Proof. The doubly pointed Heegaard diagram $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m};u,v)$ is obtained from the marked Heegaard diagram $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m};p_{n+m})$ by replacing $p_{n+m}\in \unicode[STIX]{x1D706}_{n+m}$ with the two markings $u,v$ on its two sides.

Choose a self-indexing Morse function $h:Y\rightarrow [0,3]$ with unique critical points of indices $0$ and $3$ , and compatible with the Heegaard diagram $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m};u,v)$ . Modify the Morse function so that its critical points remain unchanged, but so that the value of $h$ over the index- $2$ critical point corresponding to $\unicode[STIX]{x1D706}_{n+m}$ becomes $8/3$ . The pre-images of $3/2$ and $7/3$ under $h$ are then the surface $\unicode[STIX]{x1D6F4}$ and a torus $T$ , respectively. If we use the flow of a gradient-like vector field $\unicode[STIX]{x1D701}$ for $h$ , from the curve $\unicode[STIX]{x1D706}_{n+m}\subset \unicode[STIX]{x1D6F4}$ we obtain a curve $\unicode[STIX]{x1D706}_{n+m}^{\circ }\subset T$ which corresponds to a meridian for the knot $K_{n+m}$ . Moreover, the knot complement $Y\setminus \text{nd}(K)=Y_{n+m}\setminus \text{nd}(K_{m+n})$ may be identified with $U=h^{-1}[0,7/3]$ . For such Heegaard diagrams, the map

$$\begin{eqnarray}s_{u,v}:\mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}}\rightarrow \text{}\underline{\text{Spin}^{c}}(Y_{n+m},K_{n+m})=\text{}\underline{\text{Spin}^{c}}(Y,K)\end{eqnarray}$$

may be defined using the Morse function $h$ as follows.

For a generator $\mathbf{z}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}}$ , we may write $\mathbf{z}=\{z_{1},\ldots ,z_{g}\}$ , where $z_{i}\in \unicode[STIX]{x1D6FD}_{i}^{n+m}$ for $i=1,\ldots ,g$ . The points $z_{i}$ determine a set of $g$ flow lines for $\unicode[STIX]{x1D701}$ connecting the critical points of index $1$ to the critical points of index $2$ . One of these flow lines leaves $U$ and the rest stay inside $U$ . Modify $\unicode[STIX]{x1D701}$ in a neighbourhood of the latter $g-1$ flow lines so that the zeros of $\unicode[STIX]{x1D701}$ at the end-point critical points are cancelled against each other. After these modifications, the new vector field $\unicode[STIX]{x1D701}^{\prime }$ will have two zeros in $U$ . One of these zeros is at the unique critical point of index $0$ and the other one is at a critical point of index $1$ , which corresponds to the $\unicode[STIX]{x1D6FC}$ -curve containing $z_{g}$ , i.e. $\unicode[STIX]{x1D6FC}_{g}$ . The flow line of $\unicode[STIX]{x1D701}$ which passes through $z_{g}$ starts from the latter index- $1$ critical point and connects it to $z_{g}\in \unicode[STIX]{x1D706}_{n+m}$ to give an oriented arc $\unicode[STIX]{x1D6FF}_{1}$ . There is an oriented arc $\unicode[STIX]{x1D6FF}_{1}$ on $\unicode[STIX]{x1D706}_{n+m}$ which connects $z_{g}$ to $p_{n+m}$ in the direction of $\unicode[STIX]{x1D706}_{n+m}$ . Finally, the flow line of $-\unicode[STIX]{x1D701}$ which passes through $p_{n+m}$ determines another oriented arc $\unicode[STIX]{x1D6FF}_{3}$ from $p_{n+m}\in \unicode[STIX]{x1D706}_{n+m}$ to the critical point of index $0$ . Putting the three arcs $\unicode[STIX]{x1D6FF}_{1},\unicode[STIX]{x1D6FF}_{2}$ and $\unicode[STIX]{x1D6FF}_{3}$ together, we obtain a path $\unicode[STIX]{x1D6FF}$ in the interior of $U$ which connects the remaining two critical points. These two critical points have indices $0$ and $1$ which are of different parity. Thus, the vector field $\unicode[STIX]{x1D701}^{\prime }$ may further be modified in a neighbourhood of $\unicode[STIX]{x1D6FF}$ to arrive at a nowhere-vanishing vector field $\unicode[STIX]{x1D701}^{\prime \prime }$ which restricts to the outward normal of $T=\unicode[STIX]{x2202}(Y\setminus \text{nd}(K))$ . The restriction of this nowhere-vanishing vector field to the boundary is naturally isotopic to the translation-invariant vector field, and we thus obtain an element $\mathfrak{s}_{u,v}(\mathbf{z})$ of $\text{}\underline{\text{Spin}^{c}}(Y,K)$ .

Let us now assume that $\mathbf{z}=\mathbf{x}_{0}$ for some $\mathbf{x}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}$ . Then $\mathfrak{s}(\mathbf{x})$ is defined by modifying $\unicode[STIX]{x1D701}^{\prime }$ in a neighbourhood of a path $\unicode[STIX]{x1D6FF}^{\prime }$ , which is obtained by putting the arcs $\unicode[STIX]{x1D6FF}_{1}^{\prime },\unicode[STIX]{x1D6FF}_{2}^{\prime }$ and $\unicode[STIX]{x1D6FF}_{3}$ together. Here, $\unicode[STIX]{x1D6FF}_{1}^{\prime }$ is the flow line from $x$ to the critical point corresponding to $\unicode[STIX]{x1D6FC}_{g}$ and $\unicode[STIX]{x1D6FF}_{2}^{\prime }$ is an arc connecting $x$ to $p_{n+m}$ on $\unicode[STIX]{x1D706}_{\infty }$ . The domain containing the marking $t$ gives an isotopy between $\unicode[STIX]{x1D6FF}$ and $\unicode[STIX]{x1D6FF}^{\prime }$ which is supported away from the other arcs, and the vector fields representing $\mathfrak{s}(\mathbf{x})$ and $\mathfrak{s}(\mathbf{x}_{0})$ are thus isotopic relative to the boundary.

For an intersection point $\mathbf{x}_{i}$ with $i<0$ , the path $\unicode[STIX]{x1D6FF}^{\prime \prime }$ which corresponds to $\mathbf{x}_{i}$ connects the same critical points as $\unicode[STIX]{x1D6FF}$ does, while it differs from $\unicode[STIX]{x1D6FF}$ in a closed curve isotopic to $i$ times the curve $\unicode[STIX]{x1D706}_{\infty }$ , which represents the meridian $\unicode[STIX]{x1D707}_{K}$ of the knot $K$ . It follows that

$$\begin{eqnarray}\mathfrak{s}(\mathbf{x}_{i})=\mathfrak{s}(\mathbf{x}_{0})-i\text{PD}[\unicode[STIX]{x1D707}_{K}]=\mathfrak{s}(\mathbf{x})-i.\end{eqnarray}$$

For $\mathbf{x}_{i}$ with $i>0$ , note that $\unicode[STIX]{x1D6FF}_{1}^{\prime \prime }$ first goes out from the winding region and then returns to it from the right-hand side. It follows that the difference between $\unicode[STIX]{x1D6FF}^{\prime \prime }$ and $\unicode[STIX]{x1D6FF}$ is $-\text{PD}[\unicode[STIX]{x1D706}_{n+m}]+i\text{PD}[\unicode[STIX]{x1D707}_{K}]$ , which gives

$$\begin{eqnarray}\mathfrak{s}(\mathbf{x}_{i})=\mathfrak{s}(\mathbf{x}_{0})+\text{PD}[\unicode[STIX]{x1D706}_{n+m}]-i\text{PD}[\unicode[STIX]{x1D707}_{K}]=\mathfrak{s}(\mathbf{x})+n+m-i.\end{eqnarray}$$

