1}=P^{\prime}_{2}$.

In case $\Delta^{\prime}_{1}=\Delta^{\prime}_{2}$ and
$\Delta^{\prime\prime}_{1}=\Delta^{\prime\prime}_{2}$, by Remark 9.20 (1),
$P^{\prime}_{1}<P^{\prime}_{2}=\mu_{G^{\prime}_{l}}P^{\prime}_{1}$ for some
tile $G^{\prime}_{l}$ with diagonal labeled $\alpha^{\prime}$. Otherwise, if
$(\Delta^{\prime}_{1},\mu_{G^{\prime}_{l}}P^{\prime}_{1},\Delta^{\prime\prime}_{1})$
does not cover
$(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime\prime}_{1})$, by Lemma
4.11, we have either $\Delta^{\prime}_{1}=\Delta^{\prime}_{1}(p)$ is the first
triangle in $\Delta(T^{\prime o},p)$ and $G^{\prime}_{l}$ is the first tile of
$G_{T^{\prime o},\widetilde{\beta}}$
$\Delta^{\prime\prime}_{1}=\Delta^{\prime}_{1}(q)$ is the first triangle in
$\Delta(T^{\prime o},q)$ and $G^{\prime}_{l}$ is the last tile of
$G_{T^{\prime o},\widetilde{\beta}}$.

Suppose that $\Delta^{\prime}_{1}=\Delta^{\prime}_{1}(p)$ is the first
triangle in $\Delta(T^{\prime o},p)$ and $G^{\prime}_{l}$ is the first tile of
$G_{T^{\prime o},\widetilde{\beta}}$. It follows that the diagonal
$G^{\prime}_{l}=G^{\prime}_{1}$ is labeled $\alpha^{\prime}$ and
$m_{0}(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime\prime}_{1})=1$. Thus
$m_{0}({\bf P})=-1$ and $m_{1}({\bf P})=1$, thus $(0,1)$ is an ${\bf m}({\bf
P};\alpha)$-pair. From the construction of $\pi$, we see that
$(\Delta^{\prime}_{1},\mu_{G^{\prime}_{l}}P^{\prime}_{1},\Delta^{\prime\prime}_{1})$
and $(\Delta^{\prime}_{1},P^{\prime}_{1},\Delta^{\prime\prime}_{1})$ can not
belong to $\pi({\bf P})$, a contradiction.

Suppose that $\Delta^{\prime\prime}_{1}=\Delta^{\prime}_{1}(q)$ is the first
triangle in $\Delta(T^{\prime o},q)$ and $G^{\prime}_{l}$ is the last tile of
$G_{T^{\prime o},\widetilde{\beta}}$. Similarly, $m_{\eta_{\alpha}}({\bf
P})=1$, $m_{\eta_{\alpha}+1}({\bf P})=-1$ and
$(\eta_{\alpha},\eta_{\alpha}+1)$ can not be an ${\bf m}({\bf
P};\alpha)$-pair. Thus, $\widetilde{\beta}=\alpha$, $\eta_{\alpha}=1$ and
$m_{0}({\bf P})=-1$ with $(0,1=\eta_{\alpha})$ is an ${\bf m}({\bf
P};\alpha)$-pair. It follows that $r=\sum m_{a}({\bf P})=-1+1-1=-1$,
contradicts to $r=\sum m_{a}({\bf P})\geq 0$.

The proof is complete. $\square$

In summary of Propositions 9.4, 9.8, 9.17, 9.12, 9.21 and Remarks 9.5, 9.10,
9.20, we obtain the following theorem.

