Subsection 7.4. (0PHH) · ScalingStacks

7.4.1. Braids

Let $Z$ be a curve. Let $I$ and $J$ be two finite subsets of $Z$ .

Definition 7.4.1. A parametrized braid $I\to J$ is a family $\vartheta=(\vartheta_{s})_{s\in I}$ where $\vartheta_{s}$ is an admissible path in $Z$ with $\vartheta_{s}(0)=s$ and such that $s\mapsto\vartheta_{s}(1)$ defines a bijection $\chi(\vartheta):I\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}J$ . A braid $I\to J$ is a homotopy class of parametrized braids, i.e., a family of admissible homotopy classes of paths.

0PAM

Definition 7.4.2. We define the pre-strand category ${\mathcal{P}}^{\bullet}(Z)=S({\mathcal{S}}^{\bullet}(Z,1))$ (cf §2.4).

The objects of this pointed category are the finite subsets of $Z$ and $\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}(Z)}(I,J)$ is the set of braids $I\to J$ , together with a $0$ -element. Given $\theta:I\to J$ and $\theta^{\prime}:J\to K$ two braids, we have $\theta^{\prime}\circ\theta=(\theta^{\prime}_{\theta_{s}(1)}\circ\theta_{s})_{s\in I}$ if $\theta^{\prime}_{\theta_{s}(1)}\circ\theta_{s}$ is admissible for all $s\in I$ , and we have $\theta^{\prime}\circ\theta=0$ otherwise. If $\theta^{\prime}\circ\theta\neq 0$ , we have $\chi(\theta^{\prime}\circ\theta)=\chi(\theta^{\prime})\circ\chi(\theta)$ .

We put ${\mathcal{P}}(Z)={\mathbf{F}}_{2}[{\mathcal{P}}^{\bullet}(Z)]$ .

Note that there is a decomposition ${\mathcal{P}}^{\bullet}(Z)=\bigvee_{n\geq 0}{\mathcal{P}}^{\bullet}(Z,n)$ , where ${\mathcal{P}}^{\bullet}(Z,n)$ is the full subcategory of ${\mathcal{P}}^{\bullet}(Z)$ with objects subsets with $n$ elements. We have ${\mathcal{P}}^{\bullet}(Z,1)={\mathcal{S}}^{\bullet}(Z,1)$ .

Given $M$ a subset of $Z$ , we denote by ${\mathcal{P}}^{\bullet}_{M}(Z)$ the full subcategory of ${\mathcal{P}}^{\bullet}(Z)$ with objects the finite subsets of $M$ .

Given $\theta:I\to J$ a braid and $I^{\prime}$ a subset of $I$ , we denote by $\theta_{|I^{\prime}}$ the braid $(\theta_{s})_{s\in I^{\prime}}$ .

Let $f:Z\to Z^{\prime}$ be a morphism of curves. We denote by ${\mathcal{P}}^{\bullet}_{f}(Z)$ the full subcategory of ${\mathcal{P}}^{\bullet}(Z)$ with objects those finite subsets $I$ of $Z$ such that $|f(I)|=|I|$ .

The next proposition follows immediately from Lemma 7.3.14 and §2.4.

0PAN

Proposition 7.4.3. The functor $f:{\mathcal{S}}^{\bullet}(Z,1)\to{\mathcal{S}}^{\bullet}(Z^{\prime},1)$ defines a faithful pointed functor

f:{\mathcal{P}}^{\bullet}_{f}(Z)\to{\mathcal{P}}^{\bullet}(Z^{\prime}),\ I\mapsto f(I),\ \theta\mapsto(f(\theta_{s}))_{f(s)}.

In particular if $f:Z\to Z^{\prime}$ is injective then we have a faithful pointed functor $f:{\mathcal{P}}^{\bullet}(Z)\to{\mathcal{P}}^{\bullet}(Z^{\prime})$ .

We define a non-multiplicative $f^{\#}:\operatorname{add}\nolimits({\mathcal{P}}(Z^{\prime}))\to\operatorname{add}\nolimits({\mathcal{P}}(Z))$ that commutes with coproduct. Given $I^{\prime}$ a finite subset of $Z^{\prime}$ , we put

f^{\#}(I^{\prime})=\coprod_{p:I^{\prime}\to Z,\ fp=\operatorname{id}\nolimits_{I^{\prime}}}p(I^{\prime}).

Consider now $\theta^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}(Z^{\prime})}(I^{\prime},J^{\prime})$ non-zero. Given $s^{\prime}\in I^{\prime}$ , we have a decomposition $f^{\#}(\theta^{\prime}_{s^{\prime}})=\sum_{s\in f^{-1}(s^{\prime})}f^{\#}(\theta^{\prime}_{s^{\prime}})_{s}$ along the decomposition $f^{\#}(s^{\prime})=\bigoplus_{s\in f^{-1}(s^{\prime})}s$ (cf §7.3.4). Given $p:I^{\prime}\to Z$ with $fp=\operatorname{id}\nolimits_{I^{\prime}}$ , we put $f^{\#}_{p}(\theta^{\prime})=\bigl(f^{\#}(\theta^{\prime}_{f(s)})_{s}\bigr)_{s\in p(I^{\prime})}$ , a map in ${\mathcal{P}}(Z)$ with source $p(I^{\prime})$ .

We define

f^{\#}(\theta^{\prime})=\sum_{p:I^{\prime}\to Z,\ fp=\operatorname{id}\nolimits_{I^{\prime}}}f^{\#}_{p}(\theta^{\prime}).

Note that $f^{\#}(\theta^{\prime})=\sum_{\theta\in f^{-1}(\theta^{\prime})}\theta$ , where $f^{-1}(\theta^{\prime})$ is the set of braids in $Z$ lifting $\theta$ .

Given $f^{\prime}:Z^{\prime}\to Z^{\prime\prime}$ a morphism of curves, we have $(f^{\prime}f)^{\#}=f^{\#}f^{\prime\#}$ .

The next two propositions are immediate consequences of Propositions 7.3.18 and 7.3.19 (cf §2.4).

0PAP

Proposition 7.4.4. If $f$ is strict, then $f^{\#}$ defines a functor $\operatorname{add}\nolimits({\mathcal{P}}(Z^{\prime}))\to\operatorname{add}\nolimits({\mathcal{P}}_{f}(Z))$ commuting with coproducts.

0PAQ

Proposition 7.4.5. Let $Z$ be a curve with a finite admissible relation $\sim$ and let $q:Z\to Z/\!\!\sim$ be the quotient map. The functor $q^{\#}:\operatorname{add}\nolimits({\mathcal{P}}(Z/\!\!\sim))\to\operatorname{add}\nolimits({\mathcal{P}}_{q}(Z))$ is faithful and every map in ${\mathcal{P}}^{\bullet}(Z/\!\!\sim)$ is in the image of the functor $q:{\mathcal{P}}_{q}^{\bullet}(Z)\to{\mathcal{P}}^{\bullet}(Z/\!\!\sim)$ .

Note that the construction $Z\mapsto\operatorname{add}\nolimits({\mathcal{P}}(Z))$ and $f\mapsto f^{\#}$ defines a contravariant functor from the category of curves with strict morphisms to the category of ${\mathbf{F}}_{2}$ -linear categories.

Let $Z_{1},\ldots,Z_{r}$ be the connected components of $Z$ . The isomorphism (7.3.4) induces an isomorphism of pointed categories

(7.4.1)

{\mathcal{P}}^{\bullet}(Z_{1})\wedge\cdots\wedge{\mathcal{P}}^{\bullet}(Z_{r})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{P}}^{\bullet}(Z).

Note that the inverse functor sends a braid $\theta:I\to J$ in $Z$ to $(\theta_{1},\ldots,\theta_{r})$ , where $\theta_{i}$ is the restriction of $\theta$ to $I\cap Z_{i}$ .

0PAR

Example 7.4.6. We describe below an example of product in ${\mathcal{P}}^{\bullet}(Z)$ .

7.4.2. Degree

Consider $\theta:I\to J$ a braid. We put

i(\theta)=\frac{1}{2}\sum_{\Omega\in\pi_{0}(Z)}\sum_{\begin{subarray}{c}s\neq s^{\prime}\in I\cap\Omega\end{subarray}}i(\theta_{s},\theta_{s^{\prime}})e_{\Omega}\in({\mathbf{Z}}_{\geq 0})^{\pi_{0}(Z)}

We define $\llbracket\theta\rrbracket=\sum_{s\in I}\llbracket\theta_{s}\rrbracket\in R(Z)$ and

m(\theta)=\sum_{s\in I}\sum_{c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))}m_{c}(\llbracket\theta\rrbracket)e_{c}\in L(Z).

Finally, we define $\deg^{\prime}(\theta)\in\Gamma(Z)$ by

\deg^{\prime}(\theta)=(-i(\theta),(-m(\theta),-\llbracket\theta\rrbracket)).

Given $D\subset T(Z)$ with $D\cap\iota(D)=\emptyset$ , we denote by $\deg_{D}(\theta)$ the image of $\deg^{\prime}(\theta)$ in $\Gamma(Z,D)$ . Note that if $D^{\prime}\subset D$ , then $\deg_{D}(\theta)$ is the image of $\deg_{D^{\prime}}(\theta)\in\Gamma(Z,D^{\prime})$ in $\Gamma(Z,D)$ .