Finally, if we choose $p_{0}$ and $p_{n+m}$ close to each other, the difference between the path $\unicode[STIX]{x1D6FF}_{n+m}$ associated with the intersection point $\widehat{\mathbf{y}}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}}$ and the path associated with $\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{0}}$ is in $\lceil m/2\rceil \text{PD}[\unicode[STIX]{x1D707}_{K}]$ , implying the last claim.◻

Proof. Let us first assume that $\mathbf{x}=\mathbf{z}_{i}^{(n)}$ for some $\mathbf{z}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}$ and some $i=1,\ldots ,n$ . Then there is a small triangle class missing $u,v$ and $a$ connecting $\mathbf{x}=\mathbf{z}_{i}^{(n)},\unicode[STIX]{x1D6E9}_{g}$ and $\mathbf{z}_{i}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}}$ . The intersection of the domain ${\mathcal{D}}^{i}$ of this triangle with the winding region is the small triangle with vertices $x_{i}^{n},q$ and $x_{i}$ , which is shaded in Figure 1 (for $i=3$ ). If $\mathbf{y}$ is a generator in the image of $g_{b}^{\mathfrak{s}}(\mathbf{x})$ , then there is a domain ${\mathcal{D}}$ (corresponding to the difference between ${\mathcal{D}}^{i}$ and the domain of a holomorphic triangle contributing to $g_{b}^{\mathfrak{s}}$ ) connecting $\mathbf{y}$ and $\mathbf{z}_{i}$ with boundary on $\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n}$ and $\unicode[STIX]{x1D737}_{n+m}$ and missing the marked points $u,v$ and $a$ . Subtracting an appropriate multiple of the periodic domain bounded by $\unicode[STIX]{x1D6FD}_{i}^{n}$ and $\unicode[STIX]{x1D6FD}_{i}^{n+m}$ (for $i=1,\ldots ,g-1$ ), we may assume that the boundary of ${\mathcal{D}}$ is on $\unicode[STIX]{x1D6FC}\cup \unicode[STIX]{x1D737}^{n+m}\cup \unicode[STIX]{x1D706}_{n}$ . Moreover, since we know that

$$\begin{eqnarray}n_{a}({\mathcal{D}})=n_{u}({\mathcal{D}})=n_{v}({\mathcal{D}})=0,\end{eqnarray}$$

it follows that $n_{b}({\mathcal{D}})=0$ . In particular, ${\mathcal{D}}$ does not have any boundary on $\unicode[STIX]{x1D706}_{n}$ and is thus the domain of a Whitney disc in the Heegaard diagram $H_{n+m}$ which misses $u$ and $v$ , and connects $\mathbf{y}$ and $\mathbf{z}_{i}$ . In particular,

$$\begin{eqnarray}\displaystyle \mathfrak{s}(\mathbf{y}) & = & \displaystyle \mathfrak{s}(\mathbf{z}_{i})\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}(\mathbf{z})+m+n-i\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}(\mathbf{z}_{i}^{(n)})+m\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}+m.\nonumber\end{eqnarray}$$

Let us now assume that $\mathbf{x}$ is an arbitrary generator and that $\mathbf{y}$ appears in $g_{b}^{\mathfrak{s}}(\mathbf{x})$ . Let ${\mathcal{D}}_{1}$ denote the domain of a Whitney disc for the Heegaard diagram $H_{n}$ connecting $\mathbf{x}$ to a generator of the form $\mathbf{z}_{i}^{(n)}$ with $i\in \{1,\ldots ,n\}$ and missing the markings $u,v$ , and thus the markings $a,b$ . Let ${\mathcal{D}}_{2}$ denote the domain of a Whitney disc for $H_{n+m}$ connecting $\mathbf{y}$ to a generator of the form $\mathbf{z}_{j}$ and missing $u,v$ (and thus $a,b$ ). If ${\mathcal{D}}_{0}$ is the domain of the triangle which corresponds to the contribution of $\mathbf{y}$ in $g_{b}^{\mathfrak{s}}(\mathbf{x})$ , and ${\mathcal{D}}_{3}$ denotes the domain of a triangle class connecting $\mathbf{z}_{i}^{(n)}$ to $\mathbf{z}_{i}$ , the domain

$$\begin{eqnarray}{\mathcal{D}}={\mathcal{D}}_{0}-{\mathcal{D}}_{1}+{\mathcal{D}}_{2}-{\mathcal{D}}_{3}\end{eqnarray}$$

is a domain connecting $\mathbf{z}_{i}$ to $\mathbf{z}_{j}$ which misses $u,v$ and $a$ , and has its boundary on $\unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}^{n}\cup \unicode[STIX]{x1D737}_{n+m}$ . Subtracting appropriate multiples of the periodic domains bounded by $\unicode[STIX]{x1D6FD}_{i}^{n}$ and $\unicode[STIX]{x1D6FD}_{i}^{n+m}$ (for $i=1,\ldots ,g-1$ ), we may assume that the boundary of ${\mathcal{D}}$ is on $\unicode[STIX]{x1D6FC}\cup \unicode[STIX]{x1D737}^{n+m}\cup \unicode[STIX]{x1D706}_{n}$ . Again, since $n_{a}({\mathcal{D}})=n_{u}({\mathcal{D}})=n_{v}({\mathcal{D}})=0$ , we conclude that $n_{b}({\mathcal{D}})=0$ and ${\mathcal{D}}$ has no boundary on $\unicode[STIX]{x1D706}_{n}$ . It follows that ${\mathcal{D}}$ is the domain of a Whitney disc which corresponds to the punctured Heegaard diagram $H_{n+m}$ . It is implied that $\mathfrak{s}(\mathbf{z}_{i})=\mathfrak{s}(\mathbf{z}_{j})$ and, consequently, $i=j$ . We thus find that

$$\begin{eqnarray}\displaystyle \mathfrak{s}(\mathbf{y}) & = & \displaystyle \mathfrak{s}(\mathbf{z}_{j}^{(n)})\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}(\mathbf{z})+m+n-i\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}(\mathbf{z}_{i}^{(n)})+m\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}+m,\nonumber\end{eqnarray}$$

and the proof of the second claim is complete. For the first claim, let $\mathbf{y}_{a}$ denote a generator which appears in $g_{a}^{\mathfrak{s}}(\mathbf{x})$ while $\mathbf{y}_{b}$ denotes a generator which appears in $g_{b}^{\mathfrak{s}}(\mathbf{x})$ . Let ${\mathcal{D}}_{a}$ and ${\mathcal{D}}_{b}$ denote the domains of the corresponding triangle classes. Then ${\mathcal{D}}={\mathcal{D}}_{a}-{\mathcal{D}}_{b}$ is a domain with coefficient $0$ at $u,v$ , which connects $\mathbf{y}_{b}$ to $\mathbf{y}_{a}$ and has boundary on $\unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{n}\cup \unicode[STIX]{x1D737}_{n+m}$ . Once again, subtracting appropriate multiples of the periodic domains bounded by $\unicode[STIX]{x1D6FD}_{i}^{n}$ and $\unicode[STIX]{x1D6FD}_{i}^{n+m}$ (for $i=1,\ldots ,g-1$ ), we may assume that the boundary of ${\mathcal{D}}$ is on $\unicode[STIX]{x1D6FC}\cup \unicode[STIX]{x1D737}^{n+m}\cup \unicode[STIX]{x1D706}_{n}$ . Furthermore, $n_{b}({\mathcal{D}})=-n_{a}({\mathcal{D}})=1$ . It follows that

$$\begin{eqnarray}\displaystyle \mathfrak{s}(\mathbf{y}_{a}) & = & \displaystyle \mathfrak{s}(\mathbf{y}_{b})+(n_{v}({\mathcal{D}})-n_{b}({\mathcal{D}}))\text{PD}[\unicode[STIX]{x1D706}_{n+m}]+(n_{b}({\mathcal{D}})-n_{u}({\mathcal{D}}))\text{PD}[\unicode[STIX]{x1D706}_{n}]\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{s}+m-(n+m)+n=\mathfrak{s}.\nonumber\end{eqnarray}$$

