Finally, we check item (5) from the statement of <Ref>: that there are no linked pairs arising from arcs which lie in $S_1 \cup S_2 \cup S_3$. This is straightforward: <Ref> tells us that there is no pairwise linking between arcs in $S_1$ or in $S_3$, and one directly checks that $\alpha_{2,1}$ does not link with any arcs in $S_1$ or $S_3$. This concludes the construction of $B$ in the case where $d(2;1,2)=0$.

[size=fbox, colback=gray!30]The case where $d(2;1,2) \geq 2$.

Next, we detail the construction of $B$ in the case that $d(2;1,2) \geq 2$. As suggested by <Ref>, this requires some finesse – namely, we will start with our construction of $B$ in $\Gamma_1 \cup \Gamma_2$, and then attempt to build $B$ in $\Gamma_3$ in a naive way. When we do this, it is possible that we built something with a sink disk. In the event that we produced a sink disk, we modify the co-orientation of the arcs to eliminate the offending sector. Throughout this remainder of the construction, we let $\delta = d(2;1,2)$ and $\epsilon = d(1; 1,2)$. In particular, since $\beta$ is standardized, $\delta \geq 2$ and $\epsilon \geq 2$.

We already specified the construction of $B$ in $\Gamma_1 \cup \Gamma_2$ at the start of this proof. We now explain how to build $B$ in $\Gamma_3$: we first pivot $\beta$ about $\alpha_{3,\delta}$, and then apply our template to $\Gamma_3$. For convenience of labelling bands, we draw $B$ and $\beta$ in the original/pre-pivoted presentation.

With this construction in place, there are two possibilities at this stage: either

(B) $\mathcal{S}_3$ becomes a sink, or

(C) $\mathcal{S}_3$ is not a sink.

We analyze each of these cases in turn.

[size=fbox, colback=gray!10]Case (B): $\mathcal{S}_3$ became a sink.

Suppose Case (B) occurs, and $\mathcal{S}_3$ is a sink. This can happen for two reasons; either

$(B.1)$ $S_3$ contains no right pointing $\alpha_{3,i}$ arcs, as in <Ref>, or

$(B.2)$ $S_3$ contains some right pointing $\alpha_{3,i}$ arcs, but they co-bound a branch sector with the unique image arc in $S_3$; see <Ref>.

$\mathcal{P}$ at 44 100
$\mathcal{P}$ at 204 100

$\Gamma_1$ at 35 170
$\Gamma_2$ at 64 170
$\Gamma_3$ at 90 170

$\Gamma_1$ at 195 170
$\Gamma_2$ at 225 170
$\Gamma_3$ at 252 170

$\mathcal{P}$ at 44 310
$\mathcal{P}$ at 204 310

$\Gamma_1$ at 35 370
$\Gamma_2$ at 64 370
$\Gamma_3$ at 90 370

$\Gamma_1$ at 195 370
$\Gamma_2$ at 225 370
$\Gamma_3$ at 252 370

swap smoothing at 142 86
of $\alpha_{2,1}$ at 142 72
swap smoothing at 142 300
of $\alpha_{2,1}$ at 142 286

$\delta$ at 90 307
$\delta$ at 250 307
$\delta$ at 90 108
$\delta$ at 250 107

Understanding the $(B.1)$ case. Top Row: an example where $c_1=2$ and $\delta = d(2; 1,2) \geq 3$. We see that when $\alpha_{2,1}$ is a right pointer (as in the top row, left), $\mathcal{S}_3$ becomes a sink. So, we change the co-orientation of $\alpha_{2,1}$ and make it a left pointer (as in top row, right). Now, $\mathcal{S}_3$ is not a sink, because $\varphi(\alpha_{2,1})$ is a right pointer exiting $\mathcal{S}_3$. Bottom Row: an example where $c_1\geq 3$, and $\delta = d(2; 1,2) =2$. As in the top row, before doing any modifications, $\mathcal{S}_3$ is a sink, so we change the co-orientation of $\alpha_{2,1}$. Since we have not included any crossings from $\Gamma_4$, we dotted the vertical line corresponding to the $4^{\text{th}}$ strand of the braid. After the modification, the branch sector $\mathcal{P}$ is not a sink.

(Note: if $(B.1)$ occurs, then by the definition of our template, we must have that in the pre-pivot presentation of $\beta$, $d(3; \delta-2, \delta)=c_4$; in particular, we have that $d(3; \delta, 1) + d(3; 1, \delta-2) = 0$.)

$\mathcal{P}$ at 44 100
$\mathcal{P}$ at 204 100

$\Gamma_1$ at 35 170
$\Gamma_2$ at 64 170
$\Gamma_3$ at 90 170

$\Gamma_1$ at 195 170
$\Gamma_2$ at 225 170
$\Gamma_3$ at 252 170

$\mathcal{P}$ at 44 310
$\mathcal{P}$ at 204 310

$\Gamma_1$ at 35 370
$\Gamma_2$ at 64 370
$\Gamma_3$ at 90 370

$\Gamma_1$ at 195 370
$\Gamma_2$ at 225 370
$\Gamma_3$ at 252 370

swap smoothing at 142 86
of $\alpha_{2,1}$ at 142 72
swap smoothing at 142 300
of $\alpha_{2,1}$ at 142 286

$\delta$ at 90 307
$\delta$ at 250 307
$\delta$ at 90 100
$\delta$ at 250 100

Understanding the $(B.2)$ case. Top Row: an example where $c_1=2$, $\epsilon = d(1; 1,2) = 3$, and $\delta = d(2; 1,2) \geq 3$. The arc $\varphi(\alpha_{2,1})$ encloses the only right pointing plumbing arcs in $S_3$. We see that when $\alpha_{2,1}$ is a right pointer (as in the top row, left), $\mathcal{S}_3$ becomes a sink. Changing the co-orientation of $\alpha_{2,1}$ yields (top row, right). Now, $\mathcal{S}_3$ is not a sink. Bottom Row: an example where $c_1\geq 3$, and $\delta = d(2; 1,2) =2$. As in the top row, before doing any modifications, $\mathcal{S}_3$ is a sink, so we change the co-orientation of $\alpha_{2,1}$.

Regardless of whether $(B.1)$ or $(B.2)$ holds, we can quickly eliminate the existing sink disk sector $\mathcal{S}_3$: we first pivot $\beta$ with respect to the unique $\sigma_1$ which is enclosed by a left pointing plumbing arc, thereby “resetting” our original presentation of $\beta$. With this presentation, the unique co-oriented arc in $S_2$ is $\alpha_{2,1}$, which is currently a right pointer. We now make $\alpha_{2,1}$ a left pointer. The results are seen in <Ref> (for Case (B.1))and <Ref> (for Case (B.2)).

It is clear that items (1), (2), and (3) of the statement of <Ref> hold. We now analyze the branch sectors, and verify that there is at most one sink disk sector whose span lies in $\Gamma_1 \cup \Gamma_2$.

* We analyze the cases $c_1=2$ and $c_1 \geq 3$ separately.

$\boldsymbol{\circ}$ If $c_1=2$, then $\mathcal{S}_1 \sim \mathbbm{b}_{1,2} \sim \mathbbm{b}_{2, 2} \sim \mathbbm{b}_{3, \delta}$; see <Ref> (top row) and <Ref> (bottom row). Since $\mathbbm{b}_{3, \delta}$ is enclosed, on the left, by a left pointing plumbing arc, and this arc points out of $\mathcal{S}_1$, we deduce this sector is not a sink.

$\boldsymbol{\circ}$ Suppose $c_1 \geq 3$. Then $\alpha_{1,2}$ points out of $\mathcal{S}_1$, and the sector is not a sink.

The sector $\mathcal{S}_2$ contains $\varphi(\alpha_{1,1})$ in its boundary. This image arc is a right pointer, pointing out of $\mathcal{S}_2$, so this sector is not a sink. See <Ref> and <Ref>. The arc $\varphi(\alpha_{2,1})$ is part of $\partial \mathcal{S}_3$, and it points out of this disk sector. Thus, $\mathcal{S}_3$ is not a sink.

* By definition, a polygon sector must lie in a Seifert disk and also contain both plumbing and image arcs in its boundary. In this case, there are no polygon sectors spanning $\Gamma_1 \cup \Gamma_2$:

$\boldsymbol{\circ}$ $S_1$ contains only plumbing arcs, so it does not contain any polygon sectors.

$\boldsymbol{\circ}$ $S_2$ contains a sector co-bounded by both plumbing and image arcs, but this sector also contains $\mathbbm{b}_{1,2}$. Therefore, it does not lie in a Seifert disk, so it is not a polygon sector.

$\boldsymbol{\circ}$ Since $d(2;1,2)\geq 2$, and we applied our template to $\Gamma_3$ at $\alpha_{3,\delta}$, the band $\mathbbm{b}_{3,\delta-1}$ is not enclosed, on the left, by a plumbing arc; see <Ref> and <Ref>. In particular, the unique sector in $S_3$ which is co-bounded by both plumbing and image arcs must contain $\mathbbm{b}_{3,\delta-1}$, so it does not lie in a Seifert disk. Thus, there are no polygon sectors in $S_3$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_1$. We analyze the cases $c_1=2$ and $c_1 \geq 3$ separately.

$\boldsymbol{\circ}$ When $c_1 = 2$, we can apply the same argument as in Case (A).

$\boldsymbol{\circ}$ Suppose $c_1 \geq 3$.
If $\mathcal{H} \sim \mathbbm{b}_{1,1}$, then $\mathcal{H} \sim \mathcal{S}_2$, which we already proved is not a sink.

Suppose instead that $\mathcal{H} \sim \mathbbm{b}_{1,2}$. In this case, $\mathbbm{H} \sim \mathbbm{b}_{1,2} \sim \mathbbm{b}_{2,2} \sim \mathbbm{b}_{3, \delta}$. Since $\alpha_{3, \delta}$ is a left pointer, this arc points out of the sector, and $\mathcal{H}$ is not a sink.

The remaining bands $\mathbbm{b}_{1,3}, \ldots, \mathbbm{b}_{1,c_1}$ are enclosed, on the right, by the left pointers $\varphi(\alpha_{1,j-1})$; see <Ref> and <Ref>. So, no horizontal sector spanning $\Gamma_1$ is a sink.

Next, suppose $\mathcal{H}$ is a horizontal sector whose span includes $\Gamma_2$. There are four possibilities, each of which we analyze in turn. We refer the reader to <Ref> and <Ref> throughout.

$\boldsymbol{\circ}$ $\mathcal{H}$ is enclosed, on the left, by $\alpha_{2,1}$. But $\alpha_{2,1}$ is a left pointer exiting $\mathcal{H}$, so it is not a sink.

$\boldsymbol{\circ}$ $\mathcal{H}$ is enclosed, on the left, by $\varphi(\alpha_{1,1})$.

By construction, if $\mathcal{H}$ is enclosed on the left, by $\varphi(\alpha_{1,1})$, then $\mathcal{H} \sim \mathbbm{b}_{2,2}$. This band is enclosed, on the right, by $\alpha_{3, \delta}$, which is a left pointer. In particular, it points out of $\mathcal{H}$, so $\mathcal{H}$ is not a sink.

$\boldsymbol{\circ}$ $\mathcal{H}$ is enclosed, on the left, by $\varphi(\alpha_{1,j})$, $j \geq 2$. The arcs $\varphi(\alpha_{1,j})$, $j \geq 2$ are left pointers exiting $\mathcal{H}$, so such sectors are not sinks.

$\boldsymbol{\circ}$ $\mathcal{H}$ is not enclosed, on the left, by any arc. In this case, $\mathcal{H} \sim \mathcal{S}_2$, which is not a sink.

We deduce that there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$ in Case (B).

We now study the linking behavior between arcs in $S_1, S_2$, and $S_3$ for Case (B).

<Ref> confirms that there is no pairwise linking between arcs which are both in $S_1$ or both in $S_3$. There can be no pairwise linking between arcs in $S_1$ and $S_3$. Finally, one checks that we co-oriented $\alpha_{2,1}$ so that it does not link with arcs in either $S_1$ or $S_3$. Therefore, we have the desired linking behavior. This concludes the proof of <Ref> for Case $(B)$.

[size=fbox, colback=gray!10]Case (C): $\mathcal{S}_3$ did not become a sink when we applied our template to $\Gamma_3$.

We now turn to Case (C), and refer the reader to <Ref> throughout. Once again, we have presented $\beta$ in its original/pre-pivoted form for convinience while labelling bands. If, when we applied our template to $\Gamma_3$, we determined that $\mathcal{S}_3$ is not a sink, we must have that for some minimal $j$ with $\delta + 1 \leq j \leq c_3$, there is an arc $\alpha_{3,j}$ which is a right pointer; note that, by assumption, $\alpha_{3,j} \subset \partial \mathcal{S}_3$. Once again, items (1), (2), (3) from the conclusions of of <Ref> hold, so we need to check that there is at most potential branch sector spanning $\Gamma_1 \cup \Gamma_2$ which could be a sink.

$\mathcal{P}$ at 44 115
$\mathcal{P}$ at 198 115

$\Gamma_1$ at 35 175
$\Gamma_2$ at 64 175
$\Gamma_3$ at 90 175

$\Gamma_1$ at 190 175
$\Gamma_2$ at 215 175
$\Gamma_3$ at 245 175

$\mathcal{P}$ at 44 310
$\mathcal{P}$ at 198 310

$\Gamma_1$ at 35 370
$\Gamma_2$ at 64 370
$\Gamma_3$ at 90 370

$\Gamma_1$ at 190 370
$\Gamma_2$ at 218 370
$\Gamma_3$ at 245 370

$j$ at 86 272
$j$ at 238 237
$j$ at 85 75
$j$ at 238 58

$\delta$ at 86 300
$\delta$ at 238 300
$\delta$ at 85 108
$\delta$ at 238 108

Understanding Case (C). Let $\mathcal{H}$ denote the horizontal sector where $\mathcal{H} \sim \mathbbm{b}_{2,1} \sim \mathbbm{b}_{3,c_3}$ (this sector is orange in all frames). Top Row: In the top row, we have that $c_1=2$. On the left, we have an example where $j < c_3$, and so the horizontal sector $\mathcal{H}$ is not a sink. On the right, we have $j = c_3$, and so a priori, it is possible that the horizontal sector $\mathcal{H}$ is a sink. Bottom Row: The analogous frames when $c_1 \geq 3$.

* To see why $\mathcal{S}_1$ is not a sink, we must argue the cases where $c_1=2$ and $c_1 \geq 3$ separately.

$\boldsymbol{\circ}$  If $c_1=2$, then $\mathcal{S}_1$ contains $\mathbbm{b}_{1,2}$, and $\partial \mathcal{S}_1$ contains $\alpha_{2,1}$ in the boundary. Since this arc is a right pointer, pointing out of the sector, $\mathcal{S}_1$ is not a sink.

$\boldsymbol{\circ}$  If $c_1 \geq 3$, then $\alpha_{1,2}$ points out of $\mathcal{S}_1$, so this sector is not a sink.

As $\varphi(\alpha_{1,1}) \subset \partial \mathcal{S}_2$, and this arc is a right pointer, $\mathcal{S}_2$ is not a sink. By design, $\partial \mathcal{S}_3$ contains at least one outward pointing plumbing arc, namely $\alpha_{3,j}$.

* The same argument as in Case (B) verifies that there are no polygon sectors spanning $\Gamma_1 \cup \Gamma_2$. So, there are no polygon sectors which are sink disks.

* The same argument as Case (B) confirms that if $\mathcal{H}$ is a horizontal sector spanning exactly $\Gamma_1$, then $\mathcal{H}$ is not a sink. It remains to check that if $\mathcal{H}$ is a horizontal sector whose span includes $\Gamma_2$, then $\mathcal{H}$ is not a sink. We analyze the four possibilities in turn.

* $\mathcal{H}$ is enclosed, on the left, by $\alpha_{2,1}$. In this case, $\mathcal{H}$ meets $S_3$ in $\alpha_{3, c_3}$; see <Ref>. In particular, this means $\mathcal{H} \sim \mathbbm{b}_{3,c_3}$. This band is enclosed, on the right, by $\varphi(\alpha_{3,c_3-1})$, which can either be a right or left pointer; see <Ref>. If $j < c_3$, then $\varphi(\alpha_{3,c_3-1})$ is a left pointer (as in <Ref> (left)), and $\mathcal{H}$ is not a sink. Otherwise, $j=c_3$ (as in <Ref> (right)) and $\mathcal{H}$ is a right pointer – in this case,  $\mathcal{H}$ could be a sink disk. (This is our one potential sink disk spanning $\Gamma_1 \cup \Gamma_2$ in this case.)

* $\mathcal{H}$ is enclosed, on the left, by $\varphi(\alpha_{1,1})$. In this case, $\partial \mathcal{H}$ also contains $\alpha_{2,1}$, which points out of the sector. Thus, $\mathcal{H}$ is not a sink.

* $\mathcal{H}$ is enclosed, on the left, by $\varphi(\alpha_{1,k})$, $k \geq 2$. The arcs $\varphi(\alpha_{1,k})$, $k \geq 2$ are left pointers exiting $\mathcal{H}$, so such sectors are not sinks. See <Ref>.

* $\mathcal{H}$ is not enclosed, on the left, by any arc. In this case, $\mathcal{H} \sim \mathcal{S}_2$, which is not a sink.

Therefore, at present, there is at most one sector spanning $\Gamma_1 \cup \Gamma_2$ which could potentially be a sink – it is the horizontal sector $\mathcal{H}$ which contains $\mathbbm{b}_{2,1}$. This concludes the proof of (4) from the statement of <Ref> for the $(C)$ case. The same argument as in Case (B) confirms that we have no pairwise linking between any arcs in $S_1 \cup S_2 \cup S_3$.

This concludes the proof of <Ref>.

In <Ref>, we keep track of which sectors in $\Gamma_1 \cup \Gamma_2$ are the potential sink disks.

Case      Status of $\Gamma_3$      Potential Sink Disk contains

[fig:Generic_CmaxCodd_Gammas12]Case (A)     unknown whether homogeneous or heterogeneous     $\mathcal{S}_3 \sim \mathbbm{b}_{2,1}$

[fig:Generic_CmaxCodd_Gammas12_A]Case (B.1)     homogeneous      no sinks spanning $\Gamma_1 \cup \Gamma_2$

[fig:Generic_CmaxCodd_Gammas12_A.2]Case (B.2)     heterogeneous      no sinks spanning $\Gamma_1 \cup \Gamma_2$

[fig:Generic_CmaxCodd_Gammas12_B]Case (C)     heterogeneous     $\mathbbm{b}_{2,1} \sim \mathbbm{b}_{3, c_3}$

Bookkeeping for <Ref>, which builds $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ when $\Cmax=\Codd$. The third column details the unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$.

Next, we describe how to build $B$ in $\Gamma_1 \cup \Gamma_2$ when $\Cmax = \Ceven$.