######  Theorem 9.22.

For any ${\bf P}\in\mathcal{L}$, we have

  1. $(1)$

$|\pi({\bf P})|=2^{[\sum m_{i}({\bf P})]_{+}}$ and ${\bf m}(\bf
P^{\prime},\alpha^{\prime})=-{\bf m}(\bf P,\alpha)$ for all $\bf
P^{\prime}\in\pi(\bf P)$;

  2. $(2)$

There is a bijective map $\\{0,1\\}^{[\sum m_{i}({\bf P})]_{+}}\to\pi({\bf
P}),\vec{c}\mapsto{\bf P}(\vec{c})$ such that ${\bf P}(\vec{c})$ covers ${\bf
P}(\vec{c}\hskip 2.27621pt^{\prime})$ in $\mathcal{L}$ if and only if
$\vec{c}$ covers $\vec{c}\hskip 2.27621pt^{\prime}$ in the lattice
$\\{0,1\\}^{[\sum m_{i}({\bf P})]_{+}}$.

We need the following lemma for later use.

######  Lemma 9.23.

For any ${\bf P}\in\mathcal{L}$ with $r=\sum m_{i}({\bf P};\alpha)>0$, for any
$\ell\in\\{1,\cdots,r\\}$, assume that $\vec{c},\vec{c}\hskip
2.27621pt^{\prime}$ in $\\{0,1\\}^{r}$ satisfy
$c_{\ell}=1,c^{\prime}_{\ell}=0$ and $c_{i}=c^{\prime}_{i}$ for $i\neq\ell$,
we have

$\Omega({\bf P}(\vec{c}\hskip 2.27621pt^{\prime}),{\bf
P}(\vec{c}))=d(\alpha^{\prime})(2\ell-1-r).$

###### Proof.

We have $\Omega({\bf P}(\vec{c}\hskip 2.27621pt^{\prime}),{\bf
P}(\vec{c}))=d(\alpha^{\prime})\left(-(r-\ell)-(-\ell-1)\right)=d(\alpha^{\prime})(2\ell-1-r).$
$\square$

###### Proof.

It follows by Lemma 7.16 and Lemma 8.10. $\square$

The main result in this section is the following.

######  Theorem 9.24.

$\pi$ is a partition bijection with inverse $\pi^{\prime}$.

###### Proof.

By Theorem 9.22, $\pi({\bf P})\neq\emptyset$ for any ${\bf P}\in\mathcal{L}$.

If $\pi({\bf P})\cap\pi({\bf Q})\neq\emptyset$ for some ${\bf P},{\bf Q}$,
from the construction of $\pi$, ${\bf P}$ and ${\bf Q}$ are related by a
sequence of twists at $\alpha$. By Theorem 9.22, we have ${\bf m}(\bf
P,\alpha)={\bf m}(\bf Q,\alpha)$. Thus $\pi(P)=\pi(Q)$ by the construction of
$\pi$.

For any ${\bf P}\in\mathcal{L},{\bf P}^{\prime}\in\pi({\bf P})$, we have ${\bf
m}({\bf P},\alpha)$-pairs and ${\bf m}({\bf
P}^{\prime},\alpha^{\prime})$-pairs coincide by Lemmas 9.3, 9.7, 9.16 and
Theorem 9.22. Thus ${\bf P}\in\pi^{\prime}({\bf P}^{\prime})$ iff ${\bf
P}^{\prime}\in\pi({\bf P})$. Therefore, $\bigcup_{{\bf
P}\in\mathcal{L}}\pi({\bf P})=\mathcal{L}^{\prime}$.

Therefore, by Remark 2.5, $\pi$ is a partition bijection. Similarly,
$\pi^{\prime}$ is a partition bijection. Furthermore, $\pi^{\prime}$ is the
inverse of $\pi$ by Proposition 2.6. $\square$

##  10\. Compatibility of the partition bijections and the lattice structures

For any pair ${\bf P},{\bf Q}\in\mathcal{L}$, assume that ${\bf P}$ covers
${\bf Q}$, then we have that ${\bf P}$ and ${\bf Q}$ are either related by a
twist on a tile $G_{l}$ or related by two adjacent triangles incident to $p$
or $q$, assume $\tau$ is the label of the diagonal or the common side of the
triangles. In both cases, we say that ${\bf P}$ covers ${\bf Q}$ and _related
by $\tau$_. We denote

$\Omega({\bf P},{\bf Q})=w({\bf P})-w({\bf Q}).$

We say that ${\bf P},{\bf Q}\in\mathcal{L}$ are _related by a twist at
$\alpha$_ if either ${\bf P}$ covers ${\bf Q}$ and related by $\tau$ or ${\bf
Q}$ covers ${\bf P}$ and related by $\tau$.