This completes the proof of the lemma. ◻

Proof. The proof is almost identical to the proof used in [Reference Alishahi and EftekharyAE15, §8]. We outline the proof to set up the notation. Define the map

$$\begin{eqnarray}H_{f}^{\mathfrak{s}}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u,v)\rightarrow C_{n,m}(\mathfrak{s})\end{eqnarray}$$

by setting

$$\begin{eqnarray}H_{f}^{\mathfrak{ s}}(\mathbf{x}):=\mathop{\sum }_{\substack{ \mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}} \\ \mathfrak{s}(\mathbf{y})\equiv \mathfrak{s}~(\text{mod}~m)}}\;\mathop{\sum }_{\substack{ \Box \in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g},\mathbf{y};u,v)}}\#({\mathcal{M}}(\Box ))\cdot \mathbf{y}.\end{eqnarray}$$

The condition $\mathfrak{s}(\mathbf{y})\equiv \mathfrak{s}~(\text{mod}~m)$ implies that $\mathfrak{s}(\mathbf{y})\in \{\mathfrak{s};\mathfrak{s}+m\}$ , since $m$ is large. Considering all possible boundary degenerations of the one-dimensional moduli space corresponding to the class $\Box \in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g},\mathbf{y};u,v)$ , we find that

$$\begin{eqnarray}d\circ H_{f}^{\mathfrak{s}}+H_{f}^{\mathfrak{s}}\circ d=h^{\mathfrak{s}}\circ g^{\mathfrak{s}}.\end{eqnarray}$$

For this, one should note that the contributing boundary degenerations of the form $\Box =\unicode[STIX]{x1D6E5}\star \unicode[STIX]{x1D6E5}^{\prime }$ with

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9};u,v),\quad \unicode[STIX]{x1D6E5}^{\prime }\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9},\mathbf{y};u,v)\end{eqnarray}$$

and $\unicode[STIX]{x1D6E9}\in \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}$ come in cancelling pairs. In fact, corresponding to each $\unicode[STIX]{x1D6E5}^{\prime }$ the corresponding triangles $\unicode[STIX]{x1D6E5}$ come in cancelling pairs, where the difference between the coefficients of every cancelling pair at the marking $z$ is always a multiple of $m$ . Similarly, define

$$\begin{eqnarray}H_{g}^{\mathfrak{s}}:C_{n,m}(\mathfrak{s})\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n};u,v)\end{eqnarray}$$

by setting

$$\begin{eqnarray}H_{g}^{\mathfrak{s}}(\mathbf{x}):=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n}}}\;\mathop{\sum }_{\substack{ \Box \in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h},\mathbf{y};u,v) \\ n_{z}(\Box )\equiv 0~(\text{mod}~m)}}\#({\mathcal{M}}(\Box ))\cdot \mathbf{y}.\end{eqnarray}$$

The contributing holomorphic triangles for $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}_{n};u,v)$ which correspond to the closed top generators $\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h}$ and $\unicode[STIX]{x1D6E9}\in \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n}}$ come in cancelling pairs. The condition $n_{z}(\Box )\equiv 0~(\text{mod}~m)$ implies that if $\mathbf{y}$ appears in $H_{g}^{\mathfrak{s}}(\mathbf{x})$ , then $\mathfrak{s}(\mathbf{y})-\mathfrak{s}(\mathbf{x})$ is a multiple of $m$ . If $m$ is sufficiently large, it follows from $\mathfrak{s}(\mathbf{x})\in \{\mathfrak{s},\mathfrak{s}+m\}$ that $\mathfrak{s}(\mathbf{y})=\mathfrak{s}$ . Thus, the image of $H_{g}^{\mathfrak{s}}$ is in $\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n};u,v;\mathfrak{s})$ . Consider the degenerations of a square class

$$\begin{eqnarray}\Box \in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h},\mathbf{y};u,v)\quad \text{with}~n_{z}(\Box )\equiv 0~(\text{mod}~m).\end{eqnarray}$$

In degenerations of the form $\Box =\unicode[STIX]{x1D6E5}\star \unicode[STIX]{x1D6E5}^{\prime }$ with

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9};u,v)\quad \text{and}\quad \unicode[STIX]{x1D6E5}^{\prime }\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9},\mathbf{y};u,v),\end{eqnarray}$$

the corresponding triangles $\unicode[STIX]{x1D6E5}$ come in cancelling pairs. From these observations, we find that

$$\begin{eqnarray}d\circ H_{g}^{\mathfrak{s}}+H_{g}^{\mathfrak{s}}\circ d=h^{\mathfrak{s}}\circ f^{\mathfrak{s}}.\end{eqnarray}$$

Finally, define the homotopy map

$$\begin{eqnarray}H_{h}^{\mathfrak{s}}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n};u,v;\mathfrak{s})\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u,v)\end{eqnarray}$$

by setting

$$\begin{eqnarray}H_{h}^{\mathfrak{s}}(\mathbf{x})=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}}\;\mathop{\sum }_{\substack{ \Box \in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f},\mathbf{y};u,v)}}\#({\mathcal{M}}(\Box ))\cdot \mathbf{y}.\end{eqnarray}$$

Employ the same argument again to show that $d\circ H_{h}^{\mathfrak{s}}+H_{h}^{\mathfrak{s}}\circ d=f^{\mathfrak{s}}\circ g^{\mathfrak{s}}$ .

We next introduce the pentagon maps. Let $\unicode[STIX]{x1D737}_{n}^{\prime }$ denote a $g$ -tuple of simple closed curves which are small Hamiltonian isotopes of the curves in $\unicode[STIX]{x1D737}_{n}$ . Choosing the Hamiltonian isotopy sufficiently small, we may assume that the chain complex $\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n}^{\prime };u,v;\mathfrak{s})$ associated with $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n}^{\prime };u,v)$ and the $\text{Spin}^{c}$ class $\mathfrak{s}$ may be identified with $\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n};u,v;\mathfrak{s})$ . We assume that the top intersection point $q_{n}$ between $\unicode[STIX]{x1D706}_{n}$ and $\unicode[STIX]{x1D706}_{n}^{\prime }$ is in the winding region, and appears on the common edge between $\unicode[STIX]{x1D6E5}_{a}$ and the domain containing the marked point $u$ ; see Figure 2. Corresponding to the intersection point $p_{n}$ we obtain $p_{n}^{\prime }\in \unicode[STIX]{x1D706}_{\infty }\cap \unicode[STIX]{x1D706}_{n}^{\prime }$ . There is a top generator $\unicode[STIX]{x1D6E9}_{h}^{\prime }$ for $\widehat{\text{CF}}(\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}_{n}^{\prime };u,v)$ which uses $p_{n}^{\prime }$ and is in correspondence with $\unicode[STIX]{x1D6E9}_{h}$ .

Figure 2. The curves $\unicode[STIX]{x1D706}_{\infty }^{\prime }$ and $\unicode[STIX]{x1D706}_{n}^{\prime }$ are Hamiltonian isotopes of the curves $\unicode[STIX]{x1D706}_{\infty }$ and $\unicode[STIX]{x1D706}_{n}$ , respectively. The triangle $\unicode[STIX]{x1D6E5}_{a}$ is decomposed to two small triangles $[p_{n}q_{n}p_{n}^{\prime }]$ and $[q_{\infty }p_{n+m}^{\prime }p_{n+m}]$ and a pentagon. The union of the pentagon with the first triangle is a rectangle with vertices $q,p_{n+m},p_{n}^{\prime }$ and $q_{n}$ , which is lightly shaded, and its union with the second triangle is a rectangle with vertices $p_{n},q,p_{n+m}^{\prime },q_{\infty }$ , which is strongly shaded.