Let $\beta$ be a standardized braid on $n \geq 5$ such that $\Cmax = \Ceven$. We can build a branched surface $B$ in $\Gamma_1 \cup \Gamma_2$ such that:

* $B$ contains $1$ product disk from $\Gamma_1$,

* $B$ contains $c_2-1$ product disks from $\Gamma_2$,

* $B$ contains no product disks from $\Gamma_3$,

* there is at most one sink disk sector spanning $\Gamma_1 \cup \Gamma_2$, and

* no arcs $\alpha, \alpha'$, where $\alpha, \alpha' \subset S_1 \cup S_2$, form a linked pair in $\tau_B$.

To begin, we will adapt the presentation of our braid. This will depend on the distribution of crossings in the braid.

[size=fbox, colback=gray!30]Case (1): $c_1 \geq 3$.

Since $\beta$ is standardized, we know that for all $1 \leq j \leq c_1$, $d(1; j, j+1) = 0$ or $d(1; j, j+1) \geq 2$. We cyclically conjugate $\beta$ so that it begins with a $\sigma_1$ letter, and so that $d(1; 1,2) \geq 2$. More specifically, we can cyclically conjugate our braid to write it as follows:

\begin{equation} \label{braid_special_form}
\beta \approx \sigma_1 w_0 \ \sigma_1^{k_1} w_1 \ \sigma_1^{k_2} w_2 \ \ldots \ \sigma_1^{k_j} w_{j} \ \sigma_1^{k_{j+1}}
\end{equation}

where the $w_i$ sub-words are non-empty for all $0 \leq i \leq j$, the $k_1, \ldots, k_j$ are all positive, and for $0 \leq i \leq j$, the $w_i$ contain no $\sigma_1$ letters. (We allow the possibility that $k_{j+1}$ = 0).

Either two things are possible:

(1A) There exists some $i$, where $0 \leq i \leq j$, where $w_i = \theta_0 \ \sigma_2^{\ell_1} \ \theta_1 \ \sigma_2^{\ell_2} \ \theta_2 \ \ldots \ \sigma_2^{\ell_t} \ \theta_t$ such that

$\bullet$ $\theta_0, \ldots, \theta_{t}$ contain no $\sigma_1$ or $\sigma_2$ letters,

$\bullet$ $\theta_0$ or $\theta_t$ could be empty but $\theta_1, \ldots, \theta_{t-1}$ are non-empty,

$\bullet$ $\ell_1, \ldots, \ell_t$ are all positive,

$\bullet$ when we restrict our distance function to the subword $w_i$, we have $d(2; 1, \sum_{r=1}^t \ell_r) \geq 2$.

(1B) No such $w_i$ exists.

To start, we focus our attention to Case (1A). Let $w_\star$ denote the first subword in $\beta$ for which the conditions specified in (1A) occur. We now more concisely write the braid in <Ref> as:

$$\beta \approx \beta_1 \sigma_1 w_\star \sigma_1 \beta_2$$

After cyclically conjugating our braid, we write it as

\begin{equation} \label{braid_special_form_1}
\beta \approx \sigma_1 w_\star \sigma_1 \beta_2 \beta_1
\end{equation}

By design, this is a presentation of $\beta$ such that $d(1; 1,2) \geq 2$, and $d(2; 1, d(1; 1,2)) \geq 2$. Going forwards, we will assume that if we are in Case (1A), then $\beta$ is presented as in <Ref>.

Suppose, instead, that when we analyze the presentation of the braid as in <Ref>, we fall into Case (1B). If this occurs, then for the braid in <Ref>, for all $i$ with $1 \leq i \leq c_2$, we have that $d(2; i, i+1) = d_L(2; i, i+1) = 0$, or $d(2; i, i+1)$ and $d_L(2; i, i+1)$ are simultaneously positive. There must exist some $i$ such that the latter occurs (otherwise we have a split link); so, we rewrite our braid to spotlight those first two $\sigma_2$ letters where this occurs, and cyclically conjugate as follows to produce a new presentation of the braid:

\begin{align}
%\beta &\approx z_1 \ \sigma_2 \ (\widetilde{\theta_1} \sigma_1 \widetilde{\theta_2} \sigma_3 \widetilde{\theta_3}) \  \sigma_2 \ z_2
\beta &\approx z_1 \ \sigma_2 \ (\theta_1 \sigma_1 \theta_2 \sigma_3 \theta_3) \  \sigma_2 \ z_2 \nonumber \\
\intertext{By assumption, we know that $\theta_1, \theta_2,$ and $\theta_3$ contain no $\sigma_2$ letters. We may assume that $\theta_2$ and $\theta_3$ contain no $\sigma_1$ letters (if they did, use braid relations to commute them into $\theta_1$). We cyclically conjugate $\beta$, and perform some braid relations, to get:}
&\approx (\theta_1 \underline{\sigma_1 \theta_2} \sigma_3 \theta_3) \  \sigma_2 \ z_2 \ z_1 \ \sigma_2  \nonumber \\
&\approx (\underline{\theta_1  \theta_2} \sigma_1 \sigma_3 \theta_3) \  \sigma_2 \ z_2 \ z_1 \ \sigma_2  \nonumber \\
&\approx (\sigma_1 \sigma_3 \theta_3) \  \sigma_2 \ z_2 \ z_1 \ \sigma_2 \ \theta_1 \  \theta_2 \label{braid_special_form_2}
\end{align}

Going forwards, we assume that if $\beta$ falls under Case (1B), then $\beta$ is presented as in <Ref>.

[size=fbox, colback=gray!30]Case (2): $c_1 = 2$.
Since $\beta$ is standardized and $\widehat{\beta}$ is prime, $d(1; 1,2)$ and $d(1; 2,1)$ are both at least two. Cyclically conjugate $\beta$ so that $\beta = \sigma_1 w_1 \sigma_1 w_2$. Without loss of generality, we may assume that $w_1$ contains at least one $\sigma_3$ letter (if not, then $w_2$ contains all $c_3$-many $\sigma_3$ letters in $\beta$, so conjugate the braid to read as $\sigma_1 w_2 \sigma_1 w_1$, and then swap the labelings of $w_1$ and $w_2$). We assume that $\beta$ is presented in this form for now. We now study the distribution of the $\sigma_3$ letters in $\beta$. Let $\delta := d(1; 1,2)$. Since $\beta$ is standardized, we know that $\delta \geq 2$. We either have:

(2A) $d(2; 1, \delta) \geq 1$.

(2B) $d(2; 1, \delta) = 0$, as in <Ref>. This can happen for two reasons:

(2B.1) $\beta = \sigma_1 \sigma_3^k w_1' \sigma_1 w_2$, where $k \geq 1$, or

(2B.2) $\beta = \sigma_1 w_1'  \sigma_3^k \sigma_1 w_2$, where $k \geq 1$, and $w_1'$ contains no $\sigma_3$ letters.

$\delta$ at 47 104
$\delta$ at 200 125

$\Gamma_1$ at 23 165
$\Gamma_2$ at 48 165
$\Gamma_3$ at 75 165

$\Gamma_1$ at 175 165
$\Gamma_2$ at 204 165
$\Gamma_3$ at 230 165

Left: An example of Case (B.1). Right: An example of Case (B.2).

Claim. After applying some braid moves, a braid of the form specified in Case (2B.2) can be put into the form specified in Case (2B.1).

Proof of Claim. This is straightforward. We begin with $\beta$ with the form specified in Case (2B.2).

\begin{align*}
\beta &= \sigma_1 w_1'  \underline{\sigma_3^k \sigma_1} w_2 \\
&= \underline{\sigma_1 w_1'} \sigma_1 \sigma_3^k w_2 \\
&\approx \sigma_1 \sigma_3^k w_2 \sigma_1 w_1'
\end{align*}

But this is exactly the form specified in Case (2B.1). End of proof of Claim.

Therefore, if $\beta$ appears as in Case (2B.2), we can and will modify the presentation and write it as in Case (2B.1). In fact, we will further modify this presentation in Case (2B.1) as follows:

\begin{align}
\beta & \approx \underline{\sigma_1 \sigma_3^k} w_1' \sigma_1 w_2 \nonumber \\
& \approx \underline{\sigma_3^{k-1}} \sigma_1 \sigma_3 w_1' \sigma_1 w_2 \nonumber \\
& \approx \sigma_1 \sigma_3 w_1' \sigma_1 w_2 \sigma_3^{k-1} \nonumber\\
& \approx \sigma_1 \sigma_3 w_1' \sigma_1 w_1'' \label{braid_special_form_3}
\end{align}

In particular, this means the $c_1=2$ case has only two sub-cases to consider:

(2A) $\beta \approx \sigma_1 w_1 \sigma_1 w_2$ and $d(2; 1, \delta) \geq 1$, or

(2B) $d(2; 1, \delta) = 0$, and $\beta \approx \sigma_1 \sigma_3 w_2 \sigma_1 w_1''$.

To summarize, we have four cases total; letting $\delta = d(1; 1,2)$, we have

* Case (1A): $c_1=3$, and $d(2; 1, \delta) \geq 2$, and $\beta \approx \sigma_1 w_\star \sigma_1 \beta_2 \beta_1$

* Case (2A): $c_1=2$, $d(2; 1, \delta) \geq 2$, and $\beta \approx \sigma_1 w_1 \sigma_1 w_2$

* Case (1B): $c_1=3$, but $d(2; 1, \delta) = 0$, and $\beta \approx \sigma_1 \sigma_3 \beta''$

* Case (2B): $c_1=2$, $d(2; 1, \delta)=0$, and $\beta \approx \sigma_1 \sigma_3 w_2 \sigma_1 w_1''$

We can apply a single construction to both Case (1A) and Case (2A), and a different construction to both Case (1B) and Case (2B). We now build $B$ in $\Gamma_1 \cup \Gamma_2$.

[size=fbox, colback=gray!30]Construction of $B$ in $\Gamma_1 \cup \Gamma_2$ in Case (1A) and Case (2A).

Note that not only is $d_L(2; 1, \delta) = 0$, but additionally, $\beta$ is standardized; therefore, there exists some minimal $j$, where $1 \leq j \leq \delta-1$ such that $d(2; j, j+1) \geq 2$. In particular, $d(2; 1, j) = 0$, and $d(2; j, j+1) \geq 2$, as seen in <Ref> (we will refer to this figure throughout the proof).

We now build $B$ in $\Gamma_1 \cup \Gamma_2$:

* smooth $\alpha_{1,1}$ to the left,

* smooth $\alpha_{2,1}, \ldots, \alpha_{2,j}$ to the left,

* smooth $\alpha_{2,j+1}, \ldots, \alpha_{2, c_2-1}$ to the right, and

* we take no product disks from $\Gamma_3$.

Items (1), (2), and (3) from the statement of the lemma are immediately satisfied; we need to check that there is a unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$. Note that by design, the arc $\varphi(\alpha_{2, j})$ encloses the bands $\mathbbm{b}_{3,1}, \ldots, \mathbbm{b}_{3, d(2; j, j+1)}$ on the left.

$\mathcal{P}$ at 44 105
$\mathcal{P}$ at 174 105

$\Gamma_1$ at 36 165
$\Gamma_2$ at 62 165
$\Gamma_3$ at 90 165

$\Gamma_1$ at 165 165
$\Gamma_2$ at 193 165
$\Gamma_3$ at 220 165

$\delta$ at 58 102
$\delta$ at 188 77

$j$ at 58 126
$j$ at 188 126

Examples where we building $B$ in $\Gamma_1 \cup \Gamma_2$ in Cases (1A) and (2A). We see that the sector $\mathcal{H} \sim \mathbbm{b}_{2, j+1} \sim \mathbbm{b}_{3,1}$ is the unique potential sink disk in $\Gamma_1 \cup \Gamma_2$. On the left, an example where $j+1 = \delta$, and on the right, an example where $j+1 < \delta$.

* We see that $\mathcal{S}_2 \sim \mathbbm{b}_{1,1}$, and so $\partial \mathcal{S}_2$ contains $\alpha_{1,1}$. This arc points out of the sector, so $\mathcal{S}_2$ is not a sink.

By design, we see that $S_2$ contains both left and right pointing plumbing arcs. Since we are choosing no product disks from $\Gamma_3$, we immediately deduce that $\mathcal{S}_3$ contains both left and right pointing image arcs, so $\mathcal{S}_3$ is not a sink.

Finally, we know that $\mathcal{S}_1 \sim \mathbbm{b}_{1,2} \sim \mathbbm{b}_{2, \delta}$, as in <Ref>.

$\bullet$ If $j+1 < \delta$, then $\mathbbm{b}_{2, \delta}$ is enclosed, on the right, by a left pointing image arc, as in <Ref> (right). In particular, we see that $\mathcal{S}_1$ is not a sink.

$\bullet$ If $j+1 = \delta$, then $\mathbbm{b}_{2, \delta}$ is enclosed, on the right, by a right pointer, as in <Ref> (left). Thus, $\mathcal{S}_1$ is actually a horizontal sector $\mathcal{H}$ for which, at present, we do not know the sink disk status.

* Of $S_1, S_2,$ and $S_3$, only $S_2$ contains both plumbing and image arcs, so this is the only Seifert disk spanning $\Gamma_1 \cup \Gamma_2$ with the potential to contain a polygon sector. Moreover, any such polygon sector must contain $\varphi(\alpha_{1,1})$ in its boundary. The number of polygon sectors in $S_2$ is at least one and at most two. There is always a polygon sector $\mathcal{P}$ whose boundary contains the arcs $\varphi(\alpha_{1,1})$, $\alpha_{2,1}, \ldots, \alpha_{2,\delta}$. By design, $\alpha_{2, \delta}$ points out of $\mathcal{P}$, so this sector is not a sink; see <Ref>.

If $d_L(2; \delta, \delta+1) \geq 2$, then there is another sector $\mathcal{P}'$ in $S_2$ which contains $\varphi(\alpha_{1,1})$ in its boundary. Recall that we only took one product disk from $\Gamma_1$. Therefore, $\mathcal{P}'$ also contains $\mathbbm{b}_{1,2}, \mathbbm{b}_{1, 3}, \ldots, \mathbbm{b}_{1, d_L(2; \delta, \delta+1) + 1}$. In particular, $\mathcal{P}' \sim \mathcal{S}_1$, and we already proved the latter is not a sink. We deduce that there are no polygon sector sink disks spanning $\Gamma_1 \cup \Gamma_2$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_1$. If $\mathcal{H} \sim \mathbbm{b}_{1,1}$, then $\mathcal{S} \sim \mathcal{S}_2$, which we already argued is not a sink. Otherwise, $\mathcal{H} \sim \mathbbm{b}_{1, k}$, where $2 \leq k \leq c_1$; this means $\mathcal{H} \sim \mathcal{S}_1 \sim \mathbbm{b}_{2, \delta}$. As described above, the sink disk status of this sector is determined by whether $\delta = j+1$ or $\delta \geq j+2$. In particular, if $\delta = j+1$, $\mathcal{H} \sim \mathbbm{b}_{3,1}$ (and we do not yet know the sink disk status of the sector), but if $\delta \geq j+2$, then $\mathcal{H}$ is not a sink.

Suppose $\mathcal{H}$ is a sector spanning $\Gamma_2$. Since $\alpha_{2, 1}, \ldots, \alpha_{2, j}$ are all left pointers, if $\mathcal{H} \sim \mathbbm{b}_{2, t}$, for $1 \leq t \leq j$, then $\partial \mathcal{H}$ contains an outward pointing arc in the boundary, and hence is not a sink. If $\mathcal{H}$ contains $\mathbbm{b}_{2, j+1}$, we cannot yet determine the sink disk status of this sector. Otherwise, $\mathcal{H}$ contains $\mathbbm{b}_{2, j+2}, \ldots, \mathbbm{b}_{2, c_2}$; as these bands are enclosed, on the right, by outward pointing image arcs, such sectors cannot be sinks. See <Ref>.

Therefore, there is exactly one horizontal sector which appears to be a sink: it is the sector $\mathcal{H} \sim \mathbbm{b}_{2, j+1} \sim \mathbbm{b}_{3,1}$.

We deduce, as desired, that there is a unique sector for which the sink disk status is unknown. The proof of <Ref> ensures that there is no linking between arcs which both lie in $S_2$, and one can inspect  <Ref> to see that $\alpha_{1,1}$ does not link with any arc in $S_2$. Therefore, there is no linking between arcs in $\Gamma_1 \cup \Gamma_2$.

[size=fbox, colback=gray!30]Construction of $B$ in $\Gamma_1 \cup \Gamma_2$ for Case (1B) and Case (2B).

To build $B$ in $\Gamma_1 \cup \Gamma_2$, we: smooth $\alpha_{1,1}$ to the left, smooth $\alpha_{2,1}, \ldots, \alpha_{2,c_2-1}$ to the right, and we take no product disks from $\Gamma_3$. We refer to
<Ref> throughout.

It is clear that items (1), (2), and (3) from the statement of the lemma are satisfied; so, we study the sink disks and linked arcs in $\Gamma_1 \cup \Gamma_2$. We claim that there is only one potential sink disk sector spanning $\Gamma_1 \cup \Gamma_2$; that sector contains $\mathbbm{b}_{2,1} \sim \mathcal{S}_3 \sim \mathbbm{b}_{3,1}$.

$\mathcal{P}$ at 42 102
$\mathcal{P}$ at 180 102
$\mathcal{P}$ at 342 102

$\Gamma_1$ at 33 165
$\Gamma_2$ at 62 165
$\Gamma_3$ at 90 165

$\Gamma_1$ at 174 165
$\Gamma_2$ at 200 165
$\Gamma_3$ at 230 165

$\Gamma_1$ at 335 165
$\Gamma_2$ at 362 165
$\Gamma_3$ at 390 165

$\delta$ at 62 102
$\delta$ at 200 102
$\delta$ at 360 102

Some examples where we build $B$ in $\Gamma_1 \cup \Gamma_2$ in Cases (1B) and (2B). We see that $\mathcal{S}_2 \sim \mathbbm{b}_{1,1}$ and $\mathcal{S}_1 \sim \mathbbm{b}_{1,2}$, and that neither are sink disks. Additionally, we see that $\mathcal{S}_3$ appears to be a potential sink.

* By design, $\mathcal{S}_2 \sim \mathbbm{b}_{1,1}$. This band is enclosed, on the left, by the left pointer $\alpha_{1,1}$, so $\mathcal{S}_2$ is not a sink.

We see that $\mathcal{S}_1 \sim \mathbbm{b}_{1, t}$, where $2 \leq t \leq c_1$. In particular, $\mathbbm{b}_{1,2}$ is enclosed, on the right, by $\alpha_{2,\delta}$, so $\mathcal{S}_1 \sim \mathbbm{b}_{2,\delta}$. As we have taken no product disks from $\Gamma_3$, we know that $\partial \mathcal{S}_1$ contains $\varphi(\alpha_{2, \delta-1})$, which is a left pointer. Therefore, $\mathcal{S}_1$ is not a sink.

As $S_3$ contains only left pointing image arcs, and $\mathcal{S}_3 \sim \mathbbm{b}_{2,1}$, we see that $\partial \mathcal{S}_3$ contains only inward pointing arcs. Therefore, at present, we cannot determine the sink disk status of $\mathcal{S}_3$.