Assumption: We always assume that $T$ and $T^{\prime}$ contain no arc tagged
notched at $q$ in case
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$, $T$ and $T^{\prime}$
contain no arc tagged notched at $p$ or $q$ in case
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(p,q)})$ if there is no
other state.

######  Proposition 10.1.

We have ${\bf P}^{\prime}_{-}\in\pi({\bf P}_{-})$. In particular, we have
${\bf P}^{\prime}_{-}={\bf P}_{-}(0,\cdots,0)$.

###### Proof.

By Lemma 7.19, $P^{\prime}_{-}\in\varphi_{\alpha}^{T^{o}}(P_{-})$. By
Proposition 7.15 we may assume $\sum m_{i}(P_{-})\geq 0$. By Lemma 7.20,
$m_{i}(P_{-})\geq 0$ for any $i$. We have
$\Delta^{\prime}_{1}(q)\in\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))$,
$\Delta^{\prime}_{1}(p)\in\phi_{\alpha}^{T^{o},p}(\Delta_{1}(p))$ from Lemma
8.8.

The situation that $\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta})$ follows
by Lemma 7.19 and Lemma 7.20.

We then consider the situation that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$.

Case 1. If $m(\Delta_{1}(q);\alpha)\geq 0$, as
$\Delta^{\prime}_{1}(q)\in\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))$, we see that
$(P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(P_{-},\Delta_{1}(q))$.

Case 2. If $m(\Delta_{1}(q);\alpha)<0$, then $m(\Delta_{1}(q);\alpha)=-1$ and
$\tau_{1}(q)=\alpha$ or $\tau_{t}(q)=\alpha$. Thus we have
$\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))=\\{\Delta^{\prime}_{1}(q)\\}$.

Case 2.1. $\tau_{1}(q)=\alpha$. We may assume that $\tau_{t}(q)=\alpha_{4}$
and $\tau_{[1]}(q)=\alpha_{1}$. We have either $\widetilde{\beta}=\tau_{t}(q)$
or $\widetilde{\beta}$ crosses $\alpha_{1}$ and ending at $q$.

Case 2.1.1. $\widetilde{\beta}=\tau_{t}(q)$. Then
$\widetilde{\beta}=\alpha_{4}\in T^{o}\cap T^{\prime o}$. Thus
$P_{-}=P^{\prime}_{-}=P_{\widetilde{\beta}}$ and
$\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-}\\}$ and thus
$(P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(P_{-},\Delta_{1}(q))$.

Case 2.1.2. $\widetilde{\beta}$ crosses $\alpha_{1}$ and ending at $q$. If
$\sum m_{i}(P_{-})=0$, then
$\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-}\\}$ and thus
$(P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(P_{-},\Delta_{1}(q))$. If $\sum
m_{i}(P_{-})>0$, then $m_{i}(P_{-})=1$ for some $i$. By Lemma 7.11, one the
following happens: $\widetilde{\beta}$ crosses $\alpha_{2},\alpha,\alpha_{4}$
consecutively; or $\widetilde{\beta}$ starts from the common endpoint of
$\alpha$ and $\alpha_{1}$ then crosses $\alpha_{4}$; or $\widetilde{\beta}$
starts from $q$ then crosses $\alpha_{2}$, see Figure 16. We have either
$\widetilde{\beta}$ intersects itself or the two endpoints are $q$, a
contradiction.

Figure 16.

Case 2.2. $\tau_{t}(q)=\alpha$, we may assume that $\tau_{1}(q)=\alpha_{1}$
and $\tau_{[1]}(q)=\alpha_{4}$. We have either $\widetilde{\beta}=\alpha$ or
$\widetilde{\beta}$ crosses $\alpha_{4}$ and ending at $q$.