Define the map

$$\begin{eqnarray}P_{f}^{\mathfrak{s}}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n};u,v;\mathfrak{s})\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n}^{\prime };u,v;\mathfrak{s})=\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n};u,v;\mathfrak{s})\end{eqnarray}$$

by setting

$$\begin{eqnarray}P_{f}^{\mathfrak{s}}(\mathbf{x}):=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n}}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h}^{\prime },\mathbf{y};u,v) \\ n_{z}(\unicode[STIX]{x2B20})\equiv 0~(\text{mod}~m)}}\#({\mathcal{M}}(\unicode[STIX]{x2B20}))\cdot \mathbf{y}.\end{eqnarray}$$

The condition $n_{z}(\unicode[STIX]{x2B20})\equiv 0~(\text{mod}~m)$ implies that the image of $P_{f}^{\mathfrak{s}}$ is supported in relative $\text{Spin}^{c}$ class $\mathfrak{s}$ . Five types of the 10 possible degenerations in the boundary of the one-dimensional moduli space associated with a class $\unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h}^{\prime },\mathbf{y};u,v)$ with $n_{z}(\unicode[STIX]{x2B20})$ a multiple of $m$ , which correspond to a degeneration to a bigon and a pentagon, contribute to the coefficient of $\mathbf{y}$ in $(d\circ P_{f}^{\mathfrak{s}}+P_{f}^{\mathfrak{s}}\circ d)(\mathbf{x})$ . The remaining five types correspond to the degenerations of $\unicode[STIX]{x2B20}$ into a square $\Box$ and a triangle $\unicode[STIX]{x1D6E5}$ . Note that:

1. – there is a unique contributing class $\Box \in \unicode[STIX]{x1D70B}_{2}^{-1}(\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h}^{\prime },\unicode[STIX]{x1D6E9};u,v)$ which corresponds to the quadruple $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}_{n}^{\prime };u,v)$ . Moreover, $\unicode[STIX]{x1D6E9}=\unicode[STIX]{x1D6E9}_{n}$ is the top generator for the diagram $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n},\unicode[STIX]{x1D737}_{n}^{\prime };u,v)$ . The intersection of the domain of this class with the winding region is a rectangle with vertices $q,p_{n+m},p_{n}^{\prime }$ and $q_{n}$ , which is illustrated in Figure 2. This class has a unique holomorphic representative;

2. – the contributing triangle classes

   $$\begin{eqnarray}\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9};u,v)\quad \text{and}\quad \unicode[STIX]{x1D6E5}^{\prime }\in \unicode[STIX]{x1D70B}_{2}^{0}(\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h}^{\prime },\unicode[STIX]{x1D6E9};u,v)\end{eqnarray}$$
   corresponding to the triples $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737};u,v)$ and $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}_{n}^{\prime };u,v)$ come in cancelling pairs.

These observations imply that $d\circ P_{f}^{\mathfrak{s}}+P_{f}^{\mathfrak{s}}\circ d+J_{f}^{\mathfrak{s}}=h^{\mathfrak{s}}\circ H_{h}^{\mathfrak{s}}+H_{g}^{\mathfrak{s}}\circ g^{\mathfrak{s}}$ , where

$$\begin{eqnarray}J_{f}^{\mathfrak{s}}(\mathbf{x})=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n}^{\prime }}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{n},\mathbf{y};u,v)}}\#({\mathcal{M}}(\unicode[STIX]{x1D6E5}))\cdot \mathbf{y}.\end{eqnarray}$$

Similarly, define the holomorphic pentagon map

$$\begin{eqnarray}P_{g}^{\mathfrak{s}}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u)\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u)=\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{\prime };u)\end{eqnarray}$$

by setting

$$\begin{eqnarray}P_{g}^{\mathfrak{s}}(\mathbf{x}):=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}^{\prime }}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f}^{\prime },\mathbf{y};u,v) \\ n_{z}(\unicode[STIX]{x2B20})+\mathfrak{s}(\mathbf{x})\equiv \mathfrak{s}~(\text{mod}~m)}}\#({\mathcal{M}}(\unicode[STIX]{x2B20}))\cdot \mathbf{y}.\end{eqnarray}$$

Five types of the 10 possible degenerations in the boundary of the one-dimensional moduli space associated with a pentagon class $\unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g}^{q},\unicode[STIX]{x1D6E9}_{f}^{\prime },\mathbf{y};u,v)$ contribute to the coefficient of $\mathbf{y}$ in $(d\circ P_{g}^{\mathfrak{s}}+P_{g}^{\mathfrak{s}}\circ d)(\mathbf{x})$ . The remaining five types correspond to the degenerations of $\unicode[STIX]{x2B20}$ into a square and a triangle. The choice of the markings implies that two of these degeneration types contribute to the coefficient of $\mathbf{y}$ in $(f^{\mathfrak{s}}\circ H_{f}^{\mathfrak{s}}+H_{h}^{\mathfrak{s}}\circ h^{\mathfrak{s}})(\mathbf{x})$ . There is a unique contributing square class, corresponding to $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}_{n},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737}^{\prime };u,v)$ and the intersection points $\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f}^{\prime },\unicode[STIX]{x1D6E9}_{\infty }$ , where $\unicode[STIX]{x1D6E9}_{\infty }$ denotes the top generator for $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}^{\prime };u,v)$ , which includes the top intersection point $q_{\infty }$ of $\unicode[STIX]{x1D706}_{\infty }$ and $\unicode[STIX]{x1D706}_{\infty }^{\prime }$ . The intersection of the domain of this class with the winding region is the rectangle with vertices $p_{n},q,p_{n+m}^{\prime }$ and $q_{\infty }$ , which is illustrated in Figure 2. Moreover, the triangles which contribute in $\unicode[STIX]{x1D70B}_{2}(\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f})$ and $\unicode[STIX]{x1D70B}_{2}(\unicode[STIX]{x1D6E9}_{g},\unicode[STIX]{x1D6E9}_{f}^{\prime },\unicode[STIX]{x1D6E9}_{h}^{\prime })$ come in cancelling pairs. Thus, we obtain

$$\begin{eqnarray}d\circ P_{g}^{\mathfrak{s}}+P_{g}^{\mathfrak{s}}\circ d+J_{g}^{\mathfrak{s}}=f^{\mathfrak{s}}\circ H_{f}^{\mathfrak{s}}+H_{h}^{\mathfrak{s}}\circ h^{\mathfrak{s}},\end{eqnarray}$$

where the map $J_{g}^{\mathfrak{s}}$ is defined by

$$\begin{eqnarray}J_{g}^{\mathfrak{s}}(\mathbf{x})=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}^{\prime }}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v)}}\#({\mathcal{M}}(\unicode[STIX]{x1D6E5}))\cdot \mathbf{y}.\end{eqnarray}$$

Let $\unicode[STIX]{x1D737}_{n+m}^{\prime }$ denote a $g$ -tuple of simple closed curves which are small Hamiltonian isotopes of the curves in $\unicode[STIX]{x1D737}_{n+m}$ . Once again, we assume that the Hamiltonian isotope $\unicode[STIX]{x1D706}_{n+m}^{\prime }$ of $\unicode[STIX]{x1D706}_{n+m}$ intersects it in a pair of cancelling intersection points, and that the top intersection point is located on the common edge of $\unicode[STIX]{x1D6E5}_{a}$ with the domain containing the marking $v$ . Again, we assume that the chain complex associated with $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m}^{\prime };u,v)$ and the $\text{Spin}^{c}$ classes $\mathfrak{s},\mathfrak{s}+m$ is identified with $C_{n,m}(\mathfrak{s})$ . There is a top generator $\unicode[STIX]{x1D6E9}_{g}^{\prime }$ for $(\unicode[STIX]{x1D737}_{n},\unicode[STIX]{x1D737}_{n+m}^{\prime })$ which is in correspondence with $\unicode[STIX]{x1D6E9}_{g}$ . Define $P_{h}^{\mathfrak{s}}:C_{n,m}(\mathfrak{s})\rightarrow C_{n,m}(\mathfrak{s})$ by

$$\begin{eqnarray}P_{h}^{\mathfrak{s}}(\mathbf{x}):=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}^{\prime }}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{f},\unicode[STIX]{x1D6E9}_{h},\unicode[STIX]{x1D6E9}_{g}^{\prime },\mathbf{y};u,v) \\ n_{z}(\unicode[STIX]{x2B20})\equiv 0~(\text{mod}~m)}}\#({\mathcal{M}}(\unicode[STIX]{x2B20}))\cdot \mathbf{y}.\end{eqnarray}$$