We put $\deg(\theta)=\deg_{Z_{exc}^{+}}(\theta)$ and we denote by $\overline{\deg}(\theta)$ (resp. $\overline{\deg}_{D}(\theta)$ ) the image of $\deg(\theta)$ (resp. $\deg_{D}(\theta)$ ) in $\bar{\Gamma}(Z,Z_{exc}^{+})$ (resp. $\bar{\Gamma}(Z,D)$ ).

0PAS

Lemma 7.4.7. Let $\theta:I\to J$ be a braid in $Z$ . Let $E$ be a subset of $\{s\in I\cap Z_{o}\ |\ \theta_{s}=\operatorname{id}\nolimits_{s}\}$ and let $\bar{\theta}=(\theta_{s})_{s\in I-E}$ . We have $\deg_{E^{+}}(\theta)=\deg_{E^{+}}(\bar{\theta})$ .

0PAT

Proof. Note that $\llbracket\theta\rrbracket=\llbracket\bar{\theta}\rrbracket$ . Let $s\in E$ . We have

\sum_{c\in C(s)}m_{c}(\llbracket\theta\rrbracket)e_{c}=\sum_{c\in C(s)^{+}}\sum_{s^{\prime}\in I,\ s^{\prime}\neq s}(m_{c}-m_{\iota(c)})(\llbracket\theta_{s^{\prime}}\rrbracket)e_{c}\xrightarrow{e_{c}\to 1}2\sum_{s^{\prime}\in I,\ s^{\prime}\neq s}i(\operatorname{id}\nolimits_{s},\theta_{s^{\prime}})

by Lemma 7.3.21. The lemma follows. ∎

0PAU

Remark 7.4.8. Note that $i(\theta)=\sum_{I^{\prime}\subset I,\ |I^{\prime}|=2}i(\theta_{|I^{\prime}})$ .

The next lemma shows that the failure of multiplicativity of $\deg$ and $i$ coincide up to terms involving points in $Z_{exc}$ .

0PAV

Lemma 7.4.9. Let $\theta:I\to J$ and $\theta^{\prime}:I^{\prime}\to I$ be two braids such that $\theta\circ\theta^{\prime}$ is a braid. The element $\deg(\theta)\cdot\deg(\theta^{\prime})\cdot\deg(\theta\circ\theta^{\prime})^{-1}$ of $\Gamma(Z,Z_{exc}^{+})$ is in $\bigoplus_{\Omega}\frac{1}{2}{\mathbf{Z}}e_{\Omega}$ and it is equal to

i(\theta\circ\theta^{\prime})-i(\theta)-i(\theta^{\prime})\\ +\frac{1}{2}\sum_{\begin{subarray}{c}\Omega,\ s^{\prime}\in I^{\prime}\cap Z_{exc}\cap\Omega\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})^{+}\setminus\theta_{s^{\prime}}(0+)\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket)e_{\Omega}+\frac{1}{2}\sum_{\begin{subarray}{c}\Omega,\ s^{\prime}\in I^{\prime}\cap\Omega\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits,\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))^{+}\setminus\iota(\theta^{\prime}_{s^{\prime}}(1-))\\ \theta^{\prime}_{s^{\prime}}(1)\in Z_{exc}\end{subarray}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{\Omega}.

and is also equal to

\frac{1}{2}\sum_{\begin{subarray}{c}\Omega,(s^{\prime}_{1},s^{\prime}_{2})\in(I^{\prime}\cap\Omega)^{2}\\ (s^{\prime}_{1},s^{\prime}_{2}){\not\in}E\cup E^{\prime}\end{subarray}}\bigl(i(\theta_{s_{1}}\circ\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}},\theta_{s_{2}})-i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})\bigr)e_{\Omega}+\\ +\sum_{\Omega,\ (s^{\prime}_{1},s^{\prime}_{2})\in E\cap\Omega}\bigl(i(\theta_{s_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}},\theta_{s_{2}})-m_{\theta_{s_{1}}(0+)}^{+}(\theta^{\prime}_{s^{\prime}_{2}})\bigr)e_{\Omega}+\\ +\sum_{\Omega,\ (s^{\prime}_{1},s^{\prime}_{2})\in E^{\prime}\cap\Omega}\bigl(i(\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})-i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})-m_{\theta^{\prime}_{s^{\prime}_{1}}(1-)}^{-}(\theta_{s_{2}})\bigr)e_{\Omega}

where

•

given $(s^{\prime}_{1},s^{\prime}_{2})\in I^{\prime 2}$ , we put $s_{i}=\theta^{\prime}_{s^{\prime}_{i}}(1)$
•

$E$ is the set of pairs $(s^{\prime}_{1},s^{\prime}_{2})\in I^{\prime}\times I^{\prime}$ with $s^{\prime}_{1}\in\ Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}=\operatorname{id}\nolimits$ , $\theta_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ , $\theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits$
•

$E^{\prime}$ is the set of pairs $(s^{\prime}_{1},s^{\prime}_{2})\in I^{\prime}\times I^{\prime}$ with $s_{1}\in\ Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ , $\theta_{s_{1}}=\operatorname{id}\nolimits$ and $\theta_{s_{2}}\neq\operatorname{id}\nolimits$ .

0PAW

Proof. Given $s^{\prime}\in I^{\prime}$ and $s=\theta^{\prime}_{s^{\prime}}(1)$ , the class $\theta_{s}\circ\theta^{\prime}_{s^{\prime}}$ is admissible, hence $\theta_{s}(0+)\cup\iota(\theta_{s}(0+))=\theta^{\prime}_{s^{\prime}}(1-)\cup\iota(\theta^{\prime}_{s^{\prime}}(1-))$ unless $s\in Z_{exc}$ and one of $\theta_{s}$ and $\theta_{s^{\prime}}$ is the identity, but not the other.

Given $c\in T(Z)$ , we put

v_{c}=(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{c}=m_{c}(\llbracket\theta\rrbracket)e_{c}+m_{\iota(c)}(\llbracket\theta\rrbracket)e_{\iota(c)}=v_{\iota(c)}.

Let

a=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\cap Z_{exc}\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})\setminus\bigl((\theta_{s^{\prime}}(0+)\cup\iota(\theta_{s^{\prime}}(0+))\bigr)\end{subarray}}m_{c^{\prime}}(\llbracket\theta^{\prime}\rrbracket)e_{c^{\prime}}=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\cap Z_{exc}\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})^{+}\setminus\theta_{s^{\prime}}(0+)\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket)e_{c^{\prime}}.

We have

m(\theta\circ\theta^{\prime})-m(\theta)-m(\theta^{\prime})=

=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c^{\prime}\in(\theta\circ\theta^{\prime})_{s^{\prime}}(0+)\cup\iota((\theta\circ\theta^{\prime})_{s^{\prime}}(0+))\end{subarray}}m_{c^{\prime}}(\llbracket\theta\rrbracket)e_{c^{\prime}}-\sum_{\begin{subarray}{c}s\in I\\ c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))\end{subarray}}m_{c}(\llbracket\theta\rrbracket)e_{c}-a

=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(0+)}-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(1-)}-\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))\end{subarray}}v_{c}-a

Using (7.3.1), we find

	$\displaystyle\langle\llbracket\theta\rrbracket,\llbracket\theta^{\prime}\rrbracket\rangle$	$\displaystyle=-\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c^{\prime}\in\theta^{\prime}_{s^{\prime}}(0+)\cup\iota(\theta^{\prime}_{s^{\prime}}(0+))\end{subarray}}v_{c^{\prime}}+\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c\in\theta^{\prime}_{s^{\prime}}(1-)\cup\iota(\theta^{\prime}_{s^{\prime}}(1-))\end{subarray}}v_{c}$
		$\displaystyle=-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(0+)}+\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(1-)}.$

We deduce that

\langle\llbracket\theta\rrbracket,\llbracket\theta^{\prime}\rrbracket\rangle+m(\theta\circ\theta^{\prime})-m(\theta)-m(\theta^{\prime})=-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))^{+}\setminus\iota(\theta^{\prime}_{s^{\prime}}(1-))\\ \theta^{\prime}_{s^{\prime}}(1)\in Z_{exc}\end{subarray}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{c}-a

and the first equality of the lemma follows.

Consider $s^{\prime}_{1}\neq s^{\prime}_{2}$ in $I^{\prime}$ .

If $s^{\prime}_{1}\in Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}=\operatorname{id}\nolimits_{s^{\prime}_{1}}$ and $\theta_{s_{1}}\neq\operatorname{id}\nolimits_{s_{1}}$ , it follows from Lemma 7.3.21 that

\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits\end{subarray}}i(\operatorname{id}\nolimits_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})=\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits_{s^{\prime}_{2}}\\ c^{\prime}\in C(s^{\prime}_{1})^{+}\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\theta^{\prime}_{s^{\prime}_{2}})=\frac{1}{2}\sum_{c^{\prime}\in C(s^{\prime}_{1})^{+}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket).

Similarly, if $s_{1}\in Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ and $\theta_{s_{1}}=\operatorname{id}\nolimits$ , we have

\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta_{s_{2}}\neq\operatorname{id}\nolimits\end{subarray}}i(\operatorname{id}\nolimits_{s_{1}},\theta_{s_{2}})=\frac{1}{2}\sum_{c\in C(s_{1})^{+}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket).