Case 2.2.1. $\widetilde{\beta}=\alpha$. Thus we have
$\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))=\\{\Delta^{\prime}_{1}(q)\\}$,
$P_{-}=P_{\widetilde{\beta}}$ and
$\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-},P^{\prime}_{+}\\}$. For any
$(P^{\prime},\Delta^{\prime})\in\pi(P_{-},\Delta_{1}(q))$, from the
construction of $\pi$, the edges labeled $\alpha_{2}$ and
$\tau_{[1]}(q)=\alpha_{4}$ are in $P^{\prime}$. Thus
$(P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(P_{-},\Delta_{1}(q))$.

Case 2.2.2. $\widetilde{\beta}$ crosses $\alpha_{4}$ and ending at $q$. Then
$m_{\eta_{\alpha}}(P_{-})=1$ and thus $(\eta_{\alpha},\eta_{\alpha}+1)$ is an
${\bf m}({\bf P}_{-};\alpha)$-pair. For any
$(P^{\prime},\Delta^{\prime})\in\pi({\bf P}_{-})$, from the construction of
$\pi$, we have the edges labeled $\tau_{t-1}(q)=\alpha_{2}$ and
$\tau_{[1]}(q)=\alpha_{4}$ of $G^{\prime}(\eta_{\alpha})$ are in $P^{\prime}$.
Thus $(P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(P_{-},\Delta_{1}(q))$.

Last, we consider the situation that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(p,q)})$.

Case 1. $m(\Delta_{1}(p);\alpha),m(\Delta_{1}(q);\alpha)\geq 0$. As
$\Delta^{\prime}_{1}(p)\in\phi_{\alpha}^{T^{o},p}(\Delta_{1}(p))$ and
$\Delta^{\prime}_{1}(q)\in\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))$, we see that
$(\Delta^{\prime}_{1}(p),P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(\Delta_{1}(p),P_{-},\Delta_{1}(q))$.

Case 2. $m(\Delta_{1}(p);\alpha)\geq 0,m(\Delta_{1}(q);\alpha)<0$. Then
$m(\Delta_{1}(q);\alpha)=-1$ and $\tau_{1}(q)=\alpha$ or $\tau_{t}(q)=\alpha$.
Thus we have
$\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))=\\{\Delta^{\prime}_{1}(q)\\}$.

Case 2.1. $\tau_{1}(q)=\alpha$. We may assume that $\tau_{t}(q)=\alpha_{4}$
and $\tau_{[1]}(q)=\alpha_{1}$. We have either $\widetilde{\beta}=\tau_{t}(q)$
or $\widetilde{\beta}$ crosses $\alpha_{1}$ and ending at $q$.

Case 2.1.1. $\widetilde{\beta}=\tau_{t}(q)$. Then
$\widetilde{\beta}=\alpha_{4}\in T^{o}\cap T^{\prime o}$. Thus
$P_{-}=P^{\prime}_{-}=P_{\widetilde{\beta}}$ and $m(\Delta_{1}(p),\alpha)=0$.
Therefore, $\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-}\\}$ and thus
$(\Delta^{\prime}_{1}(p),P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi(\Delta_{1}(p),P_{-},\Delta_{1}(q))$.

Case 2.1.2. $\widetilde{\beta}$ crosses $\alpha_{1}$ and ending at $q$.

If $\sum m_{i}(P_{-})>0$, then $m_{i}(P_{-})=1$ for some $i$. From the above
discussion, we have $p=q$, $\widetilde{\beta}$ starts from $q$ then crosses
$\alpha_{2}$ and $m_{i}(P_{-})=0$ for all $i$ with $1<i<\eta_{\alpha}$. We see
that $\tau_{[1]}(p)=\alpha_{2}$, $m(\Delta_{1}(p);\alpha)=-1$ and
$m_{1}(P_{-})=1$. Contradicts to the assumption that
$m(\Delta_{1}(p);\alpha)\geq 0$.