* Of $S_1, S_2,$ and $S_3$, only $S_2$ contains both plumbing and image arcs, so this is the only Seifert disk spanning $\Gamma_1 \cup \Gamma_2$ with the potential to contain a polygon sector. Indeed, there is a polygon sector $\mathcal{P}$ whose boundary contains the arcs $\varphi(\alpha_{1,1})$, $\alpha_{2,1}, \ldots, \alpha_{2,\delta}$. By design, all the plumbing arcs $\alpha_{2,1}, \ldots, \alpha_{2, \delta}$ point out of $\mathcal{P}$, so this sector is not a sink.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_1$. As noted in our Seifert disk sector analysis, every band in $\Gamma_1$ is in the same sector as either $\mathcal{S}_1$ or $\mathcal{S}_2$. Since neither are sinks, no horizontal sector spanning $\Gamma_1$ is a sink.

Suppose $\mathcal{H}$ is a sector spanning $\Gamma_2$. We see that $\mathbbm{b}_{2,1} \sim \mathcal{S}_3$; we already described why we cannot know the sink disk status of this sector just yet. However, suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_2$ and $\mathcal{H} \sim \mathbbm{b}_{2,j}$, for some $2 \leq j \leq c_2$. Such a band $\mathbbm{b}_{2,j}$ is enclosed, on the right, by $\varphi(\alpha_{2,j-1})$, and this arc points out of $\mathcal{H}$, so such sectors are not sinks.

This confirms item (4) in the statement of the lemma. Finally, we check that there are no linked pairs between arcs in $S_1$ and $S_2$. The proof of <Ref> confirms that there can be no pairwise linking between arcs in $S_2$, and by design, $\alpha_{1,1}$ does not link with any arc in $S_2$. This completes the proof of the lemma for Case (1B) and Case (2B).

We summarize the distinguished bands and potential sink disks from the constructions of <Ref>.

Case      Distinguished Band     Potential Sink Disk Sector contains

[fig:NotGeneric_CmaxCeven_Gammas12_CaseB]Cases (1A), (2A)     $\mathbbm{b}_{2, j+1}$     $\mathbbm{b}_{2,j+1} \sim \mathbbm{b}_{3, 1}$

[fig:NotGeneric_CmaxCeven_Gammas12_CaseA]Cases (1B), (2B)     $\mathbbm{b}_{2, 1}$     $\mathbbm{b}_{2,1} \sim \mathcal{S}_3 \sim \mathbbm{b}_{3,1}$

Bookkeeping for <Ref>, which builds $B$ in $\Gamma_1 \cup \Gamma_2$ when $\Cmax=\Ceven$.

<Ref> will be vital as we construct $B$ to non-trivially spans all the columns of the braid.

§.§ Building the branched surface in interior columns

To build our laminar branched surface so that it spans many columns of the braid, we will need to fluidly move throughout the braid to decide where apply our templates. This requires the following definitions and lemmas.

Let $\beta$ be a braid word, and let $\beta'$ be some cyclic conjugation thereof.

Let $F$ and $F'$ be the Bennequin surfaces directly constructed from the braid words $\beta$ and $\beta'$, respectively. $($As remarked in <Ref>, $F$ and $F'$ are isotopic.$)$

Suppose $\sigma_{s,t}$ $($resp. $\mathbbm{b}_{s, t})$ is a distinguished letter $($resp. band$)$ in $\beta$ $($resp. $F)$ which, after cyclically conjugating to $\beta'$, becomes $\sigma_{s,t'}$ $($resp. $\mathbbm{b}_{s,t'})$ in $F'$.

We say $\beta'$ is a calibration of $\beta$ with respect to $\alpha_{s,t}$ $($or $\mathbbm{b}_{s,t})$ if in $\beta'$, the plumbing arc $\alpha_{s+1,1}$ encloses $\mathbbm{b}_{s,t'}$ on the right.

For a simple example of a calibration, see <Ref>.

$\Gamma_{s-1}$ at 52 175
$\Gamma_{s}$ at 80 175
$\Gamma_{s+1}$ at 106 175

$\Gamma_{s-1}$ at 236 175
$\Gamma_s$ at 264 175
$\Gamma_{s+1}$ at 292 175

$\mathbbm{b}_{s,t}$ at 76 126
$\mathbbm{b}_{s,t'}$ at 260 126

calibrate w.r.t. $\mathbbm{b}_{s,t}$ at 170 105

Left: We see $\beta$ and a distinguished band $\mathbbm{b}_{s,t}$ (here, the band is $\mathbbm{b}_{s,1}$).  The dotted orange line shows the axis along which we need to cyclically conjugate the braid. Right: The result after we (canonically) calibrate $\beta$ with respect to $\mathbbm{b}_{s,1}$, producing $\beta'$.  Note that in $\beta'$, the distinguished band has become $\mathbbm{b}_{s,2}$.

Fix a braid word $\beta$. For any distinguished band $\mathbbm{b}_{s,t}$ in $\beta$, where $1\leq s \leq n-2$, there exists a calibration $\beta'$ of $\beta$ with respect to $\mathbbm{b}_{s,t}$.

Let $\beta$ be a braid word and let $\sigma_{s,t}$ be a distinguished letter therein. Recall that $\mathbbm{b}_{s,t}$ is the band corresponding to $\alpha_{s,t}$ in $\beta$. In particular, we can write our braid as $\beta = \gamma_1 \sigma_s \gamma_2$, where the distinguished $\sigma_s$ is the distinguished letter $\sigma_{s,t}$. We pivot $\beta$ with respect to $\sigma_s$, producing the braid word $\widetilde{\beta}$, where $\widetilde{\beta} = \sigma_s \gamma_2 \gamma_1$. We next identify the last $\sigma_{s+1}$ letter in this presentation of the braid; that is, we write $\widetilde{\beta} = \sigma_s \ \widetilde{\gamma_2} \ \sigma_{s+1} \ \widetilde{\gamma_1}$, where there are no $\sigma_{s+1}$ letters in $\widetilde{\gamma_1}$. Pivoting $\beta$ with respect to $\sigma_{s+1}$, we produce $\beta' = \sigma_{s+1} \ \widetilde{\gamma_1} \ \sigma_s \ \widetilde{\gamma_2}$. (Note that $\widetilde{\gamma_1}$ could contain other $\sigma_s$ letters, and that is allowed.)

We claim that this is a calibration of $\beta$ with respect to $\mathbbm{b}_{s,t}$: by design, the presentation $\beta'$ of the braid is such that the first occurrence of a $\sigma_{s+1}$ letter appears before the distinguished $\sigma_s$ letter. Since $\beta$ is prime, we know that $c_{s+1} \geq 2$. In particular, a second occurrence of a $\sigma_{s+1}$ letter exists in $\beta'$, it happens after our distinguished $\sigma_s$ letter, and it somewhere in $\widetilde{\gamma_2}$. The plumbing arc $\alpha_{s+1,1}$ lies in $S_{s+1}$, has one endpoint above $\mathbbm{b}_{s+1,1}$, and the other above $\mathbbm{b}_{s+1,2}$. Moreover, $\alpha_{s+1,1}$ will enclose, on the left, all the bands $\mathbbm{b}_{s,1}, \ldots, \mathbbm{b}_{s, d_L(s+1; 1,2)}$ (as in <Ref> (right)). After calibration, our distinguished band is $\mathbbm{b}_{s,t'}$, where $1 \leq t' \leq d_L(s+1; 1,2)$. We conclude that $\alpha_{s+1,1}$ encloses our distinguished band on the right.

We remark: there are many cyclic conjugations of $\beta$ which are also calibrations of $\beta$ with respect to $\mathbbm{b}_{s,t}$.
We do not want to consider these other calibrations, so we define the following.

The canonical calibration of $\beta$ with respect to $\mathbbm{b}_{s,t}$ is the calibration of $\beta$ produced via the algorithm described in the proof of <Ref>.

Suppose we built a branched surface containing product disks from some non-trivial subset of the columns in $\Gamma_1 \cup \ldots \cup \Gamma_i$, where $4 \leq i \leq n-4$. Additionally, assume that we applied our template to $\Gamma_i$ to build $B$ in this column – in particular, assume that our branched surface meets $S_i$ in only plumbing arcs and no image arcs.
After performing some pivots and canonical calibrations, we can build our branched surface in $\Gamma_{i+1} \cup \Gamma_{i+2}$ such that: $(1)$ there are no product disks chosen from $\Gamma_{i+1}$, $(2)$ we applied our template to $\Gamma_{i+2}$ $($so there are $c_{i+2}-1$ co-oriented product disks chosen from $\Gamma_{i+2})$, $(3)$ the unique potential sink disk spanning $\Gamma_i \cup \Gamma_{i+1}$ is $\mathcal{S}_{i+2}$, and $(4)$ there is no pairwise linking between arcs in $\Gamma_i \cup \Gamma_{i+1} \cup \Gamma_{i+2}$.

Suppose we built $B$ in $\Gamma_1 \cup \ldots \cup \Gamma_i$ such that we already applied our template to $\Gamma_i$. (We refer the reader to <Ref> to recall the design of our templates.) Therefore, there is a unique band $\mathbbm{b}_{i, k}$ such that $\mathbbm{b}_{i, k-1}$ is not enclosed, on the left, by a plumbing band. We pivot the braid about $\mathbbm{b}_{i, k}$, so that $\beta \approx \sigma_i \omega$; in this new presentation, $\sigma_{i,1}$ is enclosed (on the left) by a left pointing plumbing arc, and $\sigma_{i, c_i}$ is not enclosed (on the left) by any plumbing arcs. We refer to this cyclic presentation of the braid as $P_1$, and <Ref> (left) shows some examples of $\Gamma_i \cup \Gamma_{i+1}$ in $P_1$.

There are two cases to consider:

$\boldsymbol{\circ}$ Case (1): The heterogenous case. Here, $d(i; 1, c_i - 1) > 0$, as in <Ref> (top row), or

$\boldsymbol{\circ}$ Case (2): The homogeneous case. Here, $d(i; 1, c_i-1) = 0$, as in <Ref> (bottom row).

We begin by building our branched surface in $\Gamma_{i+2}$, and then go on to show that, in both the heterogeneous and homogeneous cases, we created as most one sink disk.

We quickly establish some notation: let $\mathbbm{b}_{i,p}$ denote the first $\sigma_i$ band in $\Gamma_i$ which is followed by $\sigma_{i+1}$ bands; that is, $d(i; 1, p) = 0$, and $d(i; p, p+1) \geq 1$. Note that when $\Gamma_{i}$ is homogeneous, $p = c_{i-1}$. To build $B$, we canonically calibrate $\beta$ with respect to $\mathbbm{b}_{i+1,1}$, thereby presenting our braid as $\sigma_{i+2} \omega_1 \sigma_{i+1} \omega_2$, where the $\sigma_{i+1}$ letter written explicitly used to be $\sigma_{i+1,1}$ (in our old presentation of the braid as $P_1$). With this presentation of $\beta$, we apply our template to $\Gamma_{i+2}$. The result is seen in <Ref>, where we show an incomplete picture of what the branched surface looks like in $\Gamma_i \cup \Gamma_{i+1} \cup \Gamma_{i+2}$ with respect to the $P_1$ presentation of the braid.

$\mathcal{H}$ at 41 320
$\mathcal{H}$ at 244 320

$\Gamma_i$ at 32 410
$\Gamma_{i+1}$ at 60 410
$\Gamma_{i+2}$ at 88 410

$\Gamma_i$ at 234 410
$\Gamma_{i+1}$ at 262 410
$\Gamma_{i+2}$ at 290 410

$\mathcal{H}$ at 41 52
$\mathcal{H}$ at 244 52

$\Gamma_i$ at 32 186
$\Gamma_{i+1}$ at 60 186
$\Gamma_{i+2}$ at 88 186

$\Gamma_i$ at 234 186
$\Gamma_{i+1}$ at 262 186
$\Gamma_{i+2}$ at 290 186

calibrate w.r.t. $\mathbbm{b}_{i+1,1}$ at 158 322
calibrate w.r.t. $\mathbbm{b}_{i+1,1}$ at 158 100

$\mathbbm{b}_{i+1,1}$ at 60 352
$\mathbbm{b}_{i+1,2}$ at 258 352
$\mathbbm{b}_{i+1,1}$ at 60 90
$\mathbbm{b}_{i+1,2}$ at 258 90

$p$ at 28 370

$p$ at 25 98

Top Row: We see an example of the heterogeneous case, how to canonically calibrate $\beta$ (on the left) with respect to $\mathbbm{b}_{i+1,1}$ to product $\beta'$ (on the right). The sector $\mathcal{P}$ is not a sink, because $\alpha_{i+2,1}$ points out of the region. Bottom Row: we see an analogous example when $\Gamma_i$ is homogeneous. Note: in both rows, we do not know whether $\alpha_{i+2,t}$, where $t \neq 1$, is a left or right pointer, so we have suppressed that information.

[size=fbox, colback=gray!30]Case (1): Sink disk analysis when $\Gamma_i$ is heterogeneous.

We now verify that there are no sink disk sectors spanning $\Gamma_i \cup \Gamma_{i+1}$ in the heterogeneous case, and refer to <Ref> (top row) throughout. Note that unless stated otherwise, our labelling of bands is with respect to the $P_1$ presentation of $\beta$.

$\boldsymbol{\circ}$ By assumption, $S_i$ contains only plumbing arcs, and $\Gamma_i$ is heterogeneous. Therefore, $\partial \mathcal{S}_i$ contains $\alpha_{i, p+1}$, which points out of the sector; thus, $\mathcal{S}_i$ is not a sink.

$\boldsymbol{\circ}$ Since we take no disks from $\Gamma_{i+1}$, $\varphi(\alpha_{i, 1})$ is contained in $\partial \mathcal{S}_{i+1}$, pointing out of the sector. Thus, $\mathcal{S}_{i+1}$ is not a sink.

$\boldsymbol{\circ}$ Though we are assuming that we have applied our template to $\Gamma_{i+2}$, at present, we cannot argue that $\mathcal{S}_{i+2}$ is not a sink: we do not know whether $\Gamma_{i+2}$ is heterogeneous or homogeneous. Therefore, this is a potential sink disk sector.

* By definition, a polygon sector must lie in a Seifert disk and also contain both plumbing and image arcs in its boundary. By assumption, $S_{i}$ contains only plumbing arcs, so $S_i$ cannot contain any polygon sectors. We have taken no product disks from $\Gamma_{i+1}$, and have applied our template to $\Gamma_{i+2}$. Therefore, $S_{i+1}$ contains only image arcs, and $\Gamma_{i+2}$ contains only plumbing arcs. Therefore, none of $S_i, S_{i+1},$ and $S_{i+2}$ contain both plumbing and image arcs, and so there are no polygon sectors in $\Gamma_i \cup \Gamma_{i+1}$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_i$. There are three cases to consider:

* $\mathcal{H}$ contains $\mathbbm{b}_{i, j}$, where $1 \leq j \leq p$. Here, the bands $\mathbbm{b}_{i,j}$ are enclosed, on the left, by an outward pointing plumbing arc; so these sectors are not sinks; see <Ref> (top row).

* $\mathcal{H}$ contains $\mathbbm{b}_{i,p+1}$. This band is enclosed by inward pointing arcs on both the right and left, as in <Ref> (top row, left). When we canonically calibrated our braid with respect to $\mathbbm{b}_{i+1, 1}$, and then applied our template to $\Gamma_{i+2}$, we ensured that the sector $\mathcal{H}$ meets $S_{i+2}$ in a left pointing plumbing arc (in the new presentation of $\beta$, this arc is $\alpha_{i+2,1})$.
We deduce that $\mathcal{H}$ is not a sink.

* $\mathcal{H}$ contains $\mathbbm{b}_{i, j}$, where $p+2 \leq j \leq c_i$. By design, the band $\mathbbm{b}_{i,j}$ is enclosed, on the right, by an outward pointing image arc. Therefore, $\mathcal{H}$ is not a sink.

We now analyze the horizontal sectors spanning $\Gamma_{i+1}$. There are three cases to consider.

$\bullet$ The bands $\mathbbm{b}_{i+1, 1}, \ldots, \mathbbm{b}_{i+1, d(i; p, p+1)}$ are all part of the same branch sector $\mathcal{H}$; this sector also contains $\mathbbm{b}_{i, p+1}$. We already argued (in our analysis of the horizontal sectors spanning $\Gamma_i$) that the sector containing this band is not a sink.

$\bullet$ The bands $\mathbbm{b}_{i+1, j}$, where $j \geq d(i; p, p+1) +1$ are either enclosed (on the left) by a left pointing image arc,  or they are not enclosed (on the left) by any arc. When the former happens, the arc points out of the horizontal sector, so it is not a sink; when the latter happens, the horizontal sector also contains $\mathcal{S}_{i+1}$, which we already proved is not a sink.

We deduce that in Case (1), $\mathcal{S}_{i+2}$ is the only potential sink disk spanning $\Gamma_i \cup \Gamma_{i+1}$.

[size=fbox, colback=gray!30]Case (2): Sink disk analysis when $\Gamma_i$ is homogeneous.

We now do our sink disk analysis in Case (2), where $d(i; 1, c_i-1)=0$; a typical example appears in <Ref> (bottom row). Since we are assuming that $\beta$ is not a connected sum, we must have that $d(i; c_i - 1, c_i) := \delta_1 >0$ and $d(i; c_i, 1) =: \delta_2 >0$. Notice that in this case, the Seifert disk $S_i$ does not contain any right pointing plumbing arcs, which is why this analysis differs from Case (1).

* In this case, since $\mathcal{S}_i$ contains only plumbing arcs, $S_i \sim \mathbbm{b}_{i, c_i} \sim \mathbbm{b}_{i+1,1}$. By design, we canonically calibrated our braid with respect to $\mathbbm{b}_{i+1,1}$, and then applied our template to $\Gamma_{i+2}$. So, the band known as $\mathbbm{b}_{i+1,1}$ in the $P_1$ presentation of $\beta$ is subsequently enclosed, on the right, by a left pointing plumbing arc; see <Ref> (bottom row, right). Therefore, $\mathcal{S}_i$ is not a sink. $S_{i+1}$ contains only outward pointing image arcs, so it is not a sink. We cannot yet conclude that $S_{i+2}$ is not a sink, as we do not yet know whether $S_{i+2}$ is heterogeneous or homogeneous.

* By the same argument from Case (1), there are no polygon sectors spanning $\Gamma_{i} \cup \Gamma_{i+1} \cup \Gamma_{i+2}$.

* No horizontal sector whose span includes $\mathbbm{b}_{i,j}$, where $1 \leq j \leq c_i-1$, is a sink: as the band $\mathbbm{b}_{i,j}$ is enclosed, on the left, by the outward pointing plumbing arc $\alpha_{i,j}$. See <Ref> (bottom row, right). The sector containing $\mathbbm{b}_{i, c_i}$ is the branch sector $\mathcal{S}_{i}$, which we already argued is not a sink.

If a horizontal sector $\mathcal{H}$ spans $\Gamma_{i+1}$, it contains some band $\mathbbm{b}_{i+1, k}$. There are two cases to consider, based on whether $1 \leq k \leq \delta_1$ or $\delta_1 + 1 \leq k \leq c_{i+1}$.