The second equality of the lemma follows. ∎

0PAX

Example 7.4.10. The left (respectively second) side of the diagram below shows a typical instance where the left (respectively right) sum of Lemma 7.4.9 is nonzero.

0PAY

Remark 7.4.11. Let $\theta:I\to J$ and $\theta^{\prime}:I^{\prime}\to I$ be two braids such that $\theta\circ\theta^{\prime}$ is a braid. By Lemma 7.3.23, the terms $i(\theta_{s_{1}},\theta_{s_{2}})+i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}}\circ\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})$ , $i(\theta_{s_{1}},\theta_{s_{2}})+m_{\theta_{s_{1}}(0+)}^{+}(\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})$ and $i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})+m_{\theta^{\prime}_{s^{\prime}_{1}}(1-)}^{-}(\theta_{s_{2}})-i(\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})$ in Lemma 7.4.9 are all non-negative.

We deduce that the following assertions are equivalent:

•

$\deg(\theta)\cdot\deg(\theta^{\prime})=\deg(\theta\circ\theta^{\prime})$
•

$\deg(\theta_{|E})\cdot\deg(\theta^{\prime}_{|E^{\prime}})=\deg(\theta_{|E}\circ\theta^{\prime}_{|E^{\prime}})$ for any two-element subset $E^{\prime}\subset I^{\prime}$ , where $E=\chi(\theta^{\prime})(E^{\prime})$ .

If given $s\in I^{\prime}$ with $\theta^{\prime}_{s}=\operatorname{id}\nolimits$ or $\theta_{\chi(\theta^{\prime})(s)}=\operatorname{id}\nolimits$ , we have $s{\not\in}Z_{exc}$ , then $\deg(\theta)\cdot\deg(\theta^{\prime})=\deg(\theta\circ\theta^{\prime})$ if and only if $i(\theta_{\theta^{\prime}_{s}(1)},\theta_{\theta^{\prime}_{s^{\prime}}(1)})+i(\theta^{\prime}_{s},\theta^{\prime}_{s^{\prime}})=i((\theta\circ\theta^{\prime})_{s},(\theta\circ\theta^{\prime})_{s^{\prime}})$ for all $s\neq s^{\prime}$ in $I^{\prime}$ .

0PAZ

Lemma 7.4.12. Let $f:Z\to Z^{\prime}$ be a morphism of curves. Let $I$ and $J$ be two finite subsets of $Z$ such that $|f(I)|=|f(J)|=|I|=|J|$ . Let $\theta:I\to J$ be a braid in $Z$ . Let $E=\{s\in I\cap Z_{f}\ |\ \theta_{s}=\operatorname{id}\nolimits_{s}\}$ .

We have $f(\deg_{f^{-1}(f(E))^{+}}(\theta))=\deg_{f(E)^{+}}(f(\theta))$ .

0PB0

Proof. Assume first $E=\emptyset$ . Given $s\in I$ with $\theta_{s}=\operatorname{id}\nolimits_{s}$ , we have a bijection $C(s)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}C(f(s))$ . It follows that

	$\displaystyle f(m(\theta))$	$\displaystyle=\sum_{\begin{subarray}{c}s\in I\\ \theta_{s}\neq\operatorname{id}\nolimits_{s}\end{subarray}}\sum_{c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))}m_{c}(\llbracket\theta\rrbracket)f(e_{c})+\sum_{\begin{subarray}{c}s\in I\\ \theta_{s}=\operatorname{id}\nolimits_{s}\end{subarray}}\sum_{c\in C(s)}m_{c}(\llbracket\theta\rrbracket)f(e_{c})$
		$\displaystyle=\sum_{\begin{subarray}{c}s^{\prime}\in f(I)\\ f(\theta)_{s^{\prime}}\neq\operatorname{id}\nolimits_{s^{\prime}}\end{subarray}}\sum_{c^{\prime}\in f(\theta)_{s^{\prime}}(0+)\cup\iota(f(\theta)_{s^{\prime}}(0+))}m_{c^{\prime}}(\llbracket f(\theta)\rrbracket)e_{c^{\prime}}+\sum_{\begin{subarray}{c}s^{\prime}\in f(I)\\ f(\theta)_{s^{\prime}}=\operatorname{id}\nolimits_{s^{\prime}}\end{subarray}}\sum_{c^{\prime}\in C(s^{\prime})}m_{c^{\prime}}(\llbracket f(\theta)\rrbracket)e_{c^{\prime}}$
		$\displaystyle=m(f(\theta))$

by Lemma 7.1.24.

Given $s^{\prime}\in f(I)$ such that $f(\theta)_{s^{\prime}}=\operatorname{id}\nolimits_{s^{\prime}}$ , we have $s^{\prime}{\not\in}Z^{\prime}_{f}$ . We deduce that $i(\theta_{s},\theta_{t})=i(f(\theta)_{f(s)},f(\theta)_{f(t)})$ for all $s\neq t\in I$ by Lemma 7.3.22. So $f(i(\theta))=i(f(\theta))$ . We deduce that the lemma holds for $\theta$ .

Consider now the case where $E\neq\emptyset$ . Let $\bar{\theta}=(\theta_{s})_{s\in I-E}$ . We have $\deg_{E^{+}}(\theta)=\deg_{E^{+}}(\bar{\theta})$ by Lemma 7.4.7; taking quotients, we obtain $\deg_{f^{-1}(f(E))^{+}}(\theta)=\deg_{f^{-1}(f(E))^{+}}(\bar{\theta})$ . Since $f(\bar{\theta})=(f(\theta)_{t})_{t\in f(I)-f(E)}$ , it follows again from Lemma 7.4.7 that $\deg_{f(E)^{+}}(f(\theta))=\deg_{f(E)^{+}}(f(\bar{\theta}))$ . Since the lemma holds for $\bar{\theta}$ , we deduce that the lemma holds for $\theta$ . ∎

As a consequence of Lemma 7.4.12, we have the following result.

0PB1

Proposition 7.4.13. Let $f:Z\to Z^{\prime}$ be a morphism of curves and let $\theta^{\prime}$ be a non-zero map in ${\mathcal{P}}^{\bullet}(Z^{\prime})$ . Then $f^{\#}(\theta^{\prime})$ is a sum of maps $\theta$ such that $f(\deg_{Z_{f}^{+}}(\theta))=\deg_{f(Z_{f})^{+}}(\theta^{\prime})$ .

Let $Z_{1},\ldots,Z_{r}$ be the connected components of $Z$ . The isomorphism (7.4.1) is compatible with the degree function in the following sense. Given $\theta:I\to J$ a braid in $Z$ , let $\theta_{i}$ be the restriction of $\theta$ to $I\cap Z_{i}$ . The image of $(\deg(\theta_{1}),\ldots,\deg(\theta_{r}))$ in $\Gamma(Z)$ by the map of (7.3.3) is $\deg(\theta)$ .

Let $I$ and $J$ be two finite subsets of $Z$ and let $\theta:I\to J$ be a braid in $Z$ . We define

L(\theta)=\coprod_{i_{1}\neq i_{2}\in I}I(\theta_{i_{1}},\theta_{i_{2}}).

Note that $\zeta\mapsto\zeta^{-1}$ induces a fixed-point free involution $\mathrm{inv}$ on $L(\theta)$ .

Let $\zeta\in L(\theta)$ . Put $i_{1}=\zeta(0)$ and $i_{2}=\zeta(1)$ . We define $\theta^{\zeta}$ by $(\theta^{\zeta})_{i}=\theta_{i}$ if $i\in I-\{i_{1},i_{2}\}$ , $(\theta^{\zeta})_{i_{1}}=\theta_{i_{2}}\circ\zeta=\bar{\zeta}\circ\theta_{i_{1}}$ and $(\theta^{\zeta})_{i_{2}}=\theta_{i_{1}}\circ\zeta^{-1}=\bar{\zeta}^{-1}\circ\theta_{i_{2}}$ . Note that $\theta^{\zeta^{-1}}=\theta^{\zeta}$ .

Let $D(\theta)$ be the set of classes $\zeta$ in $L(\theta)$ such that

(a)

given a class of smooth paths $\zeta^{\prime}:\zeta(0)\to\zeta(1)$ such that $\zeta\circ\zeta^{\prime-1}$ and $\zeta^{\prime-1}\circ\zeta$ are smooth and have the same orientation as $\zeta$ and $\zeta^{\prime}$ , and given a class of smooth paths $\zeta^{\prime\prime}:\bar{\zeta}(0)\to\bar{\zeta}(1)$ such that $\bar{\zeta}\circ\zeta^{\prime\prime-1}$ and $\zeta^{\prime\prime-1}\circ\bar{\zeta}$ are smooth and have the same orientation as $\bar{\zeta}$ and $\zeta^{\prime\prime}$ , then $\zeta^{\prime}=\zeta$ or $\zeta^{\prime\prime}=\bar{\zeta}$ .
(b)

given $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ in $L(\theta)$ with $\zeta=\zeta^{\prime}\circ\zeta^{\prime\prime}$ , then $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations.

0PB2

Remark 7.4.14. Condition (a) above is automatically satisfied if the component of the support of $\zeta$ is not isomorphic to $S^{1}$ .

The subset $D(\theta)$ of $L(\theta)$ is stable under the involution $\mathrm{inv}$ .