A similar argument implies that $d\circ P_{h}^{\mathfrak{s}}+P_{h}^{\mathfrak{s}}\circ d+J_{h}^{\mathfrak{s}}=g^{\mathfrak{s}}\circ H_{g}^{\mathfrak{s}}+H_{f}^{\mathfrak{s}}\circ f^{\mathfrak{s}}$ , where

$$\begin{eqnarray}J_{h}^{\mathfrak{s}}(\mathbf{x})=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{n+m}^{\prime }}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{n+m},\mathbf{y};u,v)}}\#({\mathcal{M}}(\unicode[STIX]{x1D6E5}))\cdot \mathbf{y},\end{eqnarray}$$

and $\unicode[STIX]{x1D6E9}_{n+m}$ is the top generator of $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737}_{n+m}^{\prime };u,v)$ . Since $J_{f}^{\mathfrak{s}},J_{g}^{\mathfrak{s}},J_{h}^{\mathfrak{s}}$ are quasi-isomorphisms, [Reference Alishahi and EftekharyAE15, Lemma 3.3] completes the proof.◻

Proof. Note that the aforementioned Heegaard moves consist of $2g-2$ handle slides (composed with isotopies) on $\unicode[STIX]{x1D737}$ , supported away from the markings $z,t$ . Denote the corresponding $g$ -tuples of curves by $\unicode[STIX]{x1D737}^{0}=\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}^{1},\ldots ,\unicode[STIX]{x1D737}^{2g-2}$ , where $\widehat{\text{CF}}(\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{2g-2};z)$ may be identified with $\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u)$ . The triple $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{i-1},\unicode[STIX]{x1D737}^{i};z)$ and the top generator $\unicode[STIX]{x1D6E9}^{i}$ of the diagram $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}^{i-1},\unicode[STIX]{x1D737}^{i};z,t)$ determine a chain map $\unicode[STIX]{x1D6EF}^{i}:\widehat{\text{CF}}(\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{i-1};z)\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{i};z)$ . The triple $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737}^{i};z,t)$ and the top generator $\unicode[STIX]{x1D6E9}_{f}^{i}$ of $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737}^{i};z,t)$ determine $f^{i}:\widehat{\text{CF}}(\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m};z,t)\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{i};z)$ . Finally, the quadruple $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m},\unicode[STIX]{x1D737}^{i-1},\unicode[STIX]{x1D737}^{i};z,t)$ together with $\unicode[STIX]{x1D6E9}_{f}^{i-1}$ and $\unicode[STIX]{x1D6E9}^{i}$ determines a homomorphism $H^{i}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{n+m};z,t)\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{i};z)$ . Considering different boundary degenerations of the one-dimensional moduli space associated with a square class of index $0$ , we find that

(4) $$\begin{eqnarray}d\circ H^{i}+H^{i}\circ d=f^{i}+\unicode[STIX]{x1D6EF}^{i}\circ f^{i-1},\quad i=1,\ldots ,2g-2.\end{eqnarray}$$

Let us define $\unicode[STIX]{x1D6EF}=\unicode[STIX]{x1D6EF}^{2g-2}\circ \cdots \circ \unicode[STIX]{x1D6EF}^{1}$ and set

$$\begin{eqnarray}H=H^{2g-2}+\unicode[STIX]{x1D6EF}^{2g-2}\circ H^{2g-3}+\unicode[STIX]{x1D6EF}^{2g-2}\circ \unicode[STIX]{x1D6EF}^{2g-3}H^{2g-4}+\cdots +(\unicode[STIX]{x1D6EF}^{2g-2}\circ \ldots \circ \unicode[STIX]{x1D6EF}^{2})\circ H^{1}.\end{eqnarray}$$

Using (4), $d\circ H+H\circ d=f^{2g-2}+\unicode[STIX]{x1D6EF}\circ f^{0}$ . To complete the proof, note that $f^{\mathfrak{s}}$ and $\overline{f}^{\mathfrak{s}}$ are the restrictions of $f^{2g-2}$ and $f^{0}$ to $C_{n,m}(\mathfrak{s})$ , respectively.◻

Proof. Choose $\mathbf{w}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}$ so that $\mathfrak{s}(\mathbf{w})=\mathfrak{s}$ . There are a particular square class

$$\begin{eqnarray}\Box _{0}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{w}_{0}^{(0)},\unicode[STIX]{x1D6E9}_{0,1},\unicode[STIX]{x1D6E9}_{g_{1}}\mathbf{w}_{0};u,v,w,b)\end{eqnarray}$$

and a corresponding domain ${\mathcal{D}}_{0}={\mathcal{D}}(\Box _{0})$ which intersects the winding region in a $4$ -gon with one obtuse angle. This $4$ -gon, which is shaded in Figure 3, has vertices at $q_{0},q_{m},x_{0}$ and $x_{0}^{0}$ , and its obtuse angle is at $q_{m}$ . Since the relative $\text{Spin}^{c}$ class associated with both $\mathbf{x}$ and $\mathbf{w}_{0}^{(0)}$ is $\mathfrak{s}$ , there is a Whitney disc $\unicode[STIX]{x1D719}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\mathbf{w}_{0}^{(0)};u,v)$ , which corresponds to a domain ${\mathcal{D}}_{1}={\mathcal{D}}(\unicode[STIX]{x1D719})$ . Since $b,w$ and $u$ are not separated by $\unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{0}$ , it follows that $n_{b}({\mathcal{D}}_{1})=n_{w}({\mathcal{D}}_{1})=0$ . Finally, associated with every square class

$$\begin{eqnarray}\Box \in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{0,1},\unicode[STIX]{x1D6E9}_{g_{1}},\mathbf{z};b,u,v,w)\end{eqnarray}$$

which contributes to $H_{a}^{\mathfrak{s}}(\mathbf{x})$ we obtain a domain ${\mathcal{D}}_{2}={\mathcal{D}}(\Box )$ . From these three domains, we obtain the domain ${\mathcal{D}}={\mathcal{D}}_{2}-{\mathcal{D}}_{1}-{\mathcal{D}}_{0}$ with boundary on the union $\unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{0}\cup \unicode[STIX]{x1D737}_{1}\cup \unicode[STIX]{x1D737}_{m}$ . After subtracting suitable multiples of periodic domains bounded between the curves in $\unicode[STIX]{x1D737}_{\bullet }\setminus \{\unicode[STIX]{x1D706}_{\bullet }\},~\bullet \in \{0,1,m\}$ , we may in fact assume that

$$\begin{eqnarray}\unicode[STIX]{x2202}{\mathcal{D}}\subset \unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{m}\cup \{\unicode[STIX]{x1D706}_{0},\unicode[STIX]{x1D706}_{1}\}.\end{eqnarray}$$

Since ${\mathcal{D}}$ connects $\mathbf{w}_{0}$ to $\mathbf{y}$ , the boundary of ${\mathcal{D}}$ on each one of $\unicode[STIX]{x1D706}_{0}$ and $\unicode[STIX]{x1D706}_{1}$ is an integer multiple of these two closed curves. From $n_{v}({\mathcal{D}})=n_{w}({\mathcal{D}})=0$ , it follows that ${\mathcal{D}}$ has no boundary on $\unicode[STIX]{x1D706}_{0}$ and, from $n_{w}({\mathcal{D}})=n_{b}({\mathcal{D}})=0$ , it follows that ${\mathcal{D}}$ has no boundary on $\unicode[STIX]{x1D706}_{1}$ . In other words, ${\mathcal{D}}$ is the domain of a Whitney disc $\unicode[STIX]{x1D713}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{w}_{0},\mathbf{y};u,v,w,a)$ . In particular, it follows that $\mathfrak{s}(\mathbf{y})=\mathfrak{s}(\mathbf{w}_{0})=\mathfrak{s}$ .