If $\sum m_{i}(P_{-})=0$ and $m(\Delta_{1}(p);\alpha)=1$, then
$\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-}\\}$ and $\widetilde{\beta}$
starts from the common endpoint of $\alpha_{2}$ and $\alpha_{3}$ then crosses
$\alpha$ and $\alpha_{1}$, see Figure 17. Then we have
$\tau_{i_{1}}=\alpha,\tau_{i_{2}}=\alpha_{1}$, thus $m_{1}(P_{-})=-1$, see
Figure 17, contradicts to $\sum m_{i}(P_{-})=0$.

Figure 17.

Case 2.2. $\tau_{t}(q)=\alpha$, we may assume that $\tau_{1}(q)=\alpha_{1}$
and $\tau_{[1]}(q)=\alpha_{4}$. We have either $\widetilde{\beta}=\alpha$ or
$\widetilde{\beta}$ crosses $\alpha_{4}$ and ending at $q$.

Case 2.2.1. $\widetilde{\beta}=\alpha$. Then we have
$\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))=\\{\Delta^{\prime}_{1}(q)\\}$,
$P_{-}=P_{\widetilde{\beta}}$ and
$\Delta_{1}(p)=\\{\tau_{s}(p)=\alpha,\tau_{1}(p)=\alpha_{3},\tau_{[1]}(p)=\alpha_{2}\\}$.
Thus $\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-},P^{\prime}_{+}\\}$,
$\phi_{\alpha}^{T^{o},p}(\Delta_{1}(p))=\\{\Delta^{\prime}_{1}(p)\\}$ and
$(1,2)$ is an ${\bf m}({\bf P}_{-};\alpha)$-pair. For any
$(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime})\in\pi({\bf P})$, from the
construction of $\pi$, the edges labeled $\alpha_{2}=\tau_{[1]}(p)$ and
$\alpha_{4}$ are in $P^{\prime}$. Thus
$(\Delta^{\prime}_{1}(p),P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi({\bf
P}_{-})$.

Case 2.2.2. $\widetilde{\beta}$ crosses $\alpha_{4}$ and ending at $q$. Then
$m_{\eta_{\alpha}}(P_{-})=1$ and thus $\sum m_{i}(P_{-})>0$. As
$m(\Delta_{1}(p);\alpha)\geq 0$, we have $(\eta_{\alpha},\eta_{\alpha}+1)$ is
an ${\bf m}({\bf P}_{-};\alpha)$-pair. For any
$(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime})\in\pi({\bf P}_{-})$, from
the construction of $\pi$, we have the edges labeled
$\tau_{t-1}(q)=\alpha_{2}$ and $\tau_{[1]}(q)=\alpha_{4}$ of
$G^{\prime}(\eta_{\alpha})$ are in $P^{\prime}$. Thus
$(\Delta^{\prime}_{1}(p),P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi({\bf
P}_{-})$.

Case 3. $m(\Delta_{1}(p);\alpha)<0,m(\Delta_{1}(q);\alpha)\geq 0$. This case
is similar to case 2.

Case 4. $m(\Delta_{1}(p);\alpha),m(\Delta_{1}(q);\alpha)<0$. Thus we have
$\phi_{\alpha}^{T^{o},q}(\Delta_{1}(q))=\\{\Delta^{\prime}_{1}(q)\\}$ and
$\phi_{\alpha}^{T^{o},p}(\Delta_{1}(p))=\\{\Delta^{\prime}_{1}(p)\\}$.

If $\sum m_{i}(P_{-})\leq 0$, then
$\varphi_{\alpha}^{T^{o}}(P_{-})=\\{P^{\prime}_{-}\\}$. Thus
$\\{(\Delta^{\prime}_{1}(p),P^{\prime}_{-},\Delta^{\prime}_{1}(q))\\}=\pi({\bf
P}_{-})$.