The next lemma restricts the cases where condition (b) above needs to be checked.

0PB3

Lemma 7.4.15. Let $\zeta,\zeta^{\prime},\zeta^{\prime\prime}\in L(\theta)$ such that $\zeta=\zeta^{\prime}\circ\zeta^{\prime\prime}$ . If $\zeta^{\prime}(0)\in Z_{o}$ and $\theta_{\zeta^{\prime}(0)}=\operatorname{id}\nolimits$ , then $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations.

0PB4

Proof. Let $z=\zeta^{\prime}(0)=\zeta^{\prime\prime}(1)$ . We have $\zeta^{\prime}\in I(\operatorname{id}\nolimits_{z},\theta_{\zeta(1)})$ . Since $\bar{\zeta}^{\prime}=\theta_{\zeta(1)}\circ\zeta^{\prime}$ is smooth and has opposite orientation to $\zeta^{\prime}$ , it follows that $\zeta^{\prime}(0+)\in\iota(C(z)^{+})$ . Similarly, $\zeta^{\prime\prime}(1-)\in\iota(C(z)^{+})$ . We deduce that $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations. ∎

0PB5

Lemma 7.4.16. Let $I^{\prime}$ be a subset of $I$ such that $I-I^{\prime}\subset Z_{o}$ and $\theta_{i}=\operatorname{id}\nolimits$ for $i\in I-I^{\prime}$ .

We have $D(\theta_{|I^{\prime}})\subset D(\theta)$ .

0PB6

Proof. We have $L(\theta_{|I^{\prime}})\subset L(\theta)$ and Lemma 7.4.15 shows that $D(\theta_{|I^{\prime}})\subset D(\theta)$ . ∎

0PB7

Example 7.4.17. In the picture below, the left side shows a valid braid $\theta$ , for which the conclusion of Lemma 7.4.15 holds. For contrast, the right side shows a braid $\theta$ that is disallowed since $\theta_{i_{2}}$ is not oriented, and the conclusion of Lemma 7.4.15 fails.

7.4.3. Strands on $S^{1}$

Let $Z=S^{1}$ , viewed as an unoriented manifold. Fix a family $M=\{a_{1},\ldots,a_{n}\}$ of cyclically ordered points on $S^{1}$ , i.e., $a_{j}=e^{ie_{j}}$ for some real numbers $e_{1}<\cdots<e_{n}$ with $e_{n}-e_{1}<2\pi$ .

Fix $r^{\prime},r\in\{1,\ldots,n\}$ . There is a bijection

F_{r^{\prime},r}:r^{\prime}-r+n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{\Pi(S^{1})}(a_{r},a_{r^{\prime}}):

it sends $l$ to the homotopy class of paths going in the positive direction and winding $\lfloor\frac{l}{n}\rfloor$ times around $S^{1}$ , if $l\geq 0$ , and to the homotopy class of paths going in the negative direction and winding $\lfloor\frac{-l}{n}\rfloor$ times around $S^{1}$ , otherwise.

We put

F_{r}=\sum_{r^{\prime}}F_{r^{\prime},r}:{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\coprod_{r^{\prime}}\operatorname{Hom}\nolimits_{\Pi(S^{1})}(a_{r},a_{r^{\prime}}).

Given $r,r^{\prime}\in\{1,\ldots,n\}$ and $l,l^{\prime}\in{\mathbf{Z}}$ with $r^{\prime}-r=l\pmod{n}$ , we have $F_{r}(l+l^{\prime})=F_{r^{\prime}}(l^{\prime})\circ F_{r}(l)$ . Note also that given $j\in\{1,\ldots n\}$ and $j^{\prime}\in{\mathbf{Z}}$ , we have

\operatorname{supp}\nolimits(F_{j}(j^{\prime}-j))=\begin{cases}S^{1}&\text{ if }|j^{\prime}-j|\geq n\\ \{e^{iu}\ |\ e_{j}\leq u\leq e_{j^{\prime\prime}}+2\pi\delta_{j^{\prime}>n}&\text{ if }j^{\prime}-j\in\{0,\ldots,n-1\}\\ \{e^{iu}\ |\ e_{j^{\prime\prime}}-2\pi\delta_{j^{\prime}\leq 0}\leq u\leq e_{j}&\text{ if }j-j^{\prime}\in\{0,\ldots,n-1\}\end{cases}

where $j^{\prime\prime}\in\{1,\ldots,n\}$ and $j^{\prime\prime}-j^{\prime}\in n{\mathbf{Z}}$ .

We denote by $\vec{S}^{1}$ the oriented curve $S^{1}$ . Fix $z=e^{ix}\in S^{1}$ with $x<e_{1}$ and $e_{n}-x<2\pi$ and $\Omega$ a connected open neighbourhood of $z$ in $S^{1}$ containing no $a_{i}$ . Let $I=S^{1}-\{z\}$ unoriented and $\vec{I}=S^{1}-\{z\}$ oriented. We define $\dot{S}^{1}$ to be the curve $S^{1}$ with $(\dot{S}^{1})_{o}=\Omega$ with its standard orientation.

0PB8

Proposition 7.4.18. There is an isomorphism of pointed categories $F:({\mathcal{S}}_{n})_{+}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{P}}^{\bullet}_{M}(S^{1})$ given by $F(J)=\{a_{j}\}_{j\in\tilde{J}\cap[1,n]}$ and $F(\sigma)_{a_{j}}=F_{j}(\sigma(j)-j)$ for $\sigma$ a map of ${\mathcal{S}}_{n}$ .

It restricts to isomorphisms of pointed categories

({\mathcal{S}}_{n}^{+})_{+}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{P}}^{\bullet}_{M}(\dot{S}^{1}),\ ({\mathcal{S}}_{n}^{++})_{+}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{P}}^{\bullet}_{M}(\vec{S}^{1}),\ ({\mathcal{S}}_{n}^{f})_{+}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{P}}^{\bullet}_{M}(I)\text{ and }({\mathcal{S}}_{n}^{f++})_{+}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{P}}^{\bullet}_{M}(\vec{I}).

0PB9

Proof. Consider $J,J^{\prime}\subset{\mathbf{Z}}/n$ . We have an injective map $f:\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,J^{\prime})\to{\mathbf{Z}}^{J},\ \sigma\mapsto(\sigma(j)-j))_{b}$ , where $j\in\{1,\ldots,n\}$ and $b=j+n{\mathbf{Z}}$ . The image of that map is the set of those $c\in{\mathbf{Z}}^{J}$ such that $\{c_{b}+b\}_{b}=J^{\prime}$ and we obtain a bijection

	$\displaystyle\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,J^{\prime})$	$\displaystyle\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}_{M}(\vec{S}^{1})}(\{a_{j}\}_{j\in\tilde{J}\cap[1,n]},\{a_{j^{\prime}}\}_{j^{\prime}\in\tilde{J}^{\prime}\cap[1,n]})$
	$\displaystyle\sigma$	$\displaystyle\mapsto\bigl(F_{j^{\prime},j}(f(\sigma)_{j+n{\mathbf{Z}}})\bigr)_{j\in\tilde{J}\cap[1,n],\ j^{\prime}\in\tilde{J}^{\prime}\cap[1,n],\ \sigma(j)-j^{\prime}\in n{\mathbf{Z}}}.$

We deduce that $F$ induces a bijection on pointed $\operatorname{Hom}\nolimits$ -sets. Consider now $\sigma:J\to J^{\prime}$ and $\sigma^{\prime}:J^{\prime}\to J^{\prime\prime}$ two maps in ${\mathcal{S}}_{n}$ . Given $j\in\tilde{J}\cap[1,n]$ , we have

F(\sigma^{\prime}\sigma)_{a_{j}}=F_{j}(\sigma^{\prime}\sigma(j)-j)=F_{j}(\sigma^{\prime}(\sigma(j))-\sigma(j)+\sigma(j)-j)=F_{\sigma(j)}(\sigma^{\prime})_{a_{\sigma(j)}}\circ F_{j}(\sigma)_{a_{j}}.

We deduce that $F$ is a functor and the first statement of the proposition follows.

Consider now $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,J^{\prime})$ .

The map $F(\sigma)$ is in ${\mathcal{P}}^{\bullet}_{M}(\dot{S}^{1})$ if and only if $\sigma(j)\geq 0$ for all $j\in[1,n]\cap\tilde{J}$ , hence if and only if $\sigma$ is in ${\mathcal{S}}_{n}^{+}$ .

The map $F(\sigma)$ is in ${\mathcal{P}}^{\bullet}_{M}(\vec{S}^{1})$ if and only if $\sigma(j)-j\geq 0$ for all $j\in\tilde{J}$ , hence if and only if $\sigma$ is in ${\mathcal{S}}_{n}^{++}$ .

The map $F(\sigma)$ is in ${\mathcal{P}}^{\bullet}_{M}(I)$ if and only if $\sigma(j)\in[1,n]$ for all $j\in\tilde{J}\cap[1,n]$ , hence if and only if $\sigma$ is in ${\mathcal{S}}_{n}^{f}$ .