Suppose $\mathcal{H}$ contains $\mathbbm{b}_{i+1, k}$ with $1 \leq k \leq \delta_1$. These bands are all enclosed, on the left, by $\varphi(\alpha_{i, c_i-1})$, and so they are all in the same sector as $\mathcal{S}_i$; we already proved this sector is not a sink. Otherwise, $\mathcal{H}$ contains $\mathbbm{b}_{i+1, k}$ where $\delta_1 + 1 \leq k \leq c_{i+1}$, then $\mathcal{H}$ is in the same horizontal sector as $\mathcal{S}_{i+1}$, which we already argued is not a sink. See <Ref> (bottom row).

We deduce that in Case (2), $\mathcal{S}_{i+2}$ is the only potential sink disk spanning $\Gamma_i \cup \Gamma_{i+1}$. <Ref> confirms that there is no pairwise linking between arcs which are both in $S_1$ or both in $S_3$, and there can be no pairwise linking between arcs in $S_1$ and $S_3$.

We summarize our findings in <Ref>.

Case      Distinguished Band     Potential Sink Disk Sector contains

$\Gamma_i$ is heterogeneous     $\mathbbm{b}_{i+1, 1}$     $\mathcal{S}_{i+2}$

$\Gamma_i$ is homogeneous     $\mathbbm{b}_{i+1, 1}$     $\mathcal{S}_{i+2}$

Bookkeeping for <Ref>, where we build $B$ in the “interior” columns of the braid.

In <Ref>, we assumed that we had already built our branched surface in $\Gamma_1 \cup \ldots \Gamma_i$, where $i \geq 4$. We now prove an analogous statement to <Ref> for the case where $i=3$ and $\Cmax = \Codd$.

Suppose $\Cmax = \Codd$, and $n \geq 7$. Further suppose that we followed <Ref> to construct a branched surface for $\widehat{\beta}$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ $($so, in particular, we added $c_1-1$ $($resp. $c_3-1$$)$ co-oriented product disks from $\Gamma_1$ $($resp. $\Gamma_3$$)$ into $B$, and exactly one co-oriented product disk from $\Gamma_2$ into $B$$)$. Suppose we have not yet added any product disks from $\Gamma_4 \cup \ldots \cup \Gamma_{n-1}$ into $B$. Then, after performing some pivots and canonical calibrations, we can build our branched surface in $\Gamma_4 \cup \Gamma_5$ so that: $(1)$ $B$ contains no co-oriented product disks from $\Gamma_4$, $(2)$ we applied our template to $\Gamma_5$ $($so there are $c_5-1$ co-oriented product disks from $\Gamma_5$ in our branched surface$)$, $(3)$ there is at most one potential sink disk spanning $\Gamma_1 \cup \ldots \cup \Gamma_4$, and it is $\mathcal{S}_5$, and $(4)$ there is no pairwise linking between arcs in $\Gamma_3 \cup \Gamma_{4} \cup \Gamma_{5}$.

In <Ref>, there were four possible ways that we constructed $B$ in $\Gamma_3$. To continue building $B$ in $\Gamma_3 \cup \Gamma_4 \cup \Gamma_5$, we will analyze the four cases in turn. To begin, we use the labeling of bands and presentation from <Ref>.

[size=fbox, colback=gray!30]Construction of $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ is inherited from Case (A) of <Ref>.

As summarized in <Ref>, there is one potential sink disk spanning $\Gamma_1 \cup \Gamma_2$, and it is $\mathcal{S}_3$. Additionally, in $S_3$, $\partial \mathcal{S}_3$ contains $c_3-1$ arcs and one image arc; see <Ref>. In this case, we can apply the same procedure as in the proof of <Ref> to pivot $\beta$ and then build $B$ in $\Gamma_4 \cup \Gamma_5$; for an example of the result, see <Ref>. We claim that the sink disk analysis from <Ref> carries through without issue: this follows from the fact that the unique image arc on $S_3$ does not enclose any bands on the left (i.e. the arc $\varphi(\alpha_{2,1})$ does not enclose any other plumbing arcs on $S_3$), so the branch sector $\mathcal{S}_3$ is homeomorphic to the sector seen in <Ref> (Upper Right, Lower Right). We deduce that the unique potential sink disk spanning $\Gamma_1 \cup \ldots \cup \Gamma_4$ is $\mathcal{S}_5$.

$\Gamma_1$ at 32 240
$\Gamma_{2}$ at 60 240
$\Gamma_{3}$ at 88 240
$\Gamma_{4}$ at 116 240
$\Gamma_{5}$ at 146 240

$\Gamma_1$ at 262 240
$\Gamma_{2}$ at 290 240
$\Gamma_{3}$ at 318 240
$\Gamma_{4}$ at 346 240
$\Gamma_{5}$ at 374 240

Assume we build $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ according to Case (A) of <Ref>. Since $\partial \mathcal{S}_3 \cap S_3$ is $\varphi(\alpha_{2,1}), \alpha_{3,1}, \ldots, \alpha_{3,c_3-1}$, we can directly pivot our braid and then apply the construction – and proof – from <Ref> to build $B$ in $\Gamma_3 \cup \Gamma_4 \cup \Gamma_5$. On the left, $\Gamma_3$ is heterogeneous, and on the right, $\Gamma_3$ is homogeneous. We deduce that there is only one potential sink disk in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$, and it is $\mathcal{S}_5$. As we do not want to focus on the distribution of bands in $\Gamma_1 \cup \Gamma_2$, we have dotted the vertical lines representing the first and second strands of the braid.

[size=fbox, colback=gray!30]Construction of $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ is inherited from Case (B.1) of <Ref>.

We need to do some new analysis for this and subsequent cases. We use the labelings from <Ref> throughout, and we refer to <Ref> throughout. As summarized in <Ref>, we know that the unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$ is the sector $\mathcal{H}$ containing $\mathbbm{b}_{3, \delta-1}$; we zoom in on this sector in <Ref>.

We recall: in Case (B.1) of <Ref>, we let $\delta = d(2; 1, 2)$; additionally, we knew that $d(3; \delta-2, \delta) = c_4$, and that $d(3; \delta-2, \delta-1)$ and $d(3; \delta-1, \delta)$ were both positive. Let $\epsilon = d(3; \delta-2, \delta-1)$. To build $B$ in $\Gamma_5$, we canonically calibrate $\beta$ with respect to $\mathbbm{b}_{4, \epsilon}$, and then apply our template to $\Gamma_5$; see <Ref>. We claim there is a unique sink disk spanning $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$, and it is $\mathcal{S}_5$.

* In <Ref>, we proved that $\mathcal{S}_1, \mathcal{S}_2$, and $\mathcal{S}_3$ are not sinks. By assumption (from being in Case (B.1)), we know that $\Gamma_3$ is homogeneous; it follows that all the image arcs in $S_4$ are right pointers, so $\mathcal{S}_4$ is not a sink. Since we do not know whether $\Gamma_5$ is homogeneous or heterogeneous, we cannot yet argue that $\mathcal{S}_5$ is not a sink.

* The only possible Seifert disks with polygon sectors are $S_2$ and $S_3$, since these are the only Seifert disks spanning $\Gamma_1 \cup \ldots \cup \Gamma_4$ that contain both plumbing and image arcs. Our proof of <Ref> tells us that the unique polygon sector in $S_2$ is not a sink. By design, the polygon sector $\mathcal{P}$ such that $\mathcal{P} \cap S_3 \neq \varnothing$ is not a sink: we constructed $B$ so that $\mathcal{P} \sim \mathbbm{b}_{3, \delta-1} \sim \mathbbm{b}_{4, \epsilon}$, and we know that $\mathbbm{b}_{4, \epsilon}$ is enclosed, on the right, by $\alpha_{5,1}$, which is a left pointer (see <Ref>). We deduce that $\mathcal{P}$ is not a sink, and that there are no sink disk polygon sectors spanning $\Gamma_1 \cup \ldots \cup \Gamma_4$.

* Our proof of <Ref> ensures that the unique horizontal sector spanning $\Gamma_1 \cup \Gamma_2$ that could be a sink is the sector containing $\mathbbm{b}_{3,\delta-1}$. As argued in our polygon sector analysis above, the branch sector containing $\mathbbm{b}_{3, \delta-1}$ is not a sink. Moreover, since $\Gamma_3$ is homogeneous, then for all $i \neq \delta-1$, the horizontal sector containing $\mathbbm{b}_{3,i}$ is not a sink because $\alpha_{3, i}$ points out of the sector. Therefore, at this point, we know that no horizontal sector spanning $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ is a sink.

It remains to show that no sector spanning $\Gamma_4$ is a sink. This is straightforward: recall that we have labelled our bands in $\Gamma_4$ such that $d(3; \delta-2, \delta-1)= \epsilon \geq 1$ and $d(3; \delta-2, \delta) = c_4$. We know that the bands $\mathbbm{b}_{4, 1}, \ldots, \mathbbm{b}_{4, \epsilon}$ are all enclosed, on the left by $\varphi(\alpha_{3, \delta-2})$, and in particular, this means $\mathbbm{b}_{4, 1} \sim \ldots \sim \mathbbm{b}_{4, \epsilon}$; see <Ref>.

We already proved that the sector containing $\mathbbm{b}_{4, \epsilon}$ is not a sink, so it follows that no horizontal sector containing $\mathbbm{b}_{4, j}$, where $j \in \{1, \ldots, \epsilon\}$ is a sink. Finally, notice that the remaining bands in $\Gamma_4$ are all concentrated between $\mathbbm{b}_{3, \delta-1}$ and $\mathbbm{b}_{3, \delta}$ (this is because $d(3; \delta-1, \delta) = c_4 - \epsilon$). Since we did not include the product disk swept out by $\alpha_{3, \delta-1}$ into our branched surface, the bands $\mathbbm{b}_{4, \epsilon+1}, \ldots, \mathbbm{b}_{c_4}$ are not enclosed, on the left, by any arcs. In particular, these bands are all in the same branch sector as $\mathcal{S}_4$, which we already proved is not a sink. We deduce that there are no sink disk horizontal sectors spanning $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$.

We deduce that in Case (B.1), we can build a branched surface which contains product disks from $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_5$ such that the unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$ is $\mathcal{S}_5$.

$\Gamma_{2}$ at 40 174
$\Gamma_{3}$ at 68 174
$\Gamma_{4}$ at 96 174
$\Gamma_{5}$ at 123 174

$1$ at 38 146
$1$ at 60 94
$1$ at 123 128
$\delta$ at 60 60
$\epsilon$ at 92 94

$\Gamma_{2}$ at 235 174
$\Gamma_{3}$ at 262 174
$\Gamma_{4}$ at 290 174
$\Gamma_{5}$ at 318 174

$1$ at 235 146
$1$ at 254 96
$1$ at 318 128
$\delta$ at 254 60
$\epsilon$ at 286 94

Assume we build $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ according to Case (B.1) of <Ref>. Zooming in on the bands enclosed, on the left, by $\varphi(\alpha_{2,1})$ from <Ref> (top right, bottom right); to be consistent, we have used the same cyclic conjugated presentation of the braid from that figure. On the left, the case where $d(2; 1,2) = 2$, and on the right, we have $d(2; 1, 2) \geq 3$. For emphasis, we have partially drawn $\alpha_{3, \alpha_{c_3}}$ on $S_3$. We see that the shaded sector, which contains $\mathbbm{b}_{3, \delta-1}$, is not a sink, and that $\mathcal{S}_4 \sim \mathbbm{b}_{4, \epsilon+1} \sim \ldots \sim \mathbbm{b}_{4, c_4}$.

[size=fbox, colback=gray!30]Construction of $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ is inherited from Case (B.2) of <Ref>.

As summarized in <Ref>, we can build $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ such that there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$. We recall: in this case, we know that $\Gamma_3$ is heterogeneous, but that all the right pointing plumbing arcs co-bound a subsurface in $\Gamma_3$; see <Ref>. Assume the same presentation of the braid as in <Ref>, and now let $\mathbbm{b}_{3, p}$ denote the first right pointing plumbing arc in $\Gamma_3$; in particular, this means $\mathbbm{b}_{3, p-1}$ is a left pointer and $1 \leq p \leq \delta-2$. In <Ref>, we see examples where $p = 1$ and where $p = 2$. By the definition of our template, we know that $d(3; \delta, c_3) = 0$, $d(3; c_3, p-1) = 0$, and $d(3; p-1, p) := \epsilon \geq 1$. To build $B$ in $\Gamma_4 \cup \Gamma_5$, we canonically calibrate $\beta$ with respect to $\mathbbm{b}_{4, \epsilon}$, and then apply our template to $\Gamma_5$; see <Ref>. We claim that the resulting branched surface contains at most one sink disk in $\Gamma_1 \cup \ldots \cup \Gamma_4$, and it is $\mathcal{S}_5$.

$\Gamma_{2}$ at 40 174
$\Gamma_{3}$ at 68 174
$\Gamma_{4}$ at 96 174
$\Gamma_{5}$ at 123 174

$1$ at 38 144
$p$ at 64 118
$1$ at 123 150
$\delta$ at 60 60
$\epsilon$ at 92 122

$\Gamma_{2}$ at 235 174
$\Gamma_{3}$ at 262 174
$\Gamma_{4}$ at 290 174
$\Gamma_{5}$ at 318 174

$1$ at 232 144
$p$ at 254 94
$1$ at 318 128
$\delta$ at 254 60
$\epsilon$ at 286 94

Assume we build $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ according to Case (B.2) of <Ref>. Zooming in on the bands enclosed, on the left, by $\varphi(\alpha_{2,1})$ from <Ref>; to be consistent, we have used the same cyclic conjugated presentation of the braid from that figure. On the left, the case where $p=1$, and on right, we have $p=2$. For emphasis, we have partially drawn $\alpha_{3, c_3}$ on $S_3$. The green shaded sector, which contains $\mathbbm{b}_{3, p}$, is not a sink. We have drawn bands in $\Gamma_4$ as dashed to indicate that we do not know their precise distribution within the column.

Since we already proved that there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$, we need only prove that there is at most one sink disk spanning $\Gamma_3 \cup \Gamma_4$.

* The sector $\mathcal{S}_3$ is not a sink because $\varphi(\alpha_{2,1})$ points out of the sector. Since $\Gamma_3$ is heterogeneous and we have used no product disks from $\Gamma_4$ to build $B$, $\partial \mathcal{S}_4$ must contain an outward pointing arc. Since we do not know whether $\Gamma_5$ is heterogeneous or homogeneous, we cannot make any claims about $\mathcal{S}_5$; this is the unique potential sink disk spanning $\Gamma_3 \cup \Gamma_4$.

* By <Ref>, we know that no polygon sector in $S_3$ is a sink. Since $S_4$ (resp. $S_5$) contains only image (resp. plumbing) arcs, there are no polygon sectors in $S_4$ or $S_5$. We deduce that there are no sink disk polygon sectors in $\Gamma_3 \cup \Gamma_4$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_3$. We prove that $\mathcal{H}$ is not a sink:

$\circ$ If $\mathcal{H} \sim \mathbbm{b}_{3,i}$, where $i \in \{1, \ldots, p-1\} \cup \{\delta, \ldots, c_3\}$, then $\mathcal{H}$ is enclosed, on the left, by the left pointing plumbing arc $\alpha_{3,i}$; we deduce that $\mathcal{H}$ is not a sink.

$\circ$ If $\mathcal{H} \sim \mathbbm{b}_{3,i}$ where $i \in \{p+1, \ldots \delta-1\}$, then $\mathcal{H}$ is enclosed, on the right, by the left pointing image arc $\varphi(\alpha_{3,i-1})$, so $\mathcal{H}$ is not a sink.

$\circ$ If $\mathcal{H} \sim \mathbbm{b}_{3, p}$, then $\mathcal{H}$ is in the same branch sector as $\mathbbm{b}_{4, \epsilon}$. By construction, this sector meets $S_5$ in $\alpha_{5,1}$, which is a right pointer. We deduce that $\mathcal{H}$ is not a sink.

It remains to show that if a horizontal sector $\mathcal{H}$ spans $\Gamma_4$, then $\mathcal{H}$ is not a sink. Recall that by the definition of Case (B.2), we know that $d(3; \delta, c_3) + d(3; c_3, 1) + d(3; 1, p-1) = 0$. In particular, this means that all of our bands in $\Gamma_4$ are concentrated in between $\mathbbm{b}_{3, p-1}$ and $\mathbbm{b}_{3, \delta}$. The bands $\mathbbm{b}_{4,1}, \ldots, \mathbbm{b}_{4, \epsilon}$ are all part of the same branch sector; by design, $\mathbbm{b}_{4, \epsilon}$ is enclosed, on the right, by an outward pointing arc. So, if $\mathcal{H} \sim \mathbbm{b}_{4, i}$, where $i \in \{1, \ldots, \epsilon\}$, then $\mathcal{H}$ is not a sink. If $\mathcal{H} \sim \mathbbm{b}_{4, i}$, where $i \in \{\epsilon +1, \ldots, c_4\}$, then $\mathcal{H}$ meets $S_4$ in an outward pointing image arc, so it is not a sink. Otherwise, $\mathcal{H} \sim \mathbbm{b}_{4,i} \sim \mathcal{S}_4$, and we already proved that $\mathcal{S}_4$ is not a sink. We conclude that there are no horizontal sink disk sectors in $\Gamma_3 \cup \Gamma_4$.

Therefore, the unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$ is the branch sector $\mathcal{S}_5$.

[size=fbox, colback=gray!30]Construction of $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ is inherited from Case (C) of <Ref>.

As summarized in <Ref>, there is one potential sink disk spanning $\Gamma_1 \cup \Gamma_2$, and it is the horizontal sector $\mathcal{H} \sim \alpha_{2,1} \sim \alpha_{3, c_3}$. Therefore, to prove that we can construct $B$ in $\Gamma_1 \cup \ldots \cup \Gamma_5$ such that there are no sink disks in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$, it need to first build $B$ in $\Gamma_4 \cup \Gamma_5$ and then show that there are no sink disks spanning $\Gamma_3 \cup \Gamma_4$. We will refer to <Ref> throughout.

To build $B$ in $\Gamma_4 \cup \Gamma_4$, we first define $\epsilon := d(3; j-1, j)$, where (as in <Ref>) $\alpha_{3, j}$ is the first right pointer in $S_4$. By the definition of our template, we know that $\epsilon \geq 1$. To build $B$ in $\Gamma_5$, we canonically calibrate $\beta$ with respect to $\mathbbm{b}_{4, \epsilon}$, and then apply our template to $\Gamma_5$. We check there are no sink disks spanning $\Gamma_3 \cup \Gamma_4$.

* By being in Case (C) of <Ref>, we know that $\mathcal{S}_3$ is not a sink. Since $\Gamma_3$ is heterogeneous and we have included no product disks from $\Gamma_4$ into $B$, $\partial \mathcal{S}_4$ contains both left and right pointing image arcs, so $\mathcal{S}_4$ is not a sink. The sink disk status of $\mathcal{S}_5$ is unknown, as we do not know whether it is heterogeneous or homogeneous.