If $\sum m_{i}(P_{-})>0$, then by Lemma 7.11 we see that
$\tau_{[1]}(p),\tau_{[1]}(q)\in\\{\alpha_{2},\alpha_{4}\\}$. For any
$(\Delta^{\prime},P^{\prime},\Delta^{\prime\prime})\in\pi({\bf P}_{-})$, from
the construction of $\pi$, we have the edges labeled $\alpha_{2}$ and
$\alpha_{4}$ of $G^{\prime}(a)$ are in $P^{\prime}$, where $a$ is in some
${\bf m}({\bf P}_{-};\alpha)$-pair. Thus
$(\Delta^{\prime}_{1}(p),P^{\prime}_{-},\Delta^{\prime}_{1}(q))\in\pi({\bf
P}_{-})$.

The proof is complete. $\square$

######  Theorem 10.1.

Under the Assumption, assume that ${\bf P}$ covers ${\bf Q}$ and related by
$\tau$. Denote ${\bf P}(0,\cdots,0)={\bf P}^{\prime}$ and ${\bf
Q}(0,\cdots,0)={\bf Q}^{\prime}$.

  1. $(1)$

If $\tau\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, then ${\bf
P}^{\prime}$ covers ${\bf Q}^{\prime}$ and related by $\tau$. Moreover, we
have

$\Omega({\bf P},{\bf Q})=\Omega({\bf P}^{\prime},{\bf Q}^{\prime}).$

  2. $(2)$

If $\tau=\alpha$, then we have either

$\bullet$ ${\bf P}^{\prime}={\bf Q}^{\prime}$ or

$\bullet$ ${\bf P}^{\prime}$ is covered by ${\bf Q}^{\prime}$ and related by
$\alpha^{\prime}$ with $\Omega({\bf P},{\bf Q})=-\Omega({\bf P}^{\prime},{\bf
Q}^{\prime}).$

###### Proof.

(1) As $\tau\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, we have
${\bf m}({\bf P};\alpha)={\bf m}({\bf Q};\alpha)$.

We first consider the situation that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta})$. Then ${\bf P}=P$ and ${\bf
Q}=\mu_{G_{l}}P$. By Proposition 7.17, there is a tile $G^{\prime}_{[l]}$ of
$G_{T^{\prime o},\widetilde{\beta}}$ with diagonal labeled $\tau$ such that
any $P^{\prime}\in\varphi_{\alpha}^{T^{o}}(P)$ can twist on
$G^{\prime}_{[l]}$, $P^{\prime}>\mu_{G^{\prime}_{[l]}}(P^{\prime})$, and
$\varphi_{\alpha}^{T^{o}}({\bf
Q})=\mu_{G^{\prime}_{[l]}}(\varphi_{\alpha}^{T^{o}}(P))$. Thus ${\bf
P}^{\prime}$ covers ${\bf Q}^{\prime}$ and related by $\tau$. As ${\bf
m}(P,\alpha)={\bf m}(\mu_{G_{l}}P,\alpha)$, we see that $\pi({\bf
Q})=\mu_{G^{\prime}_{[l]}}(\pi(P))$. By Proposition 7.17 (1), we have
$m^{\pm}(P,G_{l};\tau_{i_{l}})=m^{\pm}({\bf
P}^{\prime},G^{\prime}_{[l]};\tau_{i_{l}})$ and
$n(G_{l}^{\pm};\tau_{i_{l}})=n(G^{\prime\pm}_{[l]};\tau_{i_{l}})$, thus
$\Omega({\bf P}^{\prime},{\bf Q}^{\prime})=\Omega({\bf P},{\bf Q})$.

We then consider the situation that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$. Assume that ${\bf
P}=(P,\Delta_{j}(q))$.