Let us now assume that $\Box ^{\prime }\in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{0,1},\unicode[STIX]{x1D6E9}_{g_{1}},\mathbf{z};a,u,v,w)$ contributes to $H_{b}^{\mathfrak{s}}(\mathbf{x})$ and, correspondingly, we obtain a domain ${\mathcal{D}}_{2}^{\prime }={\mathcal{D}}(\Box ^{\prime })$ . From the above discussion, we may find a domain ${\mathcal{D}}_{0}^{\prime }$ which corresponds to a class

$$\begin{eqnarray}\Box _{0}^{\prime }\in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{0,1},\unicode[STIX]{x1D6E9}_{g_{1}},\mathbf{w}_{0};b,u,v,w),\end{eqnarray}$$

with $\mathfrak{s}(\mathbf{w})=\mathfrak{s}$ . From the domains ${\mathcal{D}}_{0}^{\prime }$ and ${\mathcal{D}}_{2}^{\prime }$ , we obtain a domain ${\mathcal{D}}^{\prime }={\mathcal{D}}_{2}^{\prime }-{\mathcal{D}}_{0}^{\prime }$ which misses the markings $u,v$ and $w$ , and connects $\mathbf{w}_{0}$ to $\mathbf{z}$ . We may assume that this domain has its boundary on $\unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{m}\cup \{\unicode[STIX]{x1D706}_{0},\unicode[STIX]{x1D706}_{1}\}$ . Moreover, from the equalities $n_{v}({\mathcal{D}}^{\prime })=n_{w}({\mathcal{D}}^{\prime })=0$ , we conclude that ${\mathcal{D}}^{\prime }$ has no boundary on $\unicode[STIX]{x1D706}_{0}$ . The coefficient of $\unicode[STIX]{x1D706}_{1}$ is, however, equal to

$$\begin{eqnarray}n_{b}({\mathcal{D}}^{\prime })-n_{w}({\mathcal{D}}^{\prime })=n_{b}({\mathcal{D}}_{2}^{\prime })=1.\end{eqnarray}$$

Next, note that there is a domain ${\mathcal{D}}^{\prime \prime }$ with boundary on $\unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{m}\cup \unicode[STIX]{x1D706}_{\infty }$ , which connects $\mathbf{w}_{1}$ to $\mathbf{w}_{0}$ while $n_{\bullet }({\mathcal{D}}^{\prime \prime })=0$ for $\bullet \in \{a,b,u,v,w\}$ and $n_{z}({\mathcal{D}}^{\prime \prime })=m-1$ . There is also a periodic domain ${\mathcal{P}}$ with boundary on $\unicode[STIX]{x1D706}_{1}\cup \unicode[STIX]{x1D706}_{m}\cup \unicode[STIX]{x1D706}_{\infty }$ such that

$$\begin{eqnarray}n_{v}({\mathcal{P}})=n_{w}({\mathcal{P}})=n_{u}({\mathcal{P}})=0,\quad n_{b}({\mathcal{P}})=-n_{a}({\mathcal{P}})=1\quad \text{and}\quad n_{z}({\mathcal{P}})=m-1.\end{eqnarray}$$

From the three domains ${\mathcal{D}}^{\prime },{\mathcal{D}}^{\prime \prime }$ and ${\mathcal{P}}$ , we obtain the domain $\widehat{{\mathcal{D}}}={\mathcal{D}}^{\prime }+{\mathcal{D}}^{\prime \prime }-{\mathcal{P}}$ which connects $\mathbf{w}_{1}$ to $\mathbf{z}$ and

$$\begin{eqnarray}\displaystyle n_{\bullet }(\widehat{{\mathcal{D}}}) & = & \displaystyle n_{\bullet }({\mathcal{D}}^{\prime })+n_{\bullet }({\mathcal{D}}^{\prime \prime })-n_{\bullet }({\mathcal{P}})=0+0-0=0,\quad \bullet \in \{u,v,w\},\nonumber\\ \displaystyle n_{a}(\widehat{{\mathcal{D}}}) & = & \displaystyle n_{a}({\mathcal{D}}^{\prime })+n_{a}({\mathcal{D}}^{\prime \prime })-n_{a}({\mathcal{P}})=-1+0-(-1)=0,\nonumber\\ \displaystyle n_{a}(\widehat{{\mathcal{D}}}) & = & \displaystyle n_{b}({\mathcal{D}}^{\prime })+n_{b}({\mathcal{D}}^{\prime \prime })-n_{b}({\mathcal{P}})=1+0-1=0,\nonumber\\ \displaystyle n_{z}(\widehat{{\mathcal{D}}}) & = & \displaystyle n_{z}({\mathcal{D}}^{\prime })+n_{z}({\mathcal{D}}^{\prime \prime })-n_{z}({\mathcal{P}})=0+(m-1)-(m-1)=0.\nonumber\end{eqnarray}$$

We thus find that $\mathfrak{s}(\mathbf{z})=\mathfrak{s}(\mathbf{w}_{1})=\mathfrak{s}+m-1$ . This completes the proof of the lemma.◻

Proof. As discussed in § 3.1, the holomorphic square map

$$\begin{eqnarray}H_{h_{0}}^{\mathfrak{ s}}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4}^{\prime },\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{0};u,v,\mathfrak{s})\rightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4}^{\prime },\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u,v),\end{eqnarray}$$

which is defined from the Heegaard $4$ -tuple $(\unicode[STIX]{x1D6F4}^{\prime },\unicode[STIX]{x1D736},\unicode[STIX]{x1D737},\unicode[STIX]{x1D737}_{0},\unicode[STIX]{x1D737}_{m-1};u,v)$ , may be written as

$$\begin{eqnarray}H_{h_{0}}^{\mathfrak{ s}}=H_{h_{0},a}^{\mathfrak{s}}+H_{h_{0},b}^{\mathfrak{s}},\end{eqnarray}$$

where $H_{h_{0},a}^{\mathfrak{s}}$ counts the squares with $n_{b}(\Box )=0$ and $H_{h_{0},b}^{\mathfrak{s}}$ counts the squares with $n_{a}(\Box )=0$ . Under the identification of $\widehat{\text{CF}}(\unicode[STIX]{x1D6F4}^{\prime },\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{\bullet };u,v)$ with $\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{\bullet };u,v,w)$ for $\bullet \in \{0,\infty \}$ , we may also identify $H_{q}^{\mathfrak{s}}$ with $H_{h_{0},a}^{\mathfrak{s}}$ and $H_{q^{\prime }}^{\mathfrak{s}}$ with $H_{h_{0},b}^{\mathfrak{s}}$ . In particular, $H_{q}^{\mathfrak{s}}+H_{q^{\prime }}^{\mathfrak{s}}=H_{h_{0}}^{\mathfrak{s}}$ . Together with the discussion preceding the lemma, this completes the proof of (6).◻

Proof. Let us assume that $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g_{0}},\mathbf{y};u,v,w)$ is a triangle class which contributes to $g_{0,q}^{\mathfrak{s}}$ and let ${\mathcal{D}}_{1}={\mathcal{D}}(\unicode[STIX]{x1D6E5})$ . Choose $\mathbf{w}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}}$ with $\mathfrak{s}(\mathbf{w})=\mathfrak{s}$ and let $\unicode[STIX]{x1D719}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{w}_{0}^{(0)},\mathbf{x};u,v,w)$ denote a Whitney disc which connects $\mathbf{w}_{0}^{(0)}$ to $\mathbf{x}$ . Set ${\mathcal{D}}_{2}={\mathcal{D}}(\unicode[STIX]{x1D719})$ . There is a triangle in the Heegaard diagram with vertices $x_{0}^{0},q$ and $x_{0}$ which is obtained from the shaded $4$ -gon in Figure 3 by subtracting the small triangle with vertices $q_{0},q_{m}$ and $q$ . This triangle may be paired with the small triangles bounded between $\unicode[STIX]{x1D6FD}_{i}^{0}$ and $\unicode[STIX]{x1D6FD}_{i}^{m}$ , $i=1,\ldots ,g-1$ , to give the domain of a distinguished triangle class