* The only polygon sector in $\Gamma_3 \cup \Gamma_4$ lies in $S_3$; our proof of <Ref> ensures that this sector is not a sink.

* First, suppose $\mathcal{H}$ spans $\Gamma_3$. Adapting our usual argument, we see that the only sector that could potentially be a sink is the sector $\mathcal{H} \sim \mathbbm{b}_{3, j}$, because this sector meets $S_3$ in a right pointer and $S_4$ in a left pointer. However, by design, this sector contains $\mathbbm{b}_{4,\epsilon}$, and this band is enclosed, on the right, by a left pointing plumbing arc. We deduce that $\mathcal{H}$ is not a sink. Instead suppose $\mathcal{H}$ spans $\Gamma_4$. In this case, as in <Ref>, every band in $\Gamma_4$ is either (1) enclosed, on the left, by a left pointing image arc, or (2) in the same branch sector as $\mathcal{S}_4$, or (3) it is in the same branch sector as $\mathbbm{b}_{4, \epsilon}$. In each of these three cases, we deduce that the horizontal sector is not a sink, hence there are no horizontal sink disk sectors in $\Gamma_3 \cup \Gamma_4$.

We deduce that, in Case (C) of <Ref>, we can build $B$ in the first five columns of the braid such that the unique potential sink disk spanning $\Gamma_1 \cup \ldots \cup \Gamma_4$ is $\mathcal{S}_5$. <Ref> confirms that there is no pairwise linking between arcs which are both in $S_3$ or both in $S_5$, and there can be no pairwise linking between arcs in $S_1$ and $S_3$.

§.§ Building $B$ in the rightmost columns.

In this section, we describe how to build our branched surface $B$ in the last few columns of the braid.

Let $\beta$ be a positive braid on $n$ strands, where $n \geq 5$. Suppose $n-2 \equiv n \equiv m \mod 2$, and suppose we constructed a branched surface in $\Gamma_1 \cup \ldots \cup \Gamma_{n-2}$, such that we applied our template to $\Gamma_{n-2}$. Further assume that $\partial \mathcal{S}_{n-2}$ contains $c_{n-2}-1$ plumbing arcs and at most one image arc, and that the Seifert disk $S_{n-2}$ does not contain any polygon sectors. If we build $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$ by taking no product disks from $\Gamma_{n-1}$, then there are no sink disks spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$.

Just as in <Ref>, there will be two cases to consider, based on whether $\Gamma_{n-2}$ is heterogeneous or homogeneous.

$\boldsymbol{\circ}$ Case (1): The heterogenous case. Here, $d(n-2; 1, c_i - 1) > 0$, as in <Ref>.

$\boldsymbol{\circ}$ Case (2): The homogeneous case. Here, $d(n-2; 1, c_i-1) = 0$, as in <Ref>.

Throughout, we assume three things: (1) $\beta$ is cyclically conjugated so that it begins with a $\sigma_{n-2}$ letter, (2) $\alpha_{n-2,1}$ is a left pointer, and (3) $\mathbbm{b}_{n-2,c_{n-2}}$ is not enclosed, on the left, by any arc. We quickly justify why we can do this:

* If $S_{n-2}$ has no image arcs, then this is immediate.

* Suppose $S_{n-2}$ contains exactly one image arc, $c_{n-2}$ plumbing arcs, and no polygon sectors. Since $S_{n-2}$ contains one image arc, <Ref> tells us that either (a) $\Cmax = \Codd$ and $\Gamma_3 = \Gamma_{n-2}$, or (b) $\Cmax = \Ceven$ and $\Gamma_{2} = \Gamma_{n-2}$. However, since we are also assuming that $n \geq 5$, we deduce that

$n=5$, $\Cmax = \Codd$, and the construction of $B$ in $\Gamma_{n-2} = \Gamma_{3}$ is inherited from Case (A) of <Ref>. In particular, this means that $\mathbbm{b}_{n-2, c_{n-2}}$ is not enclosed, on the left, by any arc, and so we can cyclically conjugate $\beta$ into the form described above.

We now move onto the constructions of $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$.

[size=fbox, colback=gray!30]Sink disk analysis for Case (1): $\Gamma_{n-2}$ is heterogeneous.
Suppose $\beta$ is cyclically conjugated as described, and $\Gamma_{n-2}$ is heterogeneous; see <Ref>. Additionally, let $\mathbbm{b}_{n-2, p}$ denote the first band which is followed by $\mathbbm{b}_{n-1}$ bands. (That is, $2 \leq p \leq c_{n-2}-2$,  $d(n-2; 1, p) = 0$, and $d(n-2; p, 1) = c_{n-1}$.) We now perform our sink disk analysis.

$\Gamma_{n-2}$ at 35 190
$\Gamma_{n-1}$ at 62 190

$\Gamma_{n-2}$ at 148 190
$\Gamma_{n-1}$ at 178 190

$\Gamma_{n-2}$ at 262 190
$\Gamma_{n-1}$ at 290 190

$p$ at 30 152
$p$ at 142 152
$p$ at 258 152

Building $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$ when $\Gamma_{n-2}$ is heterogeneous and $n-2 \equiv m \mod 2$. We see that $\mathcal{S}_n$ contains all the bands in $\Gamma_{n-1}$, as well as $\mathbbm{b}_{n-2,p+1}$ and $\mathbbm{b}_{n-1, 1}$.

* $\mathcal{S}_{n-2}$ is not a sink: the arc $\alpha_{n-2, p+1}$ is in $\partial \mathcal{S}_{n-2}$, and it points out of this Seifert disk. Since $\alpha_{n-2,1}$ is a left pointer, and we have taken no disks from $\Gamma_{i+1}$, then $\varphi(\alpha_{n-2,1})$ is in $\partial \mathcal{S}_{n-1}$. It points out of the sector, so $\mathcal{S}_{n-1}$ is not a sink.

Finally, we argue that $\mathcal{S}_n$ is not a sink. Since $B$ includes no product disks from $\Gamma_{n-1}$, there are no arcs on $S_n$, hence $\mathcal{S}_n \sim \mathbbm{b}_{n-1,1} \sim \ldots \sim \mathbbm{b}_{n-1, c_{n-1}}$, as seen in <Ref>.

Since $\Gamma_{n-2}$ is heterogeneous, we know that there are both left and right pointing image arcs in $S_{n-1}$. The distribution of the bands in $\Gamma_{n-1}$ determines whether or not some or none of the bands will be enclosed, on the left, by a left pointing image arc. In particular, either (a) there is some band $\mathbbm{b}_{n-1, j}$ which is enclosed, on the left, by a left pointing image arc (as in <Ref> (left and middle)), or (b) there is no such band (as in <Ref> (right)). If (a) occurs, then $\mathcal{S}_{n-1}$ contains an outward pointing arc in its boundary, hence it is not a sink; however, if (b) occurs, then $\mathcal{S}_{n-1} \sim \mathcal{S}_{n-1}$, and we already argued that the latter is not a sink. Therefore, none of the Seifert disk sectors are sinks.

* By definition, a polygon sector must lie in a Seifert disk and also contain both plumbing and image arcs in its boundary. By assumption, $\mathcal{S}_{n-2}$ does not contain a polygon sector. As $S_{n-1}$ and $S_n$ do not contain both co-oriented plumbing and image arcs, there are no polygon sectors spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_{n-2}$. There are three possibilities:

* $\mathcal{H} \sim \mathbbm{b}_{n-2,j}$ where $1 \leq j \leq p$. In this case, $\mathcal{H}$ meets $S_{n-2}$ in an outward pointing plumbing arc, so $\mathcal{H}$ is not a sink.

* $\mathcal{H} \sim \mathbbm{b}_{n-2,p+1}$. In this case, by design, $\mathcal{H} \sim \mathbbm{b}_{n-1,1}$. We previously argued that the sector $\mathcal{S}_n$ contains all the bands in $\Gamma_{n-1}$, so in particular, $\mathcal{H} \sim \mathcal{S}_n$. We already showed that $\mathcal{S}_n$ is not a sink, so neither is $\mathcal{H}$.

* $\mathcal{H} \sim \mathbbm{b}_{n-2,j}$ where $p+2 \leq j \leq c_{n-2}$. In this case, $\mathcal{H}$ meets $S_{n-1}$ in an outward pointing image arc, so this sector is not a sink.

Therefore, no horizontal sector $\mathcal{H}$ spanning $\Gamma_{n-2}$ is a sink. Suppose $\mathcal{H}$ spans $\Gamma_{n-1}$. Since $\mathcal{S}_{n}$ contains all the bands in $\Gamma_{n-1}$, then such an $\mathcal{H}$ must be in the same sector as $\mathcal{S}_{n}$, which we already argued is not a sink. We deduce that there are no horizontal sectors spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$.

So, when $\Gamma_{n-2}$ is heterogenous, there are no sink disks spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$. We turn to Case (2).

[size=fbox, colback=gray!30]Sink disk analysis for Case (2):  $\Gamma_{n-2}$ is homogeneous.

Throughout, we refer to  <Ref>.

$\Gamma_{n-2}$ at 35 190
$\Gamma_{n-1}$ at 62 190

$\Gamma_{n-2}$ at 148 190
$\Gamma_{n-1}$ at 178 190

$\Gamma_{n-2}$ at 262 190
$\Gamma_{n-1}$ at 290 190

$\delta$ at 60 54
$\delta$ at 174 72
$\delta$ at 288 101

Building $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$ when $\Gamma_{n-2}$ is homogeneous and $n-2 \equiv m \mod 2$. We see that $\mathcal{S}_{n-2} \sim \mathcal{S}_{n-1} \sim \mathcal{S}_n$, and that this sector contains all the bands in $\Gamma_{n-1}$. We note that we have included the case where $c_{n-2} \geq 3$ and $c_{n-1}=2$, and also the case where $c_{n-2} \geq 3$ and $c_{n-1}=2$.

* We will prove that the Seifert disk sectors $\mathcal{S}_{n-2}$, $\mathcal{S}_{n-1}$, and $\mathcal{S}_n$ are all part of the same branch sector. Let $\mathcal{S}$ denote the branch sector containing $\mathcal{S}_{n-2}$.

We already argued that we could cyclically conjugate $\beta$ so that $\mathbbm{b}_{n-2,c_{n-2}}$ is not enclosed, on the left, by any arc. This automatically tells us that regardless of whether $\partial \mathcal{S}_{n-2}$ contains an image arc or not, we must have that $\mathcal{S} \sim \mathbbm{b}_{n-2, c_{n-2}}$.

But this band is enclosed, on the right, by $\varphi(\alpha_{n-2, c_{n-2}-1})$, as seen in <Ref>. Let $\delta:= d(n-2; c_{n-2}-1, c_{n-2})$. The arc $\varphi(\alpha_{n-2, c_{n-2}-1})$ encloses the bands $\mathbbm{b}_{n-1, 1}, \ldots, \mathbbm{b}_{n-1, \delta}$ on the left, and so $\mathcal{S} \sim \mathcal{S}_{n-2} \sim \mathbbm{b}_{n-1,1} \sim \ldots \sim \mathbbm{b}_{n-1, \delta}$. However, these bands are not enclosed on the right by any arcs, so $\mathcal{S} \sim \mathcal{S}_n$.

In fact, none of the bands in $\Gamma_{n-1}$ are enclosed, on the right, by arcs. Therefore, we have that $\mathcal{S} \sim \mathbbm{b}_{n-1, \delta+1} \sim \ldots \sim \mathbbm{b}_{n-1, c_{n-1}}$. But this is the homogenous case, so we know that these bands are not enclosed, on the left, by any arcs, as in <Ref>. Therefore, not only does $\mathcal{S}$ contain all the bands in $\Gamma_{n-1}$, it also contains $\mathcal{S}_{n-1}$. Therefore, $\mathcal{S} \sim \mathcal{S}_{n-2} \sim \mathcal{S}_{n-1} \sim \mathcal{S}_n$, proving the claim.

We now argue that $\mathcal{S}$ is not a sink. By construction, $\mathcal{S} \sim \mathbbm{b}_{n-2,1}$, as in <Ref>. In particular, $\alpha_{n-2,1}$ is contained in $\partial \mathcal{S}$; this arc is smoothed to the left, and it points out of $\mathcal{S}$, so this sector is not a sink. Therefore, none of the Seifert disk sectors spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$ are sinks.

* The same argument as in the heterogeneous case allows us to deduce that there are no polygon sectors spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_{n-2}$. We already argued that $\mathbbm{b}_{n-2,1}$ and $\mathbbm{b}_{n-2,c_{n-2}}$ are part of $\mathcal{S}$, so it remains to study the other bands in $\Gamma_{n-2}$. For all $2 \leq j \leq c_{n-2}-1$, the band $\mathbbm{b}_{n-2,j}$ is enclosed, on the left, by the outward pointing plumbing arc $\alpha_{n-2,j}$. Thus, no sector containing these bands is a sink.

We already argued that the all the bands in $\Gamma_{n-1}$ are part of the sector $\mathcal{S}$, which is not a sink. Therefore, no horizontal sector spanning $\Gamma_{n-1}$ is a sink. We deduce that no horizontal sectors spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$ are sink disks.

Therefore, there are no sink disks spanning $\Gamma_{n-2} \cup \Gamma_{n-1}$ in the homogeneous case.

Let $\beta$ be a positive braid in $\mathcal{B}_n$, where either $n \geq 6$, or $n=5$ with $\Cmax = \Ceven$ and $\Gamma_2$ heterogeneous. Suppose $n \not \equiv m \mod 2$. Suppose we build $B$ in $\Gamma_1 \cup \ldots \cup \Gamma_{n-3}$, such that $S_{n-3}$ contains $c_{n-3}-1$ plumbing arcs and at most one image arc. Then we can build $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$ such that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$, and such that there is no pairwise linking between arcs in $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

Proof of <Ref>.
First, we establish the presentation of $\beta$ that we will use for the proof of this lemma.
We conjugate $\beta$ so that:

* $\beta$ begins with $\sigma_{n-3}$, and

* $\mathbbm{b}_{n-3, c_{n-3}}$ is not enclosed, on the right, by a plumbing arc.

In particular, this means that $\beta = \sigma_{n-3} \ w_1 \ \sigma_{n-2} \ w_2 \ \sigma_{n-1} \ w_3$, where the identified $\sigma_{n-2}$ is the first letter of its kind in $\beta$. Without loss of generality, we may assume that $w_1$ contains no $\sigma_{n-1}$ letters (if it did, we may use the braid relations and then cyclically conjugate $\beta$ to move them into $w_3$). This is the presentation of $\beta$ that we use going forward: $\beta = \sigma_{n-3} w_1 \sigma_{n-2} w_2$, where $w_1$ contains no $\sigma_{n-2}$ or $\sigma_{n-1}$ letters. See <Ref> for an example.

To prove the lemma, we will need to consider two cases:

$\boldsymbol{\circ}$ Case (1): $\Gamma_{n-3}$ is heterogenous, i.e. when $d(n-3; 1, c_{n-3} - 1) > 0$.

$\boldsymbol{\circ}$ Case (2): $\Gamma_{n-3}$ is homogeneous, i.e. when $d(n-3; 1, c_{n-3}-1) = 0$.

[size=fbox, colback=gray!30]Sink disk analysis for Case (1): $\Gamma_{n-3}$ is heterogeneous.

We begin with the heterogeneous case, and establish some preliminaries.

Observation #1: Suppose we build $B$ in $\Gamma_{n-3}$ by applying our template (as in <Ref>) to $\Gamma_{n-3}$. Since we build $B$ in $\Gamma_{n-3}$ in a standard way, we can label bands in this column accordingly: we let $\alpha_{n-3,p}$ denote the last left pointing plumbing arc in $S_{n-3}$, and let $\delta = d(n-3; p,p+1)$. (Note that since $\Gamma_{n-3}$ is heterogeneous, $p \leq c_{n-3}-2$, and that $\delta \geq 1$; see <Ref> for some examples.) Therefore, the arcs $\alpha_{n-3, 1}, \ldots, \alpha_{n-3, p}$ are all left pointers, and $\alpha_{n-3, p+1}, \ldots, \alpha_{n-3, c_{n-3}-1}$ are all right pointers.

This setup tells us information about how the bands in $\Gamma_{n-3}$ are enclosed on the left and right:

$\boldsymbol{\circ}$ The bands $\mathbbm{b}_{n-3,2}, \ldots, \mathbbm{b}_{n-3, p+1}$ are all enclosed, on the right, by right pointing image arcs.

$\boldsymbol{\circ}$ The bands $\mathbbm{b}_{n-3, p+2}, \ldots, \mathbbm{b}_{n-3, c_{n-3}}$ are all enclosed, on the right, by left pointing image arcs.

$\boldsymbol{\circ}$ The arc $\varphi(\alpha_{n-3, p})$ encloses the bands $\mathbbm{b}_{n-2,1}, \ldots, \mathbbm{b}_{n-2,\delta}$ on the left.

$\boldsymbol{\circ}$ The bands $\mathbbm{b}_{n-2,\delta+1}, \ldots, \mathbbm{b}_{n-2,c_{n-2}}$ are either enclosed, on the left, by left pointing image arcs, or not enclosed by image arcs at all.

See <Ref> for some examples.

Observation #2: Regardless of the design of $B$ in $\Gamma_{n-3}$, we know something about the distribution of bands in $\Gamma_{n-1}$ with respect to $\Gamma_{n-2}$. Namely, there can be no $i$, where $1 \leq i \leq c_{n-2}$, such that $d(c_{n-2}; i, i+1) = c_{n-1}$ (if there were, that would mean that $\beta$ is a connected sum, which contradictions our assumption that $\widehat{\beta}$ is a prime knot). Therefore, we know that there exists some minimal $j$ such that $d_L(n-1; j, j+1) \geq 1$. Said differently, we let $\mathbbm{b}_{n-1, j}$ denote the first band in $\Gamma_{n-1}$ which is followed by a $\mathbbm{b}_{n-2}$ band, and so that $d_L(n-1; 1, j) = 0$. There are two sub-cases to consider:

* Case (1a): $\Gamma_{n-3}$ is heterogeneous and $j=1$.

* Case (1b): $\Gamma_{n-3}$ is heterogeneous and $j \geq 2$.