If ${\bf Q}=(\mu_{G_{l}}P,\Delta_{j}(q))$, then by Proposition 7.17 there is a
tile $G^{\prime}_{[l]}$ of $G_{T^{\prime o},\widetilde{\beta}}$ with diagonal
labeled $\tau$ such that
$\pi(Q)=\\{(\mu_{G^{\prime}_{[l]}}(P^{\prime}),\Delta^{\prime})\mid(P^{\prime},\Delta^{\prime})\in\pi({\bf
P})\\}$ and
$(P^{\prime},\Delta^{\prime})>(\mu_{G^{\prime}_{[l]}}(P^{\prime}),\Delta^{\prime})$.
In particular, ${\bf P}^{\prime}>{\bf Q}^{\prime}$. Next, we show ${\bf
P}^{\prime}$ covers ${\bf Q}^{\prime}$. Otherwise, by Lemma 4.4 we have
$G^{\prime}_{[l]}$ is the last tile of $G_{T^{\prime o},\widetilde{\beta}}$
and $\Delta^{\prime}=\Delta^{\prime}_{1}(q)$ is the first triangle incidents
to $q$ in $T^{\prime o}$. As
$\tau\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, we see that
$G_{l}$ is the last tile of $G_{T^{o},\widetilde{\beta}}$ and
$m(\Delta^{\prime};\alpha)=0$ and thus
$\Delta_{j}(q)=\Delta^{\prime}=\Delta_{1}(q)$ is the first triangle incidents
to $q$ in $T^{o}$. By Lemma 4.4, ${\bf P}=(P,\Delta_{j}(q))$ does not cover
${\bf Q}=(\mu_{G_{l}}P,\Delta_{j}(q))$, a contradiction. Therefore, ${\bf
P}^{\prime}$ covers ${\bf Q}^{\prime}$. Assume that ${\bf
P}^{\prime}=(P^{\prime},\Delta^{\prime})$. Then ${\bf
Q}^{\prime}=(\mu_{G^{\prime}_{[l]}}(P^{\prime}),\Delta^{\prime})$. From the
previous case, we have
$\Omega(P,\mu_{G_{l}}P)=\Omega(P^{\prime},\mu_{G^{\prime}_{[l]}}(P^{\prime}))$.
By Lemma 8.10, we see that $\Omega({\bf P}^{\prime},{\bf
Q}^{\prime})=\Omega({\bf P},{\bf Q})$.

If ${\bf Q}=(P,\Delta_{j-1}(q))$, as
$\tau\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, we have
$m(\Delta_{j-1}(q);\alpha)=m(\Delta_{j}(q);\alpha)=0$. Thus ${\bf
P}^{\prime}=(P^{\prime},\Delta_{j}(q))$ for some $P^{\prime}$ and ${\bf
Q}^{\prime}=(P^{\prime},\Delta_{j-1}(q))$. In case neither $\Delta_{j-1}(q)$
nor $\Delta_{j}(q)$ is the first triangle incident to $q$ in $T^{\prime o}$,
by Lemma 4.5 ${\bf P}^{\prime}$ covers ${\bf Q}^{\prime}$. In case either
$\Delta_{j-1}(q)$ or $\Delta_{j}(q)$ is the first triangle incident to $q$ in
$T^{\prime o}$, as
$\tau=\tau_{j-1}(q)\neq\alpha,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$, we
see the last tiles of $G_{T^{o},\widetilde{\beta}}$ and $G_{T^{\prime
o},\widetilde{\beta}}$ are the same and $E_{1}(p)\in P$ iff
$E^{\prime}_{1}(p)\in P^{\prime}$. As ${\bf P}$ coves ${\bf Q}$, by Lemma 4.5
we have ${\bf P}^{\prime}$ covers ${\bf Q}^{\prime}$. By Lemma 7.16, we have
$\Omega({\bf P}^{\prime},{\bf Q}^{\prime})=\Omega(P,Q)$.

The situation that $\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(p,q)})$
is similar to the case
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$.

(2) As $\tau=\alpha$, we have ${\bf m}({\bf P};\alpha)={\bf m}({\bf
Q};\alpha)$. Suppose that ${\bf P}^{\prime}\neq{\bf Q}^{\prime}$.