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{0}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{w}_{0}^{(0)},\unicode[STIX]{x1D6E9}_{g_{0}},\mathbf{w}_{0};u,v,w).\end{eqnarray}$$

We let ${\mathcal{D}}_{0}={\mathcal{D}}(\unicode[STIX]{x1D6E5}_{0})$ and ${\mathcal{D}}={\mathcal{D}}_{0}-{\mathcal{D}}_{1}-{\mathcal{D}}_{2}$ . It is then clear that ${\mathcal{D}}$ is a domain missing $u,v$ and $w$ , which connects $\mathbf{y}$ to $\mathbf{w}_{0}$ . After subtracting appropriate multiples of the periodic domains in the Heegaard diagram $(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D737}_{0},\unicode[STIX]{x1D737}_{m};u,v,w)$ , we may also assume that $\unicode[STIX]{x2202}{\mathcal{D}}\subset \unicode[STIX]{x1D736}\cup \unicode[STIX]{x1D737}_{m}\cup \unicode[STIX]{x1D706}_{0}$ . The condition $n_{v}({\mathcal{D}})=n_{w}({\mathcal{D}})=0$ implies that ${\mathcal{D}}$ does not have any boundary on $\unicode[STIX]{x1D706}_{0}$ . It is thus the domain of a Whitney disc $\unicode[STIX]{x1D713}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{y},\mathbf{w}_{0};u,v,w)$ . It follows that $\mathfrak{s}(\mathbf{y})=\mathfrak{s}(\mathbf{w}_{0})=\mathfrak{s}.$ The proof of the second claim is quite similar.◻

Proof. With our earlier considerations in place, the first claim already follows. We thus only need to prove the second claim. If $\mathbf{x}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{m-1}}$ satisfies $\mathfrak{s}(\mathbf{x})=\mathfrak{s}$ , then the maps $f_{0}^{\mathfrak{s}}$ and $f_{1}^{\mathfrak{s}}$ are identical on $\mathbf{x}$ , by definition. Since $G_{\infty }^{\mathfrak{s}}(\mathbf{x})=\mathbf{x}$ , the claim follows. Let us now assume that $\mathfrak{s}(\mathbf{x})=\mathfrak{s}+m-1$ . It follows that $\mathbf{x}=\mathbf{z}_{i}$ for some $i>0$ , while $\mathfrak{s}(\mathbf{z})=\mathfrak{s}+i$ . Thus, $J^{\mathfrak{s}}(\mathbf{x})=\mathbf{z}_{i+1}$ and

$$\begin{eqnarray}f_{1}^{\mathfrak{s}}(G_{\infty }^{\mathfrak{s}}(\mathbf{x}))=f_{1}^{\mathfrak{s}}(\mathbf{z}_{i+1})=f_{0}^{\mathfrak{s}}(\mathbf{z}_{i})=f_{0}^{\mathfrak{s}}(\mathbf{x}).\end{eqnarray}$$

The third equality follows, since every contributing triangle in $\unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{z}_{i},\unicode[STIX]{x1D6E9}_{f_{0}},\mathbf{y};u,v)$ is in correspondence with a contributing triangle in $\unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{z}_{i+1},\unicode[STIX]{x1D6E9}_{f_{1}},\mathbf{y};u,v)$ .◻

Proof. For $\mathbf{x}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{m}}$ and $\Box \in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{z},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v,z)$ , the ends of the moduli space ${\mathcal{M}}(\Box )$ are in correspondence with degenerations of $\Box$ . Since $\unicode[STIX]{x1D6E9}_{f_{1}}$ and $\unicode[STIX]{x1D6E9}_{\infty }$ are closed, degenerations into a bigon and a square correspond to the coefficient of $\mathbf{y}$ in $(d\circ I^{\mathfrak{s}}+I^{\mathfrak{s}}\circ d)(\mathbf{x})$ . There are no holomorphic triangles $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\unicode[STIX]{x1D6E9};u,v,z)$ , implying that there are no degenerations of the form $\Box =\unicode[STIX]{x1D6E5}\star \unicode[STIX]{x1D6E5}^{\prime }$ with

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\unicode[STIX]{x1D6E9};u,v,z)\quad \text{and}\quad \unicode[STIX]{x1D6E5}^{\prime }\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9},\mathbf{y};u,v,z).\end{eqnarray}$$

Finally, by looking at local multiplicities around $x$ , we may conclude that there are no positive triangle classes $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{f_{1}},\mathbf{z};u,v,z)$ . The contribution of degenerations of the form $\Box =\unicode[STIX]{x1D6E5}\star \unicode[STIX]{x1D6E5}^{\prime }$ with

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{x},\unicode[STIX]{x1D6E9}_{f_{1}},\mathbf{z};u,v,z)\quad \text{and}\quad \unicode[STIX]{x1D6E5}^{\prime }\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{z},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v,z)\end{eqnarray}$$

is thus trivial. From these observations, we conclude that $(d\circ I^{\mathfrak{s}}+I^{\mathfrak{s}}\circ d)(\mathbf{x})=0$ for all $\mathbf{x}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{m}}$ , completing the proof of the lemma.◻

Proof. The diagram ${\mathcal{H}}$ defines a pentagon map

$$\begin{eqnarray}Q^{\mathfrak{s}}:\widehat{\text{CFK}}(K_{1};\mathfrak{s})=\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{1};u,v,z;\mathfrak{s})\longrightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737};u,v,z;\mathfrak{s})=\widehat{\text{CFK}}(K;\mathfrak{s})\end{eqnarray}$$

by setting

$$\begin{eqnarray}Q^{\mathfrak{s}}(\mathbf{x})=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}^{\prime }},\mathfrak{s}(\mathbf{y})=\mathfrak{s}}\;\mathop{\sum }_{\unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v,z)}\#({\mathcal{M}}(\unicode[STIX]{x2B20}))\cdot \mathbf{y}.\end{eqnarray}$$

For $\unicode[STIX]{x2B20}\in \unicode[STIX]{x1D70B}_{2}^{-1}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v,z)$ , the ends of the moduli space ${\mathcal{M}}(\unicode[STIX]{x2B20})$ which correspond to the degenerations of the pentagon to a bigon and a pentagon contribute to the coefficient of $\mathbf{y}$ in $(d\circ Q^{\mathfrak{s}}+Q^{\mathfrak{s}}\circ d)(\mathbf{x})$ . Other ends correspond to the degenerations of the form $\unicode[STIX]{x2B20}=\Box \star \unicode[STIX]{x1D6E5}$ of one of the following five types:

1. (1) $\Box \in \unicode[STIX]{x1D70B}_{2}(\mathbf{z},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y})$ and $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g_{1}},\mathbf{z})$ ;

2. (2) $\Box \in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9}_{f_{1}},\mathbf{z})$ and $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{z},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y})$ ;

3. (3) $\Box \in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9},\mathbf{y})$ and $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}(\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\unicode[STIX]{x1D6E9})$ ;

4. (4) $\Box \in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\unicode[STIX]{x1D6E9},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y})$ and $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}(\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9})$ ;

5. (5) $\Box \in \unicode[STIX]{x1D70B}_{2}(\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\unicode[STIX]{x1D6E9})$ and $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}(\mathbf{x},\unicode[STIX]{x1D6E9},\mathbf{y})$ .