The proposition follows. ∎

There are morphisms of groups $F_{R}:R_{n}\to R(S^{1}),\ \alpha_{j+n{\mathbf{Z}}}\mapsto\llbracket F_{j}(1)\rrbracket$ and $F_{L}:L_{n}\to L(S^{1}),\ \varepsilon_{j+n{\mathbf{Z}}}\mapsto e_{c_{j}}$ , where $j\in\{1,\ldots,n\}$ , $c_{j}=(a_{j},a_{j}e^{iu})\in C(a_{j})$ and $u\in{\mathbf{R}}_{>0}$ is small enough.

0PBA

Lemma 7.4.19. Given $\alpha,\beta\in R_{n}$ , we have $F_{L}(\langle\alpha,\beta\rangle)=\langle F_{R}(\alpha),F_{R}(\beta)\rangle$ and there is an injective morphism of groups $F_{\Gamma}:\Gamma_{n}\to\Gamma(S^{1}),\ (r,(l,\alpha))\mapsto(r,(F_{L}(l),F_{R}(\alpha)))$ .

Let $D$ be a subset of $\{1,\ldots,n\}\times\{\pm 1\}$ that embeds in its projection on $\{1,\ldots,n\}$ . Define $\partial:D\to T(S^{1})$ by $\partial((i,\nu_{i}))=c_{i}$ if $\nu_{i}=1$ and $\partial((i,\nu_{i}))=\iota(c_{i})$ otherwise. The morphism $F_{\Gamma}$ induces an isomorphism of groups $F_{D}:\Gamma_{D}\to\Gamma_{M}(S^{1},\partial(D))$ . We have $u<u^{\prime}$ if and only if $F_{D}(u)<F_{D}(u^{\prime})$ .

Let $\sigma$ be a map in ${\mathcal{S}}_{n}$ . We have $F_{R}(\llbracket\sigma\rrbracket)=\llbracket F(\sigma)\rrbracket$ , $m(F(\sigma))=F_{L}(m(\sigma))$ , $i(F(\sigma))=\ell(\sigma)$ and $\deg(F(\sigma))=F_{\Gamma}(\deg(\sigma))$ .

0PBB

Proof. Let $r,j\in\{1,\ldots,n\}$ and let $j^{\prime}\in{\mathbf{Z}}$ . We have

m_{c_{r}}(\llbracket F_{j}(j^{\prime}-j)\rrbracket)=|\{i\in r+n{\mathbf{Z}}\ |\ j\leq i<j^{\prime}\}|-|\{i\in r+n{\mathbf{Z}}\ |\ j>i\geq j^{\prime}\}|

and

m_{\iota(c_{r})}(\llbracket F_{j}(j^{\prime}-j)\rrbracket)=-|\{i\in r+n{\mathbf{Z}}\ |\ j<i\leq j^{\prime}\}|+|\{i\in r+n{\mathbf{Z}}\ |\ j\geq i>j^{\prime}\}|

In particular, $m_{c_{r}}(\llbracket F_{j}(1)\rrbracket)=\delta_{r,j}$ and $m_{\iota(c_{r})}(\llbracket F_{j}(1)\rrbracket)=-\delta_{r,j+1}$ . This shows that $F_{R}$ is injective. This shows also that given $i\in\{1,\ldots,n\}$ , we have

\langle\llbracket F_{i}(1)\rrbracket,\llbracket F_{j}(1)\rrbracket\rangle=(\delta_{i,j+1}+\delta_{i,j})F_{L}(\varepsilon_{j+1+n{\mathbf{Z}}})-(\delta_{i,j}+\delta_{i+1,j})F_{L}(\varepsilon_{j+n{\mathbf{Z}}})=F_{L}(\langle\alpha_{i+n{\mathbf{Z}}},\alpha_{j+n{\mathbf{Z}}}\rangle).

This shows the first equality and this shows that $F_{R}$ induces an injective morphism of groups $F_{\Gamma}$ .

Taking quotients, we obtain an injective morphism of groups $F_{D}:\Gamma_{D}\to\Gamma(S^{1},\partial(D))$ compatible with the order and with image $\Gamma_{M}(S^{1},\partial(D))$ .

Consider $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ . Given $d\in{\mathbf{Z}}$ , we have $\llbracket F_{r}(d)\rrbracket=F_{R}(\alpha_{r,r+d})$ , hence $F_{R}(\llbracket\sigma\rrbracket)=\llbracket F(\sigma)\rrbracket$ .

We have

m(F(\sigma))=\sum_{r,j\in\tilde{I}\cap[1.n]}(m_{c_{r}}-m_{\iota(c_{r})})(\llbracket F_{j}(\sigma(j)-j)\rrbracket)e_{c_{r}}

and

m(\sigma)=\sum_{r,j\in\tilde{I}\cap[1,n]}\alpha_{j,\sigma(j)}\cdot\varepsilon_{r+n{\mathbf{Z}}}.

Since

\alpha_{j,\sigma(j)}\cdot\varepsilon_{r+nZ}=(m_{c_{r}}-m_{\iota(c_{r})})(F(\sigma))\varepsilon_{r+nZ},

it follows that $m(F(\sigma))=F_{L}(m(\sigma))$ .

Consider $i_{1},i_{2}\in\tilde{I}$ with $0\leq i_{1}<i_{2}<n$ . We have $i(F(\sigma)_{a_{i_{1}}},F(\sigma)_{a_{i_{2}}})=i(\gamma_{1},\gamma_{2})$ for some minimal paths $\gamma_{l}$ in $F(\sigma)_{a_{i_{l}}}$ by Lemma 7.3.22. Lemma 6.2.3 shows that $i(F(\sigma)_{a_{i_{1}}},F(\sigma)_{a_{i_{2}}})=\bigl|{\lfloor\frac{\sigma(i_{2})-\sigma(i_{1})}{n}\rfloor}\bigr|$ . Lemma 6.2.2 shows now that $i(F(\sigma))=\ell(\sigma)$ . ∎

Given $i_{1},i_{2}\in{\mathbf{Z}}$ with $i_{2}{\not\in}i_{1}+n{\mathbf{Z}}$ , we put $\lambda(i_{1},i_{2})=F_{i^{\prime}_{1}}(i_{2}-i_{1})$ , where $i^{\prime}_{1}\in[1,n]\cap(i_{1}+n{\mathbf{Z}})$ .

0PBC

Lemma 7.4.20. Let $\sigma$ be a map in ${\mathcal{S}}_{n}$ . Given $(i_{1},i_{2})$ in $L(\sigma)$ (resp. $D(\sigma)$ ), the class $\lambda(i_{1},i_{2})$ is in $L(F(\sigma))$ (resp. $D(F(\sigma))$ ) and $F(\sigma)^{\lambda(i_{1},i_{2})}=F(\sigma^{i_{1},i_{2}})$ . Furthermore, $\lambda$ induces bijections

L(\sigma)/n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}L(F(\sigma))/\mathrm{inv}\text{ and }D(\sigma)/n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}D(F(\sigma))/\mathrm{inv}.

0PBD

Proof. Note first that, given $i^{\prime}_{1}$ and $i^{\prime}_{2}$ two distinct elements of $\{1,\ldots,n\}$ , then $\lambda$ induces a bijection

\bigl((i^{\prime}_{1}+n{\mathbf{Z}})\times(i^{\prime}_{2}+n{\mathbf{Z}})\bigr)/n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{\Pi(S^{1})}(a_{i^{\prime}_{1}},a_{i^{\prime}_{2}}).

Consider $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ and $i_{1},i_{2}\in\tilde{I}$ with $i_{2}{\not\in}i_{1}+n{\mathbf{Z}}$ . Note that $\lambda(i_{1},i_{2})=\lambda(i_{2},i_{1})^{-1}$ for any $i_{1},i_{2}$ .

Let $\zeta_{r}=F(\sigma)_{a_{i_{r}}}$ and $\zeta=\lambda(i_{1},i_{2})$ . We have $\bar{\zeta}=\lambda(\sigma(i_{1}),\sigma(i_{2}))$ . So, $\zeta\in L(F(\sigma))$ if and only if $i_{1}-i_{2}$ and $\sigma(i_{1})-\sigma(i_{2})$ have opposite signs. On the other hand, $(i_{1},i_{2})\in L(\sigma)$ if and only if $i_{1}<i_{2}$ and $\sigma(i_{2})<\sigma(i_{1})$ . This shows that $\lambda(L(\sigma))\subset L(F(\sigma))$ and $\lambda$ induces a bijection $L(\sigma)/n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}L(F(\sigma))/\mathrm{inv}$ .

Consider $(i_{1},i_{2})\in L(\sigma)$ . Let $r=\lfloor\frac{i_{2}-i_{1}}{n}\rfloor$ and $s=\lfloor\frac{\sigma(i_{1})-\sigma(i_{2})}{n}\rfloor$ . We have $r>0$ if and only if $\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}-rn))\subsetneq\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}))$ and $s>0$ if and only if $\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}+sn))\subsetneq\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}))$ . There is $i$ such that $(i_{1},i)$ and $(i,i_{2})$ are in $L(\sigma)$ if and only if there are $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ with the same orientations in $L(F(\sigma))$ such that $\lambda(i_{1},i_{2})=\zeta^{\prime\prime}\circ\zeta^{\prime}$ . We have $i_{2}-i_{1}>n$ if and only if there is $\zeta^{\prime}$ such that $\zeta$ , $\zeta^{\prime}$ and $\zeta\circ\zeta^{\prime-1}$ have the same orientation. We have $\sigma(i_{1})-\sigma(i_{2})>n$ if and only if there is $\zeta^{\prime\prime}$ such that $\bar{\zeta}$ , $\zeta^{\prime\prime}$ and $\bar{\zeta}\circ\zeta^{\prime\prime-1}$ have the same orientation.