We first consider the situation that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta})$. Then ${\bf P}=P$ and ${\bf
Q}=\mu_{G_{l}}P$. By Proposition 7.17 (2), we have $m_{k}(P)=-1$ and
$G_{l}=G(k)$ for some $k$. As ${\bf P}^{\prime}\neq{\bf Q}^{\prime}$, $k$ is
in some ${\bf m}(P,\alpha)={\bf m}(\mu_{G_{l}}P,\alpha)$-pair. Assume that
$(k,k^{\prime})$ is an ${\bf m}(P,\alpha)$-pair, then
$m_{k^{\prime}}(\mu_{G_{l}}P)=m_{k^{\prime}}(P)=1$. Thus the labels of ${\bf
P}^{\prime}\cap edge(G^{\prime}(k^{\prime}))$ and $P\cap edge(G(k))$ coincide,
the labels of ${\bf Q}^{\prime}\cap edge(G^{\prime}(k^{\prime}))$ and
$\mu_{G_{l}}P\cap edge(G(k))$ coincide. Therefore ${\bf
Q}^{\prime}=\mu_{G^{\prime}(k^{\prime})}{\bf P}^{\prime}$ and ${\bf
Q}^{\prime}>{\bf P}^{\prime}$. Denote
$G^{\prime}_{[l]}=G^{\prime}(k^{\prime})$. We have

$m^{+}(P,G_{l};\alpha)-n(G^{+}_{l};\alpha)=\sum_{i>k}m_{i}(P),\hskip
14.22636ptm^{-}(P,G_{l};\alpha)-n(G^{-}_{l};\alpha)=\sum_{i<k}m_{i}(P),$
$m^{+}({\bf
P}^{\prime},G^{\prime}_{[l]};\alpha^{\prime})-n(G^{\prime+}_{[l]};\alpha^{\prime})=\sum_{i>k^{\prime}}m_{i}({\bf
P}^{\prime}),\hskip 14.22636ptm^{-}({\bf
P}^{\prime},G^{\prime}_{[l]};\alpha^{\prime})-n({\bf
G}^{\prime-}_{[l]};\alpha^{\prime})=\sum_{i<k^{\prime}}m_{i}(P^{\prime}).$

It follows that

$\Omega(P,\mu_{G_{l}}P)=d(\alpha)(\sum_{i>k}m_{i}(P)-\sum_{i<k}m_{i}(P)),\hskip
14.22636pt\Omega({\bf P}^{\prime},{\bf
Q}^{\prime})=d(\alpha^{\prime})(\sum_{i>k^{\prime}}m_{i}({\bf
P}^{\prime})-\sum_{i<k^{\prime}}m_{i}({\bf P}^{\prime})).$

As ${\bf m}(P,\alpha)=-{\bf m}({\bf P}^{\prime},\alpha^{\prime})$ and
$(k,k^{\prime})$ is an ${\bf m}(P,\alpha)$-pair, we obtain
$\Omega(P,\mu_{G_{l}}P)=-\Omega({\bf P}^{\prime},{\bf Q}^{\prime})$.

We then consider the situation that
$\mathcal{L}=\mathcal{L}(T^{o},\widetilde{\beta}^{(q)})$. Assume that ${\bf
P}=(P,\Delta_{j}(q))$.

If ${\bf Q}=(\mu_{G_{l}}P,\Delta_{j}(q))$, then by Proposition 7.17 (2) we
have $m_{k}({\bf P})=-1$ and $G_{l}=G(k)$ for some $k$. As ${\bf
P}^{\prime}\neq{\bf Q}^{\prime}$, $k$ is in some ${\bf m}({\bf
P},\alpha)$-pair. Assume that $(k,k^{\prime})$ is an ${\bf m}({\bf
P},\alpha)$-pair, then $m_{k^{\prime}}({\bf Q})=m_{k^{\prime}}({\bf P})=1$.
Assume ${\bf P}^{\prime}=(P^{\prime},\Delta^{\prime})$.

In case $k^{\prime}\neq\eta_{\alpha}+1$, we have ${\bf
Q}^{\prime}=(\mu_{G^{\prime}(k^{\prime})}(P^{\prime}),\Delta^{\prime})>{\bf
P}^{\prime}$. We claim ${\bf Q}^{\prime}$ covers ${\bf P}^{\prime}$.
Otherwise, by Lemma 4.4 $G^{\prime}(k^{\prime})$ is the last tile of
$G_{T^{\prime o},\widetilde{\beta}}$ and