With these preliminary observations in place, we now construct $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$ in the case where $\Gamma_{n-3}$ is heterogeneous and show that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

(A) at 72 225
(B) at 252 225
(C) at 422 225
(D) at 72 10
(E) at 252 10
(F) at 422 10

$\Gamma_{n-3}$ at 43 415
$\Gamma_{n-2}$ at 70 415
$\Gamma_{n-1}$ at 100 415

$\Gamma_{n-3}$ at 222 415
$\Gamma_{n-2}$ at 250 415
$\Gamma_{n-1}$ at 280 415

$\Gamma_{n-3}$ at 394 415
$\Gamma_{n-2}$ at 422 415
$\Gamma_{n-1}$ at 450 415

$j=1$ at 98 335
$j=1$ at 278 335
$j=1$ at 450 335

$p$ at 38 376
$p$ at 217 376
$p$ at 390 376

$\delta$ at 70 358
$\delta$ at 250 350
$\delta$ at 422 358

$\Gamma_{n-3}$ at 43 200
$\Gamma_{n-2}$ at 70 200
$\Gamma_{n-1}$ at 100 200

$\Gamma_{n-3}$ at 222 200
$\Gamma_{n-2}$ at 250 200
$\Gamma_{n-1}$ at 280 200

$\Gamma_{n-3}$ at 394 200
$\Gamma_{n-2}$ at 422 200
$\Gamma_{n-1}$ at 450 200

$j=1$ at 98 90
$j=1$ at 278 120
$j=1$ at 450 120

$p$ at 38 160
$p$ at 217 163
$p$ at 390 163

$\delta$ at 70 135
$\delta$ at 250 143
$\delta$ at 422 143

Building $B$ in $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ in Case (1a), where $\Gamma_{n-3}$ is heterogeneous and $j=1$. We apply our template to $\Gamma_{n-3}$, and then finish building the branched surface by coherently orienting all the plumbing arcs in $\Gamma_{n-1}$ to the right. Frames (A) – (F) show some different possible configurations of $\Gamma_{n-1}$ with respect to $\Gamma_{n-2}$. Let $\delta = d(n-3; p, p+1)$, where $\alpha_{n-3,p}$ is the last left pointing plumbing arc in $S_{n-3}$. (For simplicity, in this figure, $p=2$ in all the frames.) In (A), $\delta=1$. In (B), $\delta=2$. In (C), $\mathcal{S}_n \sim \mathcal{S}_{n-2}$. In (D), the first band following $\mathbbm{b}_{n-1,1}$ is $\mathbbm{b}_{n-2, t}$, where $t > \delta+1$. In (E), $c_{n-1} = 2$ and $c_{n-2} \geq 3$. In (F), $c_{n-1} = c_{n-2} = 2$.

[size=fbox, colback=gray!10]Case (1a): $\Gamma_{n-3}$ is heterogeneous and $j = 1$.

To build $B$ in $\Gamma_{n-2} \ \cup \ \Gamma_{n-1}$, we take no product disks from $\Gamma_{n-2}$, and we co-orient the arcs $\alpha_{n-1, 1}, \alpha_{n-1, 2}, \ldots, \alpha_{n-1, c_{n-1}-1}$ to the right; see <Ref>. Note that we chose our presentation of $\beta$ so that $\mathbbm{b}_{n-2,1}$ precedes $\mathbbm{b}_{n-1,1}$. This means, in particular, that  $\mathbbm{b}_{n-2,1}$ is not enclosed, on the right, by any plumbing arcs. We claim that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$:

* we show that $S_{n-3}, S_{n-2}, S_{n-1}$, and $S_n$ are not sinks.

* To begin, we show that $\mathcal{S}_{n-3}$ is never a sink. This analysis depends upon  the value of $n$.

* Suppose $n=5$. In this case, $n-3 = 2$. By the assumptions of this lemma, we know that $\Cmax = \Ceven$ and $\Gamma_{n-3} = \Gamma_2$ is heterogeneous. In particular, the construction of $B$ in $\Gamma_1 \cup \Gamma_2$ is inherited from Case (1A) or (2A) of <Ref>; see <Ref>. In <Ref>, we proved that $\mathcal{S}_2 = \mathcal{S}_{n-3}$ is not a sink.

* Suppose $n = 6$. In this case, $n-3 = 3$, hence $\mathcal{S}_{n-3} = \mathcal{S}_3$ and $\Cmax = \Codd$. Thus, the construction of $B$ in $\Gamma_{n-3} = \Gamma_3$ is inherited from an application of <Ref> to $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$. In particular, $B \cap S_3$ contains $c_3-1$ plumbing arcs and one image arc.

As summarized in <Ref>, there were four sub-cases of <Ref>. We check that $\mathcal{S}_3$ is not a sink when $\Gamma_{n-3}$ is heterogeneous.

* Case (A) of <Ref>. Referring to <Ref>, we see that $\partial \mathcal{S}_3$ contains $\alpha_{2,1}$, $\varphi(\alpha_{2,1})$, and $\alpha_{3,1}, \ldots, \alpha_{3, c_3-1}$. In particular, we assumed that $\Gamma_3$ is heterogeneous, so $\mathcal{S}_3$ contains some outward pointing plumbing arc $\alpha_{3, k}$, so it is not a sink.

* Case (B.1) of <Ref>. We are currently assuming that $\Gamma_3$ is heterogeneous. Since Case (B.1) of <Ref> assumes that $\Gamma_3$ is homogeneous, we need not consider this case.

* Case (B.2) of <Ref>. As noted in <Ref>, there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$, so in particular, $\mathcal{S}_3$ is not a sink.

* Case (C) of <Ref>. As noted in <Ref>, the unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$ is a horizontal sector $\mathcal{H} \not \sim \mathcal{S}_3$, so as desired, $\mathcal{S}_3$ is not a sink.

* Suppose $n \geq 7$. <Ref> tells us that $\mathcal{S}_{n-3}$ contains only plumbing arcs. Since $\alpha_{n-3, p+1}$ is a right pointer, it points out of the Seifert disk, and so $\mathcal{S}_{n-3}$ is not a sink.

We deduce that $\mathcal{S}_{n-3}$ is not a sink. Next, we show that $S_{n-2}, S_{n-1}$, $S_n$ are not sinks.

* By assumption, $\Gamma_{n-3}$ is heterogeneous, so there exist both left and right pointing plumbing arcs in $S_{n-3}$. Therefore, there are both right and left pointing image arcs in $S_{n-2}$. As we include zero product disks from $\Gamma_{n-2}$ into $B$, $S_{n-2}$ contains only image arcs. Therefore, $\partial S_{n-2}$ contains an outward pointing arc, and the sector is not a sink. See <Ref>.

* By design, $S_{n-1}$ contains only right pointing plumbing arcs, so $\mathcal{S}_{n-1}$ is not a sink.

* Finally, we analyze $\mathcal{S}_n$. By design, $\mathbbm{b}_{n-1, 1}$ is not enclosed, on the right, by any arc; therefore, $\mathcal{S}_n \sim \mathbbm{b}_{n-1, 1}$. In particular, $\partial \mathcal{S}_n$ contains $\alpha_{n-1,1}$. Since $j=1$, then by assumption, $\mathbbm{b}_{n-1,1}$ is followed by some band in $\Gamma_{n-2}$; we call this band $\mathbbm{b}_{n-2, k}$. In particular, notice that $k \geq \delta +1$, and $\alpha_{n-1,1}$ encloses $\mathbbm{b}_{n-2,k}$ on the right; see <Ref>. There are now two cases to analyze separately: either $\mathbbm{b}_{n-2,k}$ is enclosed, on the left, by an image arc, or $\mathbbm{b}_{n-2,k}$   is not bounded, on the left, by any image arcs.

$\circ$ Suppose $\mathbbm{b}_{n-2,k}$ is enclosed, on the left, by an image arc (as in frames (A), (B), and (D) of <Ref>). We see that it must be a left pointer: the only bands in $\Gamma_{n-2}$ which are enclosed, on the left, by right pointers are the bands $\mathbbm{b}_{n-2,1}, \ldots, \mathbbm{b}_{n-2, \delta}$; the remaining bands are all enclosed, on the left, by left pointers. Since $k \geq \delta+1$, $\mathbbm{b}_{n-2, k}$ is enclosed, on the left, by a left pointer. We deduce that $\mathcal{S}_{n}$ is not a sink, because $\partial \mathcal{S}_n$ contains the outward pointing arc.

$\circ$ Suppose $\mathbbm{b}_{n-2, k}$ is not enclosed, on the left, by any arc (as in frames (C), (E), and (F) of <Ref>). In this case, $\mathbbm{b}_{n-2,k}$ is contained in $\mathcal{S}_{n-2}$, which we already proved is not a sink. Therefore, we can deduce that $\mathcal{S}_n$ is not a sink, as desired.

We deduce that $\mathcal{S}_n$ is not a sink, so none of the Seifert disks sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ are sink disks.

* Polygon sectors exist when a Seifert disk contains both plumbing and image arcs. Since $S_{n-2}, S_{n-1},$ and $S_n$ contain only one type of arc, none of them can contain polygon sectors. It remains to see whether or not $S_{n-3}$ contains both plumbing and image arcs. This is dependent on the value of $n$.

* Suppose $n=5$. By the assumptions of this lemma, we know that $\Cmax = \Ceven$ and $\Gamma_{n-3} = \Gamma_2$ is heterogeneous. In particular, the construction of $B$ in $\Gamma_1 \cup \Gamma_2$ is inherited from Case (2B) of <Ref>. By design, we know that there is a unique polygon sector in $S_2$, and we proved it is not a sink in <Ref>.

* Suppose $n=6$. In this case, the construction of $B$ in $\Gamma_{n-3} = \Gamma_3$ is inherited from <Ref>. We check that there are no polygon sinks in this case, based on the four sub-cases of <Ref>.

* Case (A) of <Ref>. Referring to <Ref>, we see that $\partial \mathcal{S}_3$ contains $\alpha_{2,1}$, $\varphi(\alpha_{2,1})$, and $\alpha_{3,1}, \ldots, \alpha_{3, c_3-1}$. In particular, the only branch sector co-bounded by the unique image arc in $S_3$ and the plumbing arcs in $S_3$ is $\mathcal{S}_3$. Hence, $S_3$ does not contain any polygon sectors.

* Case (B.1) of <Ref>. We are assuming that $\Gamma_3$ is heterogeneous. However, Case (B.1) of <Ref> assumes that $\Gamma_3$ is homogeneous, so we need not consider this case.

* Case (B.2) of <Ref>. As noted in <Ref>, there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$, so in particular, there are no polygon sink disk sectors in $S_3$.

* Case (C) of <Ref>. As seen in <Ref>, no polygon sector in $S_3$ is a sink.

* Suppose $n \geq 7$. As noted in <Ref>, the Seifert disk $S_{n-3}$ must contain only plumbing arcs, so it cannot contain any polygon sectors.

We deduce that there are no polygon sink disk sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

* Finally, we analyze the horizontal sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$. Throughout, we let $\mathcal{H}$ denote a horizontal sector, and we consult <Ref>.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-3}$.

* If $\mathcal{H} \sim \mathbbm{b}_{n-3,i}$, for $1 \leq i \leq p$, then $\mathcal{H}$ is not a sink, because $\partial \mathcal {H}$ contains $\alpha_{n-3, i}$, and these arcs all point into $S_{n-3}$.

* If $\mathcal{H} \sim \mathbbm{b}_{n-3, p+1}$, then $\mathcal{H} \sim \mathbbm{b}_{n-2, 1}, \ldots, \mathbbm{b}_{n-2, \delta}$. However, we know that $\mathbbm{b}_{n-2,1}$ is not enclosed, on the right, by any arc. Therefore, $\mathcal{H} \sim S_{n-1}$, and we already proved this sector is not a sink.

* Finally, if $\mathcal{H} \sim \mathbbm{b}_{n-3, i}$, for $p+2 \leq i \leq c_{n-3}$, then $\mathcal{H}$ is enclosed, on the right, $\varphi(\alpha_{n-3, i-1})$, and this arc points out of $\mathcal{H}$ and into $S_{n-2}$.

We deduce that any horizontal sector $\mathcal{H}$ spanning $\Gamma_{n-3}$ is not a sink.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-2}$.

* If $\mathcal{H} \sim \mathbbm{b}_{n-2, i}$, for $1\leq i \leq \delta$, then $\mathcal{H} \sim S_{n-1}$, which we already argued is not a sink.

* Otherwise, $\mathcal{H} \sim \mathbbm{b}_{n-2, i }$ where $ \delta+1 \leq i \leq c_{n-2}$. Either $\mathbbm{b}_{n-2,i}$ is not enclosed, on the left, by an image arc (so it is in the same sector as $\mathcal{S}_{n-2}$, which we already proved is not a sink), or it is enclosed, on the left, by an image arc $\varphi(\alpha_{n-3, j})$, where $ p+2 \leq j \leq c_{n-3}$; these arcs point out of $\mathcal{H}$ and into in $S_{n-2}$, so $\mathcal{H}$ is not a sink.

We deduce that any horizontal sector $\mathcal{H}$ spanning $\Gamma_{n-2}$ is not a sink.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-1}$.

* Suppose $\mathcal{H} \sim \mathbbm{b}_{n-1,1}$. Since this band is not enclosed, on the right, by any arc, we know that $\mathbbm{b}_{n-1,1} \sim \mathcal{S}_n$. Since $\mathcal{S}_n$ is not a sink, neither is $\mathcal{H}$.

* Instead suppose that $\mathcal{H}$ contains $\mathbbm{b}_{n-1, s}$, where $2 \leq s \leq c_n$. The band $\mathbbm{b}_{n-1, s}$ is enclosed, on the right, by $\varphi(\alpha_{n-1,s-1})$; this arc points out of $\mathbbm{b}_{n-1,s}$ and into $S_n$, so $\mathcal{H}$ is not a sink.

We deduce that no horizontal sector $\mathcal{H}$ spanning $\Gamma_{n-1}$ is a sink.

Therefore, no horizontal sector spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ is a sink.

This concludes the proof the $j=1$ sub-case of the heterogenous case. We now move onto the $j \geq 2$ case, and refer to <Ref> throughout.

[size=fbox, colback=gray!10]Case (1b): $\Gamma_{n-3}$ is heterogeneous and $j \geq 2$

First, we describe the construction of the branched surface in $\Gamma_{n-1}$: we co-orient $\alpha_{n-1, 1}, \ldots, \alpha_{n-1, j-1}$ to the left, and co-orient $\alpha_{n-1, j}, \ldots, \alpha_{n-1, c_{n-1}-1}$ to the right; see <Ref> for an example. We note: since $\widehat{\beta}$ is not a connected sum, if $j \geq 2$, then $c_{n-1} \geq j+1$, hence $c_{n-1} \geq 3$. We claim that there are no sink disk spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$:

* applying the same arguments from the $j=1$ case, we see that $\mathcal{S}_{n-3}$ and $\mathcal{S}_{n-2}$ are not sinks.

Recall that we defined $j$ so that $d_L(n-1; j, j+1) \geq 1$, and $\mathbbm{b}_{n-1,j}$ is the first band in $\Gamma_{n-1}$ for which this holds. In particular, since $j \geq 2$, we know that $c_{n-1} \geq 3$. In particular, since $\alpha_{n-1, 1}, \ldots, \alpha_{n-1, j-1}$ are all left pointers, and $\alpha_{n-1, j}, \ldots, \alpha_{n-1, c_{n-1}-1}$ are all right pointers, we see that $S_{n-1}$ and $S_n$ each contain both left and right pointing arcs. We deduce that neither $\mathcal{S}_{n-1}$ and $\mathcal{S}_n$ are sinks.

* The same argument as in the $j=1$ case shows that there are no sink disk polygon sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

$\circ$ If $\mathcal{H}$ spans $\Gamma_{n-1}$, we can apply the proof from the $j=1$ case to see that $\mathcal{H}$ is not a sink.

$\circ$ If $\mathcal{H}$ spans $\Gamma_{n-2}$, the proof from the $j=1$ case applies, and we see that $\mathcal{H}$ is not a sink.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-1}$.

$\bullet$ If $\mathcal{H} \sim \mathbbm{b}_{n-1, i}$ where $1 \leq i \leq j-1$, then $\partial \mathcal{H}$ contains $\alpha_{n-1,i}$, which points out of $\mathcal{H}$ and into $S_{n-1}$. Therefore, $\mathcal{H}$ is not a sink.

$\bullet$ If $\mathcal{H} \sim \mathbbm{b}_{n-1, j}$, then $\mathcal{H}$ also contains some band $\mathbbm{b}_{n-2, k}$, where $k \geq 2$. (Note that our presentation of $\beta$ ensures that $k \neq 1$, and recall that $\delta = d(n-3; p, p+1)$.)

$\circ$ If $2 \leq k \leq \delta$, then $\mathcal{H}$ meets $S_{n-2}$ in $\varphi(\alpha_{n-3,p})$. Therefore, we know that $\mathcal{H} \sim \mathbbm{b}_{n-2,1} \sim \mathbbm{b}_{n-2, 2} \sim \ldots \sim \mathbbm{b}_{n-2, \delta}$. But $\mathbbm{b}_{n-2,1} \sim \mathcal{S}_{n-1}$, which we already argued is not a sink. We deduce that $\mathcal{H} \sim \mathcal{S}_{n-1}$, so $\mathcal{H}$ is not a sink.

$\circ$ If $k \geq \delta+1$, then $\mathcal{H}$ meets $S_{n-2}$ in a left pointing image arc, or it is not enclosed by any arc (and $\mathcal{H} \sim \mathcal{S}_{n-2}$); see <Ref>. In both cases, we can deduce that $\mathcal{H}$ is not a sink.

$\bullet$  Finally, suppose $\mathcal{H}$ contains $\mathbbm{b}_{n-1, i}$ for $j+1 \leq i \leq c_{n}$. The band $\mathbbm{b}_{n-1, i}$ is enclosed, on the right, by $\varphi(\alpha_{n-1, j-1})$; this arc points out of $\mathbbm{b}_{n-1, i}$, and so $\mathcal{H}$ is not a sink.

Therefore, we have no horizontal sink disk sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

$\mathcal{H}$ at 94 331
$\mathcal{H}$ at 264 278

$\Gamma_{n-3}$ at 44 400
$\Gamma_{n-2}$ at 72 400
$\Gamma_{n-1}$ at 100 400

$\Gamma_{n-3}$ at 215 400
$\Gamma_{n-2}$ at 242 400
$\Gamma_{n-1}$ at 270 400

$j=2$ at 98 322
$j=2$ at 268 268

$p$ at 38 362
$p$ at 210 362

$\delta$ at 65 328
$\delta$ at 235 336

$\mathcal{H}$ at 94 70
$\mathcal{H}$ at 264 70

$\Gamma_{n-3}$ at 44 185
$\Gamma_{n-2}$ at 72 185
$\Gamma_{n-1}$ at 100 185

$\Gamma_{n-3}$ at 215 185
$\Gamma_{n-2}$ at 242 185
$\Gamma_{n-1}$ at 270 185

$j=2$ at 98 62
$j=2$ at 268 62

$p$ at 38 152
$p$ at 210 152

$\delta$ at 65 132
$\delta$ at 235 132

(A) at 72 210
(B) at 242 210
(D) at 72 5
(E) at 242 5

Building $B$ in $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ in Case (1b), where $\Gamma_{n-3}$ is heterogeneous and $j\geq 2$ (for simplicity of the figures, we let $j=2$ throughout). We apply our template to $\Gamma_{n-3}$, and then finish building the branched surface co-orienting $\alpha_{n-1,1} , \ldots, \alpha_{n-1, j-1}$ to the left, and $\alpha_{n-1,j}, \ldots, \alpha_{n-1, c_{n-1}-1}$ to the right. Frames (A) – (D) show some different possible configurations of $\Gamma_{n-1}$ with respect to $\Gamma_{n-2}$. Recall that $\delta := d(n-3; p, p+1)$, where $\alpha_{n-3,p}$ is the last left pointing plumbing arc in $S_{n-3}$. (For simplicity, in this figure, $p=2$ in all the frames.) Finally, let $\mathcal{H}$ denote the horizontal sector containing $\mathbbm{b}_{n-1,j}$. In (A), $\mathcal{H}$ meets $S_{n-2}$ in a right pointing image arc, and so $\mathcal{H} \sim \mathcal{S}_{n-1}$. In (B), $\mathcal{H}$ meets $S_{n-2}$ in a left pointing image arc. In (C), $\mathcal{H}$ is not enclosed by any arcs when it meets $S_{n-2}$, and so $\mathcal{H} \sim \mathcal{S}_{n-2}$. In (D), $c_{n-2} = 2$.