We deduce that $(i_{1},i_{2})\in D(\sigma)$ if and only if $\lambda(i_{1},i_{2})\in D(F(\sigma))$ .

Assume now $(i_{1},i_{2})\in L(\sigma)$ . Let $i^{\prime}_{r}\in[1,n]\cap(i_{r}+n{\mathbf{Z}})$ for $r\in\{1,2\}$ . We have

(F(\sigma)^{\lambda(i_{1},i_{2})})_{a_{i^{\prime}_{1}}}=F_{i^{\prime}_{2}}(\sigma(i_{2})-i_{2})\circ F_{i^{\prime}_{1}}(i_{2}-i_{1})=F_{i^{\prime}_{1}}(\sigma(i_{2})-i_{1})=(F(\sigma^{i_{1},i_{2}}))_{a_{i_{1}}}.

Similarly, $(F(\sigma)^{\lambda(i_{1},i_{2})})_{a_{i^{\prime}_{2}}}=(F(\sigma^{i_{1},i_{2}}))_{a_{i_{2}}}$ . It follows that $F(\sigma)^{\lambda(i_{1},i_{2})}=F(\sigma^{i_{1},i_{2}})$ . This completes the proof of the lemma. ∎

7.4.4. Strand category

Let $Z$ be a curve.

We have a positivity result in the setting of Lemma 7.4.9.

0PBE

Lemma 7.4.21. Let $\theta$ and $\theta^{\prime}$ be two braids such that $\theta\circ\theta^{\prime}$ is a braid. We have $\deg(\theta)\cdot\deg(\theta^{\prime})\leq\deg(\theta\circ\theta^{\prime})$ .

Given $D$ a subset of $T(Z)$ containing $Z_{exc}^{+}$ and such that $D\cap\iota(D)=\emptyset$ , the following assertions are equivalent:

•

$\deg(\theta)\cdot\deg(\theta^{\prime})=\deg(\theta\circ\theta^{\prime})$
•

$\deg(\theta)\cdot\deg(\theta^{\prime})$ and $\deg(\theta\circ\theta^{\prime})$ have the same image in $\Gamma(Z,D)$
•

$\deg(\theta)\cdot\deg(\theta^{\prime})$ and $\deg(\theta\circ\theta^{\prime})$ have the same image in $\bar{\Gamma}(Z,D)$ .

0PBF

Proof. Assume $Z=S^{1}$ unoriented. Let $M$ be a family as in §7.4.3. Assume $M$ contains $\theta_{s}(r)$ and $\theta^{\prime}_{s}(r)$ for $r\in\{0,1\}$ and all $s$ . Proposition 7.4.18 and Lemma 7.4.19 show that the inequality follows from the corresponding inequality for maps in ${\mathcal{S}}_{n}$ , which is given by Lemmas 6.2.1 and 6.2.5.

Given $Z$ a non-singular connected curve, there is an injective morphism of curves $Z\to S^{1}$ , and the lemma follows from Proposition 7.4.3 and Lemma 7.4.12. We deduce that the inequality holds for any non-singular curve $Z$ .

Consider now a general curve $Z$ and let $q:\hat{Z}\to Z$ be the non-singular cover. Since the functor $q^{\#}:\operatorname{add}\nolimits({\mathcal{P}}(Z))\to\operatorname{add}\nolimits({\mathcal{P}}(\hat{Z}))$ is compatible with degrees (Proposition 7.4.13), it follows that the inequality holds for $Z$ .

The equivalence of the three assertions follows from the fact that an element of $(\frac{1}{2}{\mathbf{Z}}_{\geq 0})^{\pi_{0}(Z)}\subset\Gamma(Z,Z_{exc}^{+})$ is zero if and only if its image in $\frac{1}{2}{\mathbf{Z}}_{\geq 0}\subset\bar{\Gamma}(Z,D)$ is zero. ∎

By Lemma 7.4.21, the degree function gives a $\Gamma(Z,Z_{exc}^{+})$ -filtration on the category ${\mathcal{P}}^{\bullet}(Z)$ .

0PBG

Definition 7.4.22. We define the strand category ${\mathcal{S}}^{\bullet}(Z)$ as the $\Gamma(Z,Z_{exc}^{+})$ -graded pointed category associated with the filtered pointed category ${\mathcal{P}}^{\bullet}(Z)$ (cf §2.3.3).

The category ${\mathcal{S}}^{\bullet}(Z)$ has the same objects and the same maps as the category ${\mathcal{P}}^{\bullet}(Z)$ . It is a pointed category with objects the finite subsets of $Z$ and with $\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(I,J)$ the set of braids $I\to J$ , together with a $0$ -element.

The product of two braids $\theta:I\to J$ with $\theta^{\prime}:J\to K$ is defined as follows:

\theta^{\prime}\cdot\theta=\begin{cases}\theta^{\prime}\circ\theta&\text{ if }\deg(\theta^{\prime}\circ\theta)=\deg(\theta^{\prime})\cdot\deg(\theta)\\ 0&\text{ otherwise.}\end{cases}

Note that the strand category decomposes as a disjoint union ${\mathcal{S}}^{\bullet}(Z)=\coprod_{n\geq 0}{\mathcal{S}}^{\bullet}(Z,n)$ , where ${\mathcal{S}}^{\bullet}(Z,n)$ is the full subcategory with objects subsets with $n$ elements.

It follows from Lemma 7.4.21 that given $D$ a subset of $T(Z)$ containing $Z_{exc}^{+}$ and such that $D\cap\iota(D)=\emptyset$ , the structure of $\Gamma(Z,D)$ -graded (resp. $\bar{\Gamma}(Z,D)$ -graded) category on ${\mathcal{S}}(Z)$ obtained from the quotient morphism $f:\Gamma(Z,Z_{exc}^{+})\to\Gamma(Z,D)$ (resp. $f:\Gamma(Z,Z_{exc}^{+})\to\bar{\Gamma}(Z,D)$ ) is the same as the graded category obtained from the structure of $\Gamma(Z,D)$ -filtered (resp. $\bar{\Gamma}(Z,D)$ -filtered) category on ${\mathcal{P}}^{\bullet}(Z)$ that is deduced from the structure of $\Gamma(Z,Z_{exc}^{+})$ -filtered category via $f$ .

0PBH

Remark 7.4.23. We leave to the reader to check the following alternate definition of the product in the strand category.

We have $\theta^{\prime}\cdot\theta\neq 0$ if and only if there are parametrized braids $\vartheta,\vartheta^{\prime}$ with $\theta=[\vartheta]$ and $\theta^{\prime}=[\vartheta^{\prime}]$ and there are $\alpha:I^{\prime}\to I$ and $\alpha^{\prime}:K\to K^{\prime}$ two parametrized braids with $I^{\prime},K^{\prime}\subset Z\setminus Z_{exc}$ such that $i(\alpha)=i(\alpha^{\prime})=0$ , $\alpha^{\prime}_{\vartheta^{\prime}\circ\vartheta\circ\alpha_{s}(1)}\circ\vartheta^{\prime}_{\vartheta\circ\alpha_{s}(1)}\circ\vartheta_{\alpha_{s}(1)}\circ\alpha_{s}$ is admissible for all $s\in I$ and $i(\alpha^{\prime}\circ\theta^{\prime}\circ\theta\circ\alpha)=i(\alpha^{\prime}\circ\theta^{\prime})+i(\theta\circ\alpha)$ .

Let ${\mathcal{S}}(Z)={\mathbf{F}}_{2}[{\mathcal{S}}^{\bullet}(Z)]$ , a $\Gamma(Z,Z_{exc}^{+})$ -graded ${\mathbf{F}}_{2}$ -linear category.

Let $f:Z\to Z^{\prime}$ be a morphism of curves.

Let ${\mathcal{S}}^{\bullet}_{f}(Z)$ be the full subcategory of ${\mathcal{S}}^{\bullet}(Z)$ with objects those finite subsets $I$ of $Z$ such that $|f(I)|=|I|$ . We deduce from Proposition 7.4.3 and Lemma 7.4.12 a faithful $\Gamma(Z^{\prime},Z_{exc}^{\prime+})$ -graded pointed functor $f:{\mathcal{S}}^{\bullet}_{f}(Z)\to{\mathcal{S}}^{\bullet}(Z^{\prime})$ . Here, the $\Gamma(Z^{\prime},Z_{exc}^{\prime+})$ -grading on ${\mathcal{S}}^{\bullet}_{f}(Z)$ comes from the $\Gamma(Z,f^{-1}(Z_{exc}^{\prime})^{+})$ -grading via the morphism $\Gamma(f)$ .

Assume $f$ is strict. Propositions 7.4.4 and 7.4.13 provide an additive ${\mathbf{F}}_{2}$ -linear $\Gamma(Z^{\prime},Z_{exc}^{\prime+})$ -graded functor $f^{\#}:\operatorname{add}\nolimits({\mathcal{S}}(Z^{\prime}))\to\operatorname{add}\nolimits({\mathcal{S}}_{f}(Z))$ , where the $\Gamma(Z^{\prime},Z_{exc}^{\prime+})$ -grading on ${\mathcal{S}}_{f}(Z)$ is deduced from the $\Gamma(Z,f^{-1}(Z_{exc}^{\prime})^{+})$ -grading via the morphism $\Gamma(f)$ .