Degenerations of types 1 and 2 correspond to the coefficient of $\mathbf{y}$ in $(I^{\mathfrak{s}}\circ g_{1}^{\mathfrak{s}})(\mathbf{x})$ and $(X^{\mathfrak{s}}\circ H_{h_{1}}^{\mathfrak{s}})(\mathbf{x})$ , respectively, where

$$\begin{eqnarray}X^{\mathfrak{s}}(\mathbf{z})=\mathop{\sum }_{\mathbf{y}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}^{\prime }},\mathfrak{s}(\mathbf{y})=\mathfrak{s}}\;\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{z},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v,z)}}\#({\mathcal{M}}(\unicode[STIX]{x1D6E5}))\cdot \mathbf{y}.\end{eqnarray}$$

Considering the local multiplicities around $\unicode[STIX]{x1D706}_{\infty }\cap \unicode[STIX]{x1D706}_{\infty }^{\prime }$ , one concludes that there are no triangle classes $\unicode[STIX]{x1D6E5}\in \unicode[STIX]{x1D70B}_{2}^{0}(\mathbf{z},\unicode[STIX]{x1D6E9}_{\infty },\mathbf{y};u,v,z)$ with positive domain. In particular, $X^{\mathfrak{s}}$ is trivial. There are no triangle classes which contribute in the degenerations of type 3. The contributing triangles in degenerations of type 4 come in cancelling pairs. Thus, the total number of boundary ends corresponding to degenerations of types 3 and 4 is zero. There is a unique square class in $\unicode[STIX]{x1D70B}_{2}^{-1}(\unicode[STIX]{x1D6E9}_{g_{1}},\unicode[STIX]{x1D6E9}_{f_{1}},\unicode[STIX]{x1D6E9}_{\infty },\unicode[STIX]{x1D6E9};u,v,z)$ with non-trivial contribution to degenerations of type 5. The intersection of the domain of this square class with the winding region is the rectangle with vertices $q_{m},p_{m},p_{\infty }$ and $p_{1}^{\prime }$ , which is shaded yellow in Figure 5. For this square class, $\unicode[STIX]{x1D6E9}\in \mathbb{T}_{\unicode[STIX]{x1D6FD}_{1}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}^{\prime }}$ is the top generator and $\Box$ has a unique holomorphic representative. Using the generator $\unicode[STIX]{x1D6E9}$ , we define the map

$$\begin{eqnarray}\overline{\mathfrak{f}}_{0}^{\mathfrak{s}}:\widehat{\text{CFK}}(K_{1};\mathfrak{s})=\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{1};u,v,z;\mathfrak{s})\longrightarrow \widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}^{\prime };u,v,z;\mathfrak{s})=\widehat{\text{CFK}}(K;\mathfrak{s}),\end{eqnarray}$$

which is again one of the maps which appeared in the splicing formula of [Reference EftekharyEft15]. The contribution of the degenerations of type $5$ thus corresponds to the coefficient of $\mathbf{y}$ in $\overline{\mathfrak{f}}_{0}^{\mathfrak{s}}(\mathbf{x})$ . These observations complete the proof of the lemma.◻

Proof. We first need to verify that $\overline{F}_{\infty }^{\mathfrak{s}}$ and $\overline{F}_{0}^{\mathfrak{s}}$ are chain maps. Note that

$$\begin{eqnarray}\displaystyle (\overline{F}_{\infty }^{\mathfrak{s}}\circ d+d\circ \overline{F}_{\infty }^{\mathfrak{s}})(\mathbf{x}_{1},\mathbf{x}_{2}) & = & \displaystyle ((G_{\infty }^{\mathfrak{s}}\circ d+d\circ G_{\infty }^{\mathfrak{s}})(\mathbf{x}_{1}),(f_{1}^{\mathfrak{s}}\circ G_{\infty }^{\mathfrak{s}}+f_{0}^{\mathfrak{s}})(\mathbf{x}_{1}))\nonumber\\ \displaystyle & = & \displaystyle (0,(f_{1}^{\mathfrak{s}}\circ G_{\infty }^{\mathfrak{s}}+f_{0}^{\mathfrak{s}})(\mathbf{x}_{1}))\nonumber\\ \displaystyle & = & \displaystyle 0,\nonumber\end{eqnarray}$$

where the last equality follows from the second part of Lemma 3.4. Since $I^{\mathfrak{s}}$ is a chain map by Lemma 3.5, it follows that $\overline{F}_{0}^{\mathfrak{s}}$ is also a chain map. Lemma 3.6 implies that

$$\begin{eqnarray}\overline{\mathfrak{f}}_{0}^{\mathfrak{s}}+\overline{F}_{0}^{\mathfrak{s}}\circ \jmath _{1}^{\mathfrak{s}}=d\circ Q^{\mathfrak{s}}+Q^{\mathfrak{s}}\circ d.\end{eqnarray}$$

This proves the commutativity of the right-hand-side square up to chain homotopy. To prove the commutativity of the left-hand-side square, define

$$\begin{eqnarray}\displaystyle & \displaystyle R^{\mathfrak{s}}:\widehat{\text{CF}}(\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{0};u,v,w;\mathfrak{s})=\widehat{\text{CF}}(\unicode[STIX]{x1D6F4}^{\prime },\unicode[STIX]{x1D736},\unicode[STIX]{x1D737}_{0};u,v;\mathfrak{s})\rightarrow M(f_{1}^{\mathfrak{s}}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle R^{\mathfrak{s}}(\mathbf{x}):=(H^{\mathfrak{s}}(\mathbf{x}),P^{\mathfrak{s}}(\mathbf{x})),\quad \forall \mathbf{x}\in \mathbb{T}_{\unicode[STIX]{x1D6FC}}\cap \mathbb{T}_{\unicode[STIX]{x1D6FD}_{0}}~\text{with}~\mathfrak{s}(\mathbf{x})=\mathfrak{s}. & \displaystyle \nonumber\end{eqnarray}$$

We thus find that

$$\begin{eqnarray}\displaystyle (d\circ R^{\mathfrak{s}}+R^{\mathfrak{s}}\circ d)(\mathbf{x}) & = & \displaystyle d(H^{\mathfrak{s}}(\mathbf{x}),P^{\mathfrak{s}}(\mathbf{x}))+(H^{\mathfrak{s}}(d(\mathbf{x})),P^{\mathfrak{s}}(d(\mathbf{x})))\nonumber\\ \displaystyle & = & \displaystyle ((G_{\infty }^{\mathfrak{s}}\circ g_{0}^{\mathfrak{s}}+g_{1}^{\mathfrak{s}}\circ \overline{\mathfrak{f}}_{\infty }^{\mathfrak{s}})(\mathbf{x}),(d\circ P^{\mathfrak{s}}+P^{\mathfrak{s}}\circ d+f_{1}^{\mathfrak{s}}\circ H^{\mathfrak{s}})(\mathbf{x}))\nonumber\\ \displaystyle & = & \displaystyle ((G_{\infty }^{\mathfrak{s}}\circ g_{0}^{\mathfrak{s}}+g_{1}^{\mathfrak{s}}\circ \overline{\mathfrak{f}}_{\infty }^{\mathfrak{s}})(\mathbf{x}),(H_{h_{1}}^{\mathfrak{s}}\circ \overline{\mathfrak{f}}_{\infty }^{\mathfrak{s}}+H_{h_{0}}^{\mathfrak{s}})(\mathbf{x}))\nonumber\\ \displaystyle & = & \displaystyle (\overline{F}_{\infty }^{\mathfrak{s}}\circ \jmath _{0}^{\mathfrak{s}}+\jmath _{1}^{\mathfrak{s}}\circ \overline{\mathfrak{f}}_{\infty }^{\mathfrak{s}})(\mathbf{x}).\nonumber\end{eqnarray}$$

The second equality follows from the first part of Lemma 3.4, while the third equality follows from Lemma 3.2. This observation completes the proof of Theorem 3.7. ◻