We see that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ in the $j\geq 2$ subcase of the heterogeneous case. This concludes the proof of the case where $\Gamma_{n-3}$ is heterogeneous, and we can proceed to study the case where $\Gamma_{n-3}$ is homogeneous.

[size=fbox, colback=gray!30]Sink disk analysis for Case (2): $\Gamma_{n-3}$ is homogeneous.

We now turn to the case where $\Gamma_{n-3}$ is homogeneous, i.e. $d(n-3; 1, c_{n-3}-1) = 0$.  As in Cases (1a) and (1b), we let $\delta := d(n-3; c_{n-3}-1, c_{n-3})$; additionally, we let  $\mathbbm{b}_{n-1,j}$ denote the first band in $\Gamma_{n-1}$ such that $d_L(n-1; j, j+1) \geq 1$.

We make a few quick observations.

Observation #1: since $\Gamma_{n-3}$ is homogeneous, the bands $\mathbbm{b}_{n-2, 1}, \ldots, \mathbbm{b}_{n-2, \delta}$ are all enclosed, on the left, by $\varphi(\alpha_{c_{n-3}, c_{n-3}-1})$, and that the bands $\mathbbm{b}_{n-2, \delta+1}, \ldots, \mathbbm{b}_{n-2, c_{n-2}}$ are not enclosed, on the left, by any arcs. (One can refer  to Figures <ref> and <ref>.)

Observation #2: since $\Gamma_{n-3}$ is homogeneous, $d(n-3; c_{n-3}-1, c_{n-3}) + d(n-3; c_{n-3}, 1) = c_{n-3}$, where each term on the left is strictly positive. In particular, all the bands in $\Gamma_{n-2}$ are concentrated between three consecutive bands in $\Gamma_{n-3}$: $\mathbbm{b}_{n-3, c_{n-3}-1}$, $\mathbbm{b}_{n-3, c_{n-3}}$, and $\mathbbm{b}_{n-3,1}$. Finally, this also means that $d_L(n-2; 1, \delta) = 0$ and $d_L(n-2; \delta+1, c_{n-2}) = 0$, as in Figures <ref> and <ref>.

With these preliminaries in place, we can proceed with building $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$. As in the case where $\Gamma_{n-3}$ is heterogeneous (i.e. Cases (1a) and (1b)), we have two sub-cases to consider:

* Case (2a): $\Gamma_{n-3}$ is homogeneous and $j=1$.

* Case (2b): $\Gamma_{n-3}$ is homogeneous and $j \geq 2$.

[size=fbox, colback=gray!10]Case (2a): $\Gamma_{n-3}$ is homogeneous and $j=1$

We begin by building $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$: we take no product disks from $\Gamma_{n-2}$, and we take $c_{n-1}-1$ disks from $\Gamma_{n-1}$ by co-orienting $\alpha_{n-1,1}, \ldots, \alpha_{n-1, c_{n-1}-1}$ to the right.

Since $j=1$, we must have that $d(n-2, 1, \delta) = 0$ (otherwise, there would exist some $s$ such that $1 \leq s \leq \delta$ such that $d(n-2; s, s+1) = 1$, and this contradicts that $\beta$ is standardized). Analogously, we see that there cannot exist some $s$ with $\delta+1 \leq s \leq c_{n-2}$ such that $d(n-2; s, s+1) = 1$. We deduce that $\mathbbm{b}_{n-1, 1}$ must occur at the same “level” as $\mathbbm{b}_{n-3, c_{n-3}}$, as seen in <Ref>. More precisely, we have that $d_L(n-2; \delta, \delta+1) = d(n-2; \delta, \delta+1) = 1$.

With this established, we can now prove that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$; we refer to <Ref> throughout.

* We first show that $\mathcal{S}_{n-3}$ is not a sink. This requires analyzing the cases $n=6$ and $n \geq 7$ separately.

* Suppose $n = 6$. In this case, $n-3 = 3$, hence $\mathcal{S}_{n-3} = \mathcal{S}_3$ and $\Cmax = \Codd$. Thus, the construction of $B$ in $\Gamma_{n-3} = \Gamma_3$ is inherited from an application of <Ref> to $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$. In particular, $B \cap S_3$ contains $c_3-1$ plumbing arcs and one image arc. Moreover, by consulting <Ref>, we see that if $\Gamma_3$ is homogeneous, then we are either in Case (A) or Case (B.1) of <Ref>. So, we must argue that for these sub-cases, $S_{n-3}$ is not a sink when we build $B$ in $\Gamma_{n-1} = \Gamma_{5}$ as above.

* In Case (A) of <Ref>, we know that $\partial \mathcal{S}_3$ contains $\varphi(\alpha_{2,1}), \alpha_{3, 1}, \ldots, \alpha_{3, c_3-1}$ (see <Ref>), and that $\mathcal{S}_3 \sim \mathbbm{b}_{3, c_3}$.

* In Case (B.1) of <Ref>, we consult <Ref> to see that $\mathcal{S}_3$ is not a sink.

Therefore, to prove that $\mathcal{S}_{n-3}$ is not a sink when $n=6$, it suffices to show that the sector containing $\mathbbm{b}_{3, c_3}$ is not a sink. In fact, this observation also holds in the $n \geq 7$ case:

* If $n \geq 7$, then by <Ref>, $S_{n-3}$ contains only plumbing arcs, so $\mathcal{S}_{n-3} \sim \mathbbm{b}_{n-3, c_{n-3}}$.

Therefore, we need to show that when $n \geq 6$, the sector containing $\mathbbm{b}_{n-3, c_{n-3}}$ is not a sink. Let $\mathcal{S}$ denote this sector. By design, the sector $\mathcal{S} \sim \mathbbm{b}_{n-2, 1}$. This band is not enclosed, on the right, by any bands, so $\mathcal{S}_{n-3} \sim \mathcal{S}_{n-1}$. As $S_{n-1}$ contains only right pointing plumbing arcs, the arc $\alpha_{{n-1},1}$ points out of $\mathcal{S}_{n-1}$. Thus, $\mathcal{S} \sim \mathcal{S}_{n-1} \sim \mathcal{S}_{n-3}$ is not a sink, as desired.

It remains to show that $\mathcal{S}_{n-2}$ and $\mathcal{S}_n$ are not sinks.  All the arcs in $S_{n-2}$ are right pointing image arcs, so $\mathcal{S}_{n-2}$ is not a sink.

Finally, by design, we see that $\mathcal{S}_{n} \sim \mathbbm{b}_{n-1, 1} \sim \mathbbm{b}_{n-2, \delta+1}$; see <Ref>. However, we already noticed that $\mathbbm{b}_{n-2, \delta+1}$ is not enclosed, on the left, by any arcs. We deduce that $\mathcal{S}_{n} \sim \mathcal{S}_{n-2}$, which we already argued is not a sink.

We deduce that none of the Seifert disk sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ are sinks.

* As above, we analyze the $n=6$ and $n \geq 7$ cases separately.

$\circ$ If $n=6$, $S_{n-3}$ contains both plumbing and image arcs, so $S_{n-3} = S_3$ could contain a polygon sector. We are assuming that $\Gamma_3$ is homogeneous, putting us in Cases (A) and (B.1) of <Ref>. However, as we saw in the proof of that lemma, there are no polygon sectors in $\Gamma_3$ in Case (A); see  <Ref>. In Case (B.1), there is a sector which appears like it could be a polygon sector, but it contains a band in $\Gamma_3$ (see <Ref>), so it does not lie in $S_{n-3}$. We deduce that there are no polygon sectors in $S_3$, regardless of which sub-case of <Ref> occurs.

As we chose no product disks from $\Gamma_{n-2}$ and $c_{n-1}-1$ product disks from $\Gamma_{n-1}$, the remaining Seifert disks cannot contain both plumbing and image arcs, and there are no polygon sectors in $S_{n-2}, S_{n-1}$, and $S_n$.

$\circ$ If $n \geq 7$, then <Ref> ensures that $S_{n-2}, S_{n-1}$, and $S_n$ all only contain only plumbing or image arcs. Therefore, none of these Seifert disks can contain a polygon sector, and in particular, there are no polygon sink disk sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$. (We recommend the reader refer to <Ref> throughout.) Note: we need not analyze the $n=6$ and $n \geq 7$ cases separately.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-3}$. Suppose $\mathcal{H}$ contains $\mathbbm{b}_{n-3, i}$, for $1 \leq i \leq c_{n-3}-1$. This means $ \partial \mathcal{H}$ contains $\alpha_{n-3, i}$, and this arc points out of $\mathbbm{b}_{n-3, i}$ and into $S_{n-3}$. Therefore, $\mathcal{H}$ is not a sink. Alternatively, if $\mathcal{H} \sim \mathbbm{b}_{n-3, c_{n-3}}$, then $\mathcal{H} \sim \mathbbm{b}_{n-2, 1} \sim \ldots \sim \mathbbm{b}_{n-2, \delta} \sim \mathcal{S}_{n-1}$; we already argued that $\mathcal{S}_{n-1}$ is not a sink.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-2}$. If $\mathcal{H}$ contains $\mathbbm{b}_{n-2, i}$, for $1 \leq i \leq \delta$, then $\mathcal{H} \sim \mathcal{S}_{n-1}$, which we already proved is not a sink. Alternatively, suppose $\mathcal{H} \sim \mathbbm{b}_{n-2, i}$, for $\delta +1 \leq i \leq c_{n-2}$. These arcs are not enclosed, on the left, by any arcs, so $\mathcal{H} \sim \mathcal{S}_{n-2}$, which we proved is not a sink. Therefore, no horizontal sector spanning $\Gamma_{n-2}$ is a sink.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-1}$. If $\mathcal{H} \sim \mathbbm{b}_{n-1, 1}$, then $\mathcal{H} \sim \mathcal{S}_{n}$, which is not a sink. Otherwise, $\mathcal{H} \sim \mathbbm{b}_{n-1, i}$, where $2 \leq i \leq c_{n-1}$. Notice that $\mathbbm{b}_{n-1, i}$ is enclosed, on the right, by $\varphi(\alpha_{n-1, i-1})$, and this arc points out of $\mathbbm{b}_{n-1, i}$. Therefore, $\mathcal{H}$ is not a sink.

Therefore, no horizontal sector whose span includes $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ is a sink.

We deduce there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ in Case (2a).

(A) at 66 5
(B) at 246 5
(C) at 418 5

$\Gamma_{n-3}$ at 38 200
$\Gamma_{n-2}$ at 65 200
$\Gamma_{n-1}$ at 95 200

$\Gamma_{n-3}$ at 218 200
$\Gamma_{n-2}$ at 245 200
$\Gamma_{n-1}$ at 275 200

$\Gamma_{n-3}$ at 390 200
$\Gamma_{n-2}$ at 417 200
$\Gamma_{n-1}$ at 445 200

$j=1$ at 94 92
$j=1$ at 274 84
$j=1$ at 444 84

$\delta$ at 65 118
$\delta$ at 245 100
$\delta$ at 416 100

Constructing $B$ in $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ when $\Gamma_{n-3}$ is homogeneous and $j=1$. All the arcs in $\Gamma_{n-1}$ are coherently oriented to the right. By design, $S_{n-3} \sim \mathcal{S}_{n-1}$ and $\mathcal{S}_{n-2} \sim \mathcal{S}_n$. In (A), $\delta := d(n-3; c_{n-3}-1, c_{n-3}) = 1$. In (B), $\delta \geq 3$. In (C), $c_{n-1} = 2$.

[size=fbox, colback=gray!10]Case (2b): $\Gamma_{n-3}$ is homogeneous and $j \geq 2$

First, we describe how to build the branched surface in $\Gamma_{n-1}$: we choose no product disks from $\Gamma_{n-2}$, and $c_{n-1}-1$ product disks from $\Gamma_{n-1}$ by co-orienting $\alpha_{n-1,1}, \ldots, \alpha_{n-1, j-1}$ to the left and we co-orienting $\alpha_{n-1, j}, \ldots, \alpha_{n-1, c_{n-1}-1}$ to the right.

We note: since $\widehat{\beta}$ is not a connected sum, if $j \geq 2$, then $c_{n-1} \geq j+1$, hence $c_{n-1} \geq 3$. We prove that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$, and refer to <Ref> throughout.

(A) at 66 5
(B) at 235 5
(C) at 406 5

$\Gamma_{n-3}$ at 38 200
$\Gamma_{n-2}$ at 65 200
$\Gamma_{n-1}$ at 95 200

$\Gamma_{n-3}$ at 208 200
$\Gamma_{n-2}$ at 235 200
$\Gamma_{n-1}$ at 265 200

$\Gamma_{n-3}$ at 380 200
$\Gamma_{n-2}$ at 408 200
$\Gamma_{n-1}$ at 435 200

$j$ at 88 74
$j$ at 258 46
$j$ at 430 74

$\delta$ at 65 118
$\delta$ at 235 100
$\delta$ at 406 118

Constructing $B$ in $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ when $\Gamma_{n-3}$ is homogeneous and $j \geq 2$. By design, $S_{n-3} \sim \mathcal{S}_{n-1}$. In (A), $\delta := d(n-3; c_{n-3}-1, c_{n-3}) = 1$. In (B), $\delta \geq 3$. In (C), $c_{n-2} = 2$.

* The same argument as in Case (2a) tells us that either $n = 6$ and $\mathcal{S}_{n-3} = \mathcal{S}_3$ is not a sink (because the construction of $B$ in $\Gamma_3$ is inherited from Case (B.1) of <Ref>), or $n \geq 6$ and $\mathcal{S}_{n-3} \sim \mathbbm{b}_{n-3, c_{n-3}} \sim \mathcal{S}_{n-1}$.  When the latter occurs, we know that $\alpha_{n-1, j}$ is a right pointing plumbing arc, pointing out of $\mathcal{S}_{n-1}$, so $\mathcal{S}_{n-3} \sim \mathcal{S}_{n-1}$ is not a sink. Additionally, all the image arcs in $S_{n-2}$ point out of $\mathcal{S}_{n-2}$, and so $\mathcal{S}_{n-2}$ is not a sink. Finally, $\varphi(\alpha_{n-1,1})$ is in $\partial \mathcal{S}_n$, and points out of $\mathcal{S}_{n-1}$ and into $\mathbbm{b}_{n-1, 2}$, so $\mathcal{S}_n$ is not a sink. We deduce that none of the Seifert disk sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ are sink disks.

* The same argument from Case (2a) tells us that there are no sink disk polygon sectors spanning these columns.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$. We argue that such an $\mathcal{H}$ cannot be a sink.

$\circ$ If $\mathcal{H}$ spans $\Gamma_{n-3}$, we apply the same argument from Case (2a) to see that $\mathcal{H}$ is not a sink.

$\circ$ If $\mathcal{H}$ spans $\Gamma_{n-2}$, we apply the same argument from Case (2a) to see that $\mathcal{H}$ is not a sink.

$\circ$ Suppose $\mathcal{H}$ spans $\Gamma_{n-1}$. If $\mathcal{H} \sim \mathbbm{b}_{n-1, i}$, where $1 \leq i \leq j-1$, then $\partial \mathcal{H}$ contains $\alpha_{n-1, i}$. This arc points out of $\mathcal{H}$ and into $S_{n-1}$, so $\mathcal{H}$ is not a sink.

If $\mathcal{H} \sim \mathbbm{b}_{n-1, j}$, then $\mathcal{H}$ also contains $\mathbbm{b}_{n-2, \delta+1}$, as in <Ref>. This band is not enclosed, on the left, by any arcs, so $\mathcal{H} \sim \mathcal{S}_{n-2}$; we already argued that this sector is not a sink.

Finally, if $\mathcal{H} \sim \mathbbm{b}_{n-1, i}$ for $j+1 \leq i \leq c_{n-1}$, then $\partial \mathcal{H}$ contains $\varphi(\alpha_{n-1, i-1})$; this arc points out of the $\mathbbm{b}_{n-1, i}$ and into $S_n$, so $\mathcal{H}$ is not a sink.

We deduce that no horizontal sector spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$ is a sink.

Therefore, there are no sink disk branch sectors spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$.

Finally, we simultaneously verify that for all four of the cases of this lemma (i.e. Cases (1a), (1b), (2a), and (2b)), there is no pairwise linking between arcs in $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$. Since we chose no disks from $\Gamma_{n-2}$, we need only show that there is no pairwise linking between arcs in $S_{n-3}$ and $S_{n-1}$. As linking only occurs between $\alpha$ and $\alpha'$ if they are in the same column or adjacent columns, it suffices to show that there is no pairwise linking if $\alpha$ and $\alpha'$ lie in the same Seifert disk. This follows from the proof of <Ref>. $\Box$

§.§ Finalé, a.k.a. sewing the constructions together.

Finally, we combine the constructions of the previous sections to build taut foliations for positive $n$-braid knots, where $n \geq 5$.

Suppose $K$ is a positive $n$-braid knot. Fix a standardized positive $n$-braid $\beta$ with $\widehat{\beta} \approx K$ and $n \geq 5$. Suppose, for this $\beta$, that $\Cmax = \Codd$. There is a laminar branched surface $B$ for $K$ such that $\tau_B$ fully carries all slopes $r < g(K)+1$.

Before we execute the construction of the branched surface, we note: at every stage of the construction (i.e. within the proofs of Lemmas <ref>, <ref>, <ref>, <ref>, and <ref>), we always conjugated our brain into a specific presentation before building our branched surface in those columns. Throughout the forthcoming proof, we will need to intermittently conjugate $\beta$.

To start, we follow the procedure outlined in <Ref>, which builds $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$. As such, the branched surface contains product disks from $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$. Our construction of the branched surface in the remaining columns depends on $n$. We first build $B$ for all $n \geq 5$, and subsequently produce a lower bound on $\TauSup$.

[size=fbox, colback=gray!30]Suppose $n=5$.
In this case, our braid has four columns, and we already constructed $B$ in $\Gamma_1\cup \Gamma_2 \cup \Gamma_3$. Therefore, it suffices to see how to build $B$ in $\Gamma_4$, and then check that no sink disks were produced. When $n=5$, we take no product disks from $\Gamma_4$. As there were four sub-cases of <Ref>, we need to analyze each separately to determine why $B$ has no sink disks. Recall that we summarized the outcomes of <Ref> in <Ref>.