If $f$ is a quotient morphism, it follows from Proposition 7.4.5 that $f^{\#}$ is faithful.

Given $M$ a subset of $Z$ , we denote by ${\mathcal{S}}^{\bullet}_{M}(Z)$ the full subcategory of ${\mathcal{S}}^{\bullet}(Z)$ whose objects are the finite subsets of $M$ . The $\Gamma(Z,Z_{exc}^{+})$ -grading on ${\mathcal{S}}^{\bullet}_{M}(Z)$ comes from a $\Gamma_{M}(Z,Z_{exc}^{+})$ -grading.

We denote by ${\mathcal{S}}_{M,f}^{\bullet}(Z)$ the full subcategory of ${\mathcal{S}}_{f}^{\bullet}(Z)$ with objects subsets contained in $M$ .

We put ${\mathcal{A}}^{\bullet}(Z)={\mathcal{S}}_{Z_{exc}}^{\bullet}(Z)$ and ${\mathcal{A}}^{\bullet}(Z,n)={\mathcal{S}}_{Z_{exc}}^{\bullet}(Z,n)$ . Let ${\mathcal{A}}(Z)={\mathbf{F}}_{2}[{\mathcal{A}}^{\bullet}(Z)]$ and ${\mathcal{A}}(Z,n)={\mathbf{F}}_{2}[{\mathcal{A}}^{\bullet}(Z,n)]$ .

Let $Z_{1},\ldots,Z_{r}$ be the connected components of $Z$ . The isomorphism (7.4.1) induces an isomorphism of $\Gamma(Z,Z_{exc}^{+})$ -graded pointed categories

(7.4.2)

{\mathcal{S}}^{\bullet}(Z_{1})\wedge\cdots\wedge{\mathcal{S}}^{\bullet}(Z_{r})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}(Z)

where the grading on the left hand term is deduced from the the $\bigl(\prod_{i=1}^{r}\Gamma(Z_{i},(Z_{i})_{exc}^{+})\bigr)$ -grading via (7.3.3) and an isomorphism of ${\mathbf{F}}_{2}$ -linear categories

(7.4.3)

{\mathcal{S}}(Z_{1})\otimes\cdots\otimes{\mathcal{S}}(Z_{r})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}(Z).

0PBI

Example 7.4.24. In the example below the first row is the product in ${\mathcal{P}}^{\bullet}(Z)$ , while the second row is the product in ${\mathcal{S}}^{\bullet}(Z)$ .

7.4.5. Generation

We equip $Z$ with a metric. Given $\xi$ a path in $Z$ , we denote by $|\xi|$ its length. Given $\zeta$ a homotopy class of paths in $Z$ , we put $|\zeta|=|\xi|$ , where $\xi$ is a minimal path in $\zeta$ . Given $\theta:I\to J$ a braid in $Z$ , we put $|\theta|=\sum_{s\in I}|\theta_{s}|$ .

Let $M$ be a finite subset of $Z$ .

0PBJ

Lemma 7.4.25. Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I,J)$ and $\theta^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I^{\prime},I)$ such that $\theta\circ\theta^{\prime}$ is a braid. Let $I_{0}$ be a finite subset of $M\setminus(I\cup I^{\prime}\cup J)$ .

If $|\theta|+|\theta^{\prime}|=|\theta\circ\theta^{\prime}|$ , then $(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})\cdot(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=(\theta\cdot\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}}$ .

0PBK

Proof. Let $s^{\prime}\in I^{\prime}$ . Since $|\theta_{\theta^{\prime}(s^{\prime})}|+|\theta^{\prime}_{s^{\prime}}|=|\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}}|$ , it follows that $i(\theta_{\theta^{\prime}(s^{\prime})},\operatorname{id}\nolimits_{i})+i(\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i})=i(\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i})$ for all $i\in I_{0}$ . As a consequence,

i((\theta\circ\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=i(\theta\circ\theta^{\prime})-i(\theta)-i(\theta^{\prime}).

The lemma follows now from Lemma 7.4.9. ∎

Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I,J)$ be a non-zero braid.

Let $I_{0}=\{i\in I\ |\ \theta_{i}=\operatorname{id}\nolimits_{i}\}$ . Let $i\in I\setminus I_{0}$ . There is a (unique) decomposition $\theta_{i}=\beta^{i}\cdot\alpha^{i}$ in ${\mathcal{S}}^{\bullet}(Z,1)$ with

•

$\alpha^{i}(1)\in M\setminus I_{0}$
•

$|\theta_{i}|=|\alpha^{i}|+|\beta^{i}|$
•

given a minimal path $\xi$ in $\alpha^{i}$ , we have $\xi((0,1))\cap M\subset I_{0}$ .

We define a quiver $\Gamma(\theta)$ with vertex set $I\setminus I_{0}$ . There is an arrow $i\to i^{\prime}$ if $\alpha^{i}(1)=i^{\prime}$ .

Note that there is at most one arrow with a given source (that arrow can be a loop).

0PBL

Lemma 7.4.26. Let $I^{\prime}$ be a non-empty finite subset of $I\setminus I_{0}$ such that

•

if there is an arrow $i\to i^{\prime}$ in $\Gamma(\theta)$ with $i\in I^{\prime}$ , then $i^{\prime}\in I^{\prime}$
•

given $i\neq i^{\prime}\in I^{\prime}$ , we have $\alpha^{i}(1)\neq\alpha^{i^{\prime}}(1)$ .

There is a (unique) decomposition $\theta=\theta^{u}\cdot u$ in ${\mathcal{S}}_{M}^{\bullet}(Z)$ , where $|\theta|=|\theta^{u}|+|u|$ and

u_{i}=\begin{cases}\alpha^{i}&\text{ if }i\in I^{\prime}\\ \operatorname{id}\nolimits_{i}&\text{ otherwise.}\end{cases}

0PBM

Proof. Note that the second assumption on $I^{\prime}$ shows that the full subquiver of $\Gamma(\theta)$ with vertex set $I^{\prime}$ is a disjoint union of oriented lines and oriented circles.

Let $i_{1}\neq i_{2}\in I$ . If $i_{1},i_{2}\in I\setminus I^{\prime}$ , then $u(i_{1})\neq u(i_{2})$ . Assume now $i_{1}\in I^{\prime}$ and $i_{2}\in I\setminus I^{\prime}$ . Since $i_{1}\to i_{2}$ is not an arrow of the quiver, we have $u(i_{1})\neq i_{2}$ , hence $u(i_{1})\neq u(i_{2})$ . Finally if $i_{1},i_{2}\in I^{\prime}$ , then $u(i_{1})\neq u(i_{2})$ . We have shown that $u$ is a braid.

Note that there is a (unique) decomposition $\theta=\theta^{u}\circ u$ with $|\theta|=|\theta^{u}|+|u|$ . In order to show that $\theta^{u}\cdot u\neq 0$ , we can replace $\theta$ by $\theta_{|I\setminus I_{0}}$ and $M$ by $M\setminus I_{0}$ , thanks to Lemma 7.4.25. So, we assume now that $I_{0}=\emptyset$ .

Let $q:\tilde{Z}\to Z$ be a non-singular cover of $Z$ . Let $\tilde{M}=q^{-1}(M)$ . Let $\tilde{\theta}:\tilde{I}\to\tilde{J}$ be the unique lift of $\theta$ to $\tilde{Z}$ . We have a decomposition $\tilde{\theta}_{i}=\tilde{\beta}^{i}\cdot\tilde{\alpha}^{i}$ for $i\in\tilde{I}$ and $q(\tilde{\alpha}^{i})=\alpha^{q(i)}$ .

Let $\tilde{I}^{\prime}=q^{-1}(I^{\prime})\cap\tilde{I}$ . Note that $q$ induces a morphism of quivers $\Gamma(\tilde{\theta})\to\Gamma(\theta)$ , hence $\tilde{I}^{\prime}$ satisfies the assumptions of the lemma and we have a decomposition $\tilde{\theta}=\tilde{\theta}^{\tilde{u}}\circ\tilde{u}$ . Since $q(\tilde{u})=u$ , it follows that if the lemma holds for $\tilde{\theta}$ , then it holds for $\theta$ .

We assume now that $Z$ is non-singular. If the lemma holds for connected components of $Z$ , it will hold for $Z$ , hence it is enough to prove the lemma for $Z$ connected. Assume now $Z$ is connected. There is an injective morphism of curves $f:Z\to S^{1}$ , where $S^{1}$ is unoriented. It the lemma holds for $S^{1}$ , it holds for $Z$ .