[size=fbox, colback=gray!10]Suppose $n=5$, and Case (A) from <Ref> holds.  In this case, as we recorded in <Ref>, we know that there is a unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$, and it is the sector $\mathcal{S} \sim \mathcal{S}_3 \sim \mathbbm{b}_{2,1}$ (we suggest the reader consult  <Ref> to “see” $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$). By design, $\partial S_{n-2} = \partial S_3$ contains exactly one image arc and $c_3-1$ plumbing arcs, and there are no polygon sectors in $S_3$. Therefore, if we take no product disks from $\Gamma_4$, we can apply <Ref> and deduce that there are no sink disks spanning $\Gamma_3 \cup \Gamma_4$. This produces a sink disk free branched surface.

[size=fbox, colback=gray!10]Suppose $n=5$, and Case (B.1) from <Ref> holds.

As we recorded in <Ref>, we know that there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$; we suggest the reader refer to <Ref> to “see” $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$.

In this case, we know that $d(3; \delta-2, \delta)=c_4$. Moreover, since $\widehat{\beta}$ is not a connected sum, then we cannot have all the bands in $\Gamma_4$ concentrated between a pair of consecutive bands in $\Gamma_3$. We deduce that $d(3; \delta-2, \delta-1) \geq 1$ and $d(3; \delta-1, \delta) \geq 2$. Let $d(3; \delta-2, \delta) = \epsilon_1$ and $d(3; \delta-1, \delta) = \epsilon_2$. We include no product disks from $\Gamma_4$ into $B$, and now argue that there are no sink disks spanning $\Gamma_3 \cup \Gamma_4$. This will allow us to deduce that we have constructed a sink disk free branched surface.

* As summarized in <Ref>, $\mathcal{S}_3$ is not a sink. Since $S_4$ contains only right pointing image arcs, $\mathcal{S}_4$ is not a sink.

We now argue that $\mathcal{S}_5 \sim \mathcal{S}_4$, hence $\mathcal{S}_5$ is not a sink. By design, we have chosen no product disks from $\Gamma_4$, so in particular, $\mathcal{S}_5 \sim \mathbbm{b}_{4,\epsilon_1 + \epsilon_2}$. Moreover, by design, $\mathbbm{b}_{4, \epsilon_1 + \epsilon_2}$ is not enclosed, on the left, by any arc; see <Ref>. Therefore, $\mathcal{S}_4 \sim \mathbbm{b}_{4, \epsilon_1 + \epsilon_2}$, and we deduce that $\mathcal{S}_4 \sim \mathcal{S}_5$.

$\Gamma_{3}$ at 42 174
$\Gamma_{4}$ at 70 174

$\delta$ at 40 62
$\delta-2$ at 40 130
$\epsilon_1$ at 70 100
$\epsilon_1 + \epsilon_2$ at 69 62

Zooming on on the bands $\mathbbm{b}_{3,\delta-2}, \mathbbm{b}_{3, \delta-1}$, and $\mathbbm{b}_{3, \delta}$ in <Ref>. We see that $\mathcal{S}_{4} \sim \mathcal{S}_5$.

* There is a unique sector $\mathcal{P}$ in $\Gamma_3 \cup \Gamma_4$ which lies in a Seifert disk and contains both plumbing and image arcs in its boundary; see <Ref>. By design, $\mathcal{P} \sim \mathbbm{b}_{3, \delta-1} \sim \mathbbm{b}_{4, \epsilon_1} \sim \mathcal{S}_5$, and we already proved that $\mathcal{S}_5$ is not a sink.

* Suppose $\mathcal{H}$ is a horizontal sector spanning $\Gamma_3$. If $\mathcal{H}$ spans $\Gamma_3$, it contains some band $\mathbbm{b}_{3, t}$. By design, all bands in $\Gamma_3$ are either enclosed, on the left, by a left pointing plumbing arc (so it is not a sink), or the band is part of the sector $\mathcal{P}$ described above (which we just proved is not a sink). Therefore, no horizontal sector spanning $\Gamma_3$ is a sink. Alternatively, if $\mathcal{H}$ spans $\Gamma_4$, then since there are no plumbing or image arcs in $S_5$, $\mathcal{H} \sim \mathcal{S}_5$, which is not a sink.

We deduce that we can construct a sink disk free branched surface in this case.

[size=fbox, colback=gray!10]Suppose $n=5$, and Case (B.2) from <Ref> holds.

As summarized in <Ref>, in this case, we know that $\Gamma_3$ is heterogeneous, and that there are no sink disks spanning $\Gamma_1 \cup \Gamma_2$. We now argue that there are no sink disks spanning $\Gamma_3 \cup \Gamma_4$:

* <Ref> confirms that $\mathcal{S}_3$ is not a sink. $S_3$ contains both left and right pointing plumbing arcs, so $\mathcal{S}_4$ contains both left and right pointing image arcs; thus, it is not a sink. We now argue that $\mathcal{S}_5$ is not a sink.

Throughout the forthcoming analysis, we use the same band labellings as in Case (B.2) and <Ref>. By the assumptions of Case (B.2), we have $d(3; c_3, \delta) = c_4$. Let $\alpha_{3,p}$ be the first right pointing plumbing arc in $\Gamma_3$; by assumption, $1 \leq p \leq \delta$. Since $\widehat{\beta}$ is not a connected sum, then we cannot have that $d(3; p, p+1) = c_4$. Therefore, $d(3; p, p+1) : = \epsilon < c_4$ and $d(3; p+1, \delta) := \eta \geq 2$. Note that $\epsilon + \eta = c_4$.

Since $d(3; p+1, \delta) \geq 2$, then there exists some band $\mathbbm{b}_{4, k}$, where $\epsilon+1 \leq k \leq \epsilon + \eta$ such that either (a) $\mathbbm{b}_{4, k}$ is enclosed, on the left, by a left pointing image arc, or (b) $\mathbbm{b}_{4, k}$ is not enclosed, on the left, by any image arcs; see

see <Ref>.
However, since we included no product disks from $\Gamma_4$ while building $B$, we know that no band in $\Gamma_4$ is enclosed, on the right, by any arcs. Therefore, $\mathcal{S}_5 \sim \mathbbm{b}_{4, 1} \sim \ldots \sim \mathbbm{b}_{4, k} \sim \ldots \sim \mathbbm{b}_{4, c_4}$.

If (a) occurs, then $\partial \mathcal{S}_5$ contains an outward pointing image arc, but if (b) happens, then $\mathcal{S}_5 \sim \mathcal{S}_4$ (and we already proved the latter is not a sink). We deduce that $\mathcal{S}_5$ is never a sink.

$\Gamma_{3}$ at 42 170
$\Gamma_{4}$ at 70 170

$p=1$ at 44 124
$\delta$ at 44 52
$\epsilon$ at 70 100

$\Gamma_{3}$ at 195 170
$\Gamma_{4}$ at 225 170

$p=2$ at 192 90
$\delta$ at 195 52
$\epsilon$ at 223 100

$\Gamma_{3}$ at 350 170
$\Gamma_{4}$ at 378 170

$p=2$ at 344 90
$\delta$ at 350 52
$\epsilon$ at 376 100

Zooming in on the bands enclosed, on the left, by $\varphi(\alpha_{2,1})$ from <Ref> (to be consistent, we have used the same cyclic conjugated presentation of the braid from that figure).
We know that $d(3; 1, \delta) \geq 1$, but we do not know the distribution of the bands within $\Gamma_4$, so the visible bands in $\Gamma_4$ are dashed. No band in $\Gamma_4$ is enclosed, on the right, by any arc. If there is at least one enclosed, on the left, by an image arc, the image arc is a left pointer, and $\mathcal{S}_5$ is not a sink. Otherwise, $\mathcal{S}_5 \sim \mathcal{S}_4$, which we already proved is not a sink.

* There is a unique sector $\mathcal{P}$ in $S_3$ which contains both plumbing and image arcs in its boundary. Notice that $\partial \mathcal{P}$ contains $\alpha_{3, \delta-2}$, which is a right pointing plumbing arc.

Therefore, $\mathcal{P}$ is not a sink.

* If $\mathcal{H}$ is a horizontal sector spanning $\Gamma_4$, then $\mathcal{H} \sim \mathbbm{b}_{4, i}$, for some $1 \leq i \leq c_4$. But these bands are not enclosed, on the right, by any arcs, so $\mathcal{H} \sim \mathcal{S}_5$, which we already proved is not a sink.

Next, suppose that $\mathcal{H}$ is a horizontal sector spanning $\Gamma_3$, so that $\mathcal{H} \sim \mathbbm{b}_{3, j}$, for some $1 \leq j \leq c_3$.

$\circ$ The bands $\mathbbm{b}_{3, \delta}, \mathbbm{b}_{3, \delta+1}, \ldots, \mathbbm{b}_{3, c_3}$, and $\mathbbm{b}_{3,1}, \ldots, \mathbbm{b}_{3, p-1}$ are all enclosed, on the left, by a left pointing image arc. Therefore, if $\mathcal{H}$ contains one of these bands, $\mathcal{H}$ is not a sink.

$\circ$ Suppose $\mathcal{H} \sim \mathbbm{b}_{3, p}$. Since $d(3; p, p+1) \geq 2$, then $\mathcal{H} \sim \mathbbm{b}_{4,1} \sim \ldots \sim \mathbbm{b}_{4, \epsilon} \sim \mathcal{S}_5$; we already proved $\mathcal{S}_5$ is not a sink.

$\circ$ The bands $\mathbbm{b}_{3, p+1}, \ldots, \mathbbm{b}_{3, \delta-1}$ are all enclosed, on the right, by a left pointing image arc. Thus, if $\mathcal{H}$ contains one of these bands, it is not a sink.

Therefore, $B$ contains no sink disk spanning $\Gamma_3 \cup \Gamma_4$, hence it is sink disk free.

[size=fbox, colback=gray!10]Suppose $n=5$, and Case (C) from <Ref> holds.

As summarized in <Ref>, in this case, after applying our template to $\Gamma_3$, the column is heterogeneous. Additionally, we know that there is a unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$, and it is the sector $\mathcal{S}$ containing $\mathbbm{b}_{2,1} \sim \mathbbm{b}_{3, c_3}$ (we refer the reader to  <Ref> to recall our labelling conventions). We include no product disks from $\Gamma_4$ into $B$, and show that the result contains no sink disks spanning $\Gamma_3 \cup \Gamma_4$:

* In <Ref>, we proved that $\mathcal{S}_3$ is not a sink. By design, $S_4$ contains both left and right pointing image arcs; since $B$ includes no product disks from $\Gamma_4$, we deduce that $\partial \mathcal{S}_4$ must contain an outward pointing arc, so it is not a sink.

We now prove that $\mathcal{S}_5$ is not a sink. Since we included no product disks from $\Gamma_4$, we know that $\mathcal{S}_5 \sim \mathbbm{b}_{4,1} \sim \ldots \sim \mathbbm{b}_{4, c_4}$, and it suffices to show that some band in $\Gamma_4$ is enclosed, on the left, by a left pointing image arc, or that $\mathcal{S}_5 \sim \mathcal{S}_4$.

We first notice: by the definition of our template (and the fact that we are in Case (C) of <Ref>), we know that the following inequalities hold (once again, we refer the reader to <Ref> to recall the labelings):

\begin{align*}
d(3; \delta, j-1) &= 0 \text{ (by the construction in \Cref{lemma:Generic_Cmax=Codd_Gammas12})}\\
d(3; j-1, j) &\geq 1 \text{ (by the definition of our template)} \\
d(3; j-1, j) &< c_4 \text{ (because $\beta$ is not a connected sum)}
\end{align*}

These three inequalities imply that $d(3; j+1, \delta) \geq 1$. This means that either (a) there is some band $\mathbbm{b}_{4, k}$ which is enclosed, on the left, by a left pointing image arc, or (b) that there is some band $\mathbbm{b}_{4,k}$ which is not enclosed, on the left, by any image arc, so $\mathcal{S}_5 \sim \mathcal{S}_4$. We deduce that $\mathcal{S}_5$ is not a sink.

* Applying the same analysis from Case (B.2) tells us that there are no sink disk polygon sectors in $\Gamma_3 \cup \Gamma_4$; see <Ref>.

* As described in <Ref>, there is a unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$, and it is the horizontal sector containing both $\mathbbm{b}_{2,1}$ and $ \mathbbm{b}_{3, c_3}$. So, to prove that $B$ is completely sink disk free, we need to argue that there are no horizontal sink disks spanning $\Gamma_3 \cup \Gamma_4$.

We quickly argue that no horizontal sector spanning $\Gamma_4$ is a sink: we already argued that $\mathcal{S}_5 \sim \mathbbm{b}_{4,1} \sim \ldots \sim \mathbbm{b}_{4, c_4}$. Since we already proved that $\mathcal{S}_5$ is not a sink, we immediately know that no horizontal sector spanning $\Gamma_4$ is a sink.

Finally, we argue that no horizontal sector spanning $\Gamma_3$ is a sink. Let $\mathcal{H}$ denote such a sector.

$\circ$ If $\mathcal{H} \sim \mathbbm{b}_{3, k}$, for $1 \leq k \leq \delta-1$, then $\partial \mathcal{H}$ contains $\varphi(\alpha_{3,k-1})$ which points out of the sector, so $\mathcal{H}$ is not a sink.

$\circ$ If $\mathcal{H} \sim \mathbbm{b}_{3, k}$, for $\delta \leq k \leq j-1$, then $\partial \mathcal{H}$ contains $\alpha_{3,k}$ which points out of the sector, so $\mathcal{H}$ is not a sink.

$\circ$ Suppose $\mathcal{H} \sim \mathbbm{b}_{3, j}$. Then by definition of our template, $\mathcal{H} \sim \mathbbm{b}_{4,s}$ for some $s$. But as discussed above, this means $\mathcal{H} \sim \mathcal{S}_5$, which we already proved is not a sink.

$\circ$ If $\mathcal{H} \sim \mathbbm{b}_{3, k}$, for $j+1 \leq k \leq c_3$, then $\partial \mathcal{H}$ contains $\varphi(\alpha_{3,k-1})$ which points out of the sector, so $\mathcal{H}$ is not a sink.

For emphasis, we remark: suppose $\mathcal{H} \sim \mathbbm{b}_{2,1} \sim \mathbbm{b}_{3, c_3}$. If $j = c_3$, then $\mathcal{H} \sim \mathcal{S}_5$ (which is not a sink), and if $j < c_3$, then $\mathcal{H}$ meets $\mathcal{S}_4$ is a left pointing image arc. Regardless, it is not a sink. Therefore, we see that no horizontal sector spanning $\Gamma_3 \cup \Gamma_4$ is a sink, and $B$ is sink disk free.

This concludes the analysis when $n=5$.

[size=fbox, colback=gray!30]Suppose $n= 6$.

In <Ref>, we constructed our branched surface so that it contained product disks from $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$. When $n=6$, our braid has five columns; to complete the construction, we must specify some presentation of our braid, and then build $B$ in the remaining two columns. We will choose no product disks from $\Gamma_4$ and $c_5-1$ product disks from $\Gamma_5$.

As recalled in <Ref>, were four sub-cases in <Ref>, and each of these four cases, there is at most one potential sink disk spanning $\Gamma_1 \cup \Gamma_2$: in Cases (A) and (C), the sector is $\mathcal{S}_3$ or $\mathbbm{b}_{3, c_3}$, respectively; in Cases (B.1) and (B.2), there were no sink disks spanning $\Gamma_1 \cup \Gamma_2$. In particular, the unique potential sink disk spanning $\Gamma_1 \cup \Gamma_2$ always spans $\Gamma_3$. <Ref> describes how to cyclically conjugate $\beta$ with respect to some distinguished band in $\Gamma_3$, and then build $B$ in $\Gamma_4 \cup \Gamma_5$. <Ref> allows us to conclude that there are no sink disks spanning $\Gamma_3 \cup \Gamma_4 \cup \Gamma_5$. We deduce that we can build a branched surface $B$ which contains co-oriented product disks from all columns except $\Gamma_4$, such that $B$ is sink disk free.

[size=fbox, colback=gray!30]Suppose $n=7$.

By assumption, we already built $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ via <Ref>; we now follow <Ref> to construct $B$ in $\Gamma_4 \cup \Gamma_5$. The conclusions of <Ref> tell us that the unique potential sink disk spanning $\Gamma_1 \cup \ldots \cup \Gamma_5$ is $\mathcal{S}_5$. Applying <Ref>, we see that there are no sinks spanning $\Gamma_5 \cup \Gamma_6$. We deduce that we have constructed a sink disk free branched surface.

[size=fbox, colback=gray!30]Suppose $n\geq 8$.
We already built $B$ in $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3$ via <Ref>, and we now follow <Ref> to construct $B$ in $\Gamma_4 \cup \Gamma_5$. We now iteratively build $B$ in the remaining columns. To simplify our exposition, we define the value $f$ as follows:

\begin{align*}
f :=
\begin{cases}
\displaystyle \frac{(n-1)-5}{2}-2 & \qquad n-1 \equiv 1 \mod 2 \\ \\
\displaystyle \frac{(n-1)-6}{2} & \qquad n-1 \equiv 0 \mod 2
\end{cases}
\end{align*}

To build $B$, we iteratively apply <Ref>, by assuming that we have build $B$ to contain co-oriented product disks in some non-trivial subset of columns $\Gamma_1 \cup \ldots \Gamma_{5+2t}$, for $t \in \{0, \ldots, f-1\}$. By construction, before each application of <Ref>, the only potential sink disk spanning $\Gamma_1 \cup \ldots \cup \Gamma_{5+2t}$ is $\mathcal{S}_{5+2t}$; after building $B$ in $\Gamma_1 \cup \ldots \cup \Gamma_{5+2t} \cup \Gamma_{5+2t+1} \cup \Gamma_{5+2t+2}$ as directed by <Ref>, the only potential sink disk spanning these columns is $\mathcal{S}_{5+2t+2}$.

Suppose $n-1 \equiv 1 \mod 2$. When $t=f-1$, our last application of <Ref> assumes we have constructed $B$ to include product disks from $\Gamma_1 \cup \ldots \cup \Gamma_{n-5}$, and then concludes by producing a branched surface which includes product disks from $\Gamma_1 \cup \ldots \cup \Gamma_{n-3}$. By design, at this stage, the only potential sink disk spanning these columns is $\mathcal{S}_{n-3}$. We now apply  <Ref> to build $B$ in $\Gamma_{n-2} \cup \Gamma_{n-1}$.  <Ref> ensures that there are no sink disks spanning $\Gamma_{n-3} \cup \Gamma_{n-2} \cup \Gamma_{n-1}$. We deduce that our resulting branched surface, which contains co-oriented product disks from $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_5 \cup \ldots \cup \Gamma_{n-1}$, is sink disk free.

Instead suppose $n-1 \equiv 0 \mod 2$. When $t=f-1$, our last application of <Ref> assumes that we have already constructed our branched surface in $\Gamma_1 \cup \ldots \cup \Gamma_{n-4}$, and then outputs a branched surface containing product disks spanning $\Gamma_1 \cup \ldots \cup \Gamma_{n-2}$. The only potential sink disk spanning these columns is $\mathcal{S}_{n-2}$. We now apply <Ref> to see that without including any product disks from $\Gamma_{n-1}$, we have constructed a sink disk free branched surface which contains co-oriented product disks from $\Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_5 \cup \ldots \cup \Gamma_{n-2}$.