We assume finally that $Z=S^{1}$ unoriented. Let $i_{1}\neq i_{2}\in I^{\prime}$ such that $i(u_{i_{1}},u_{i_{2}})\neq 0$ . Note that $u_{i_{1}}$ and $u_{i_{2}}$ have opposite directions and $i(u_{i_{1}},u_{i_{2}})=1$ . Furthermore, $\theta_{i_{r}}$ has the same direction as $u_{i_{r}}$ , hence $i(\theta_{i_{1}},\theta_{i_{2}})=i((\theta^{u})_{i_{1}},(\theta^{u})_{i_{2}})+1$ . Given $i_{1}\neq i_{2}\in I$ with $i_{1}{\not\in}I^{\prime}$ , we have $i(u_{i_{1}},u_{i_{2}})=0$ . It follows from Remark 7.4.11 that $\theta^{u}\cdot u\neq 0$ . This completes the proof of the lemma. ∎

Note that the length of a map in ${\mathcal{S}}^{\bullet}_{M}(Z)$ takes value in a finitely generated submonoid of ${\mathbf{R}}_{\geq 0}$ . So, a repeated application of the previous lemma provides a decomposition of any map $\theta$ of ${\mathcal{S}}_{M}^{\bullet}(Z)$ as a product $\theta=u_{n}\cdots u_{1}$ , where $u_{i}$ is a map $u$ as in the lemma.

7.4.6. Decomposition at a point

Let $z_{0}\in Z_{o}$ with $z_{0}{\not\in}M$ .

Given $\zeta$ a homotopy class of admissible paths in $Z$ with $\zeta\neq\operatorname{id}\nolimits_{z_{0}}$ , we put $\mu(\zeta)=i(\zeta,\operatorname{id}\nolimits_{z_{0}})$ .

Assume $\mu(\zeta)\geq 1$ . There is a unique decomposition $\zeta=\zeta^{r-}\cdot\zeta^{r}$ in ${\mathcal{S}}^{\bullet}(Z,1)$ such that $\zeta^{r}(1)=z_{0}$ and $\mu(\zeta^{r})=1$ .

Given $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{M}^{\bullet}(Z)}(I,J)$ , we put $\mu(\theta)=\sum_{s\in I}\mu(\theta_{s})$ . Given $\theta^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{M}^{\bullet}(Z)}(I^{\prime},I)$ with $\theta\cdot\theta^{\prime}\neq 0$ , we have $\mu(\theta\cdot\theta^{\prime})=\mu(\theta)+\mu(\theta^{\prime})$ .

0PBN

Lemma 7.4.27. Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{M}^{\bullet}(Z)}(I,J)$ with $\mu(\theta)\geq 2$ .

There exists a decomposition $\theta=r^{\prime}(\theta)\cdot r(\theta)$ in ${\mathcal{S}}_{M}(Z)$ with $\mu(r(\theta))=1$ and with the following property.

Let $s\in I$ such that $\mu(r(\theta)_{s})=1$ . Given $s^{\prime}\in I$ such that $\mu(\theta_{s^{\prime}})\geq 1$ and $\mathrm{supp}(\theta_{s^{\prime}}^{r})\subset\mathrm{supp}(\theta_{s}^{r})$ , then $s^{\prime}=s$ .

0PBP

Proof. We prove the lemma by induction on $|\theta|$ . Assume there is a set $I^{\prime}$ satisfying the assumptions of Lemma 7.4.26 and such that $\mu(u)=0$ . By induction, there is a decomposition $\theta^{u}=r^{\prime}(\theta^{u})\cdot r(\theta^{u})$ as in the lemma. Now $r(\theta)=r(\theta^{u})\cdot u$ and $r^{\prime}(\theta)=r^{\prime}(\theta^{u})$ satisfy the requirements of the lemma.

Assume now that given any set $I^{\prime}$ satisfying the assumptions of Lemma 7.4.26, we have $\mu(u)\geq 1$ .

Let $s\in I$ with $\mu(\theta_{s})\geq 1$ such that given $s^{\prime}\in I$ with $\mu(\theta_{s^{\prime}})\geq 1$ , we have $\mathrm{supp}(\theta_{s}^{r})\subset\mathrm{supp}(\theta_{s^{\prime}}^{r})$ . Given $s^{\prime}\in I\setminus\{s\}$ , we have $\mu(\alpha^{s^{\prime}})=0$ (notations of §7.4.5).

Let $I^{\prime}$ be the set of $s^{\prime}\in I$ such that there is a sequence $s_{0}=s,s_{1},\ldots,s_{r}=s^{\prime}$ of elements of $I$ such that $s_{i}\to s_{i+1}$ is an arrow of $\Gamma(\theta)$ for $0\leq i<r$ . Assume there exist $s_{1},\ldots,s_{d}$ in $I^{\prime}\setminus\{s_{0}\}$ such that $s_{d}=s_{1}$ and $s_{i}\to s_{i+1}$ is an arrow of $\Gamma(\theta)$ for $1\leq i<d$ . Then $I^{\prime\prime}=\{s_{1},\ldots,s_{d}\}$ satisfies the assumptions of Lemma 7.4.26. On the other hand, we have $\mu(\alpha^{s^{\prime}})=0$ for $s^{\prime}\in I^{\prime\prime}$ , hence we get a contradiction. It follows that $I^{\prime}$ is a cycle or a line and it satisfies the assumptions of Lemma 7.4.26. The braids $r^{\prime}(\theta)=\theta^{u}$ and $r(\theta)=u$ of Lemma 7.4.26 satisfy the requirements of the lemma. ∎

7.4.7. Differential

Let us start with a description of $i(\theta)$ in terms of $L(\theta)$ , using our previous analysis of $S^{1}$ .

Let $f:Z\to Z^{\prime}$ be a morphism of curves. Given $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}_{f}(Z)}(I,J)$ , the map $f$ induces an injection $f:L(\theta)\hookrightarrow L(f(\theta))$ by the discussion above Lemma 7.3.24.

0PBQ

Lemma 7.4.28. Given $\theta^{\prime}\in f(\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}_{f}(Z)}(I,J))$ , the map $f$ induces a bijection $\bigcup_{\theta\in f^{-1}(\theta^{\prime})}L(\theta)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}L(\theta^{\prime})$ . It restricts to a bijection $\bigcup_{\theta\in f^{-1}(\theta^{\prime})}D(\theta)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}D(\theta^{\prime})$ .

0PBR

Proof. Assume first $f$ is a non-singular cover of $Z^{\prime}$ .

Let $\zeta^{\prime}\in L(\theta^{\prime})$ . There are $i^{\prime}_{1}\neq i^{\prime}_{2}\in f(I)$ such that $\zeta^{\prime}\in I(\theta^{\prime}_{i^{\prime}_{i}},\theta^{\prime}_{i^{\prime}_{2}})$ . By Lemma 7.3.24, there are elements $\zeta_{r}\in f^{-1}(\theta^{\prime}_{i_{r}})$ and $\zeta\in I(\zeta_{1},\zeta_{2})$ such $\zeta^{\prime}=f(\zeta)$ . We define $\theta\in f^{-1}(\theta^{\prime})$ by setting $\theta_{\zeta_{r}(0)}=\zeta_{r}$ and by setting $\theta_{i}$ to be any lift of $\theta^{\prime}_{f(i)}$ for all $f(i)\notin\{i^{\prime}_{1},i^{\prime}_{2}\}$ . This shows the surjectivity part of the first statement of the lemma.

Consider now $\theta$ and $\hat{\theta}$ maps in ${\mathcal{P}}_{f}^{\bullet}(Z)$ such that $f(\theta)=f(\hat{\theta})=\theta^{\prime}$ . Let $\zeta\in L(\theta)$ and $\hat{\zeta}\in L(\hat{\theta})$ such that $f(\zeta)=f(\hat{\zeta})=\zeta^{\prime}$ . There are $i^{\prime}_{1}\neq i^{\prime}_{2}\in f(I)$ such that $\zeta^{\prime}\in I(\theta^{\prime}_{i^{\prime}_{1}},\theta^{\prime}_{i^{\prime}_{2}})$ . We have $\hat{\theta}_{\hat{\zeta}(t)},\theta_{\zeta(t)}\in f^{-1}(\theta^{\prime}_{i^{\prime}_{r}})$ for $t\in\{0,1\}$ . It follows from Lemma 7.3.24 that $\zeta=\hat{\zeta}$ . So, the first statement of the lemma holds.

Assume now $f$ is an open embedding. The injectivity of the first map of the lemma is clear, while the surjectivity follows from Lemma 7.3.25.

We deduce the first part of the lemma for $Z$ and $Z^{\prime}$ non-singular and the general case follows now by taking non-singular covers of $Z$ and $Z^{\prime}$ and the lift of $f$ .

Let us prove now the second statement of the lemma about $D(\theta)$ .

Consider $\theta\in f^{-1}(\theta^{\prime})$ and $\zeta\in L(\theta)$ . It is clear that if $f(\zeta)\in D(\theta^{\prime})$ , then $\zeta\in D(\theta)$ .

Assume now $\zeta\in D(\theta)$ . Fix $i_{1}\neq i_{2}\in I$ so that $\zeta\in I(\theta_{i_{1}},\theta_{i_{2}})$ .

Let $\zeta^{\prime},\zeta^{\prime\prime}\in L(\theta^{\prime})$ such that $f(\zeta)=\zeta^{\prime}\circ\zeta^{\prime\prime}$ . Let $z=\zeta^{\prime}(0)=\zeta^{\prime\prime}(1)$ . If $|f^{-1}(\theta^{\prime}_{z})|>1$ , then $z\in Z_{o}$ and $\theta^{\prime}_{z}=\operatorname{id}\nolimits$ , hence $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations by Lemma 7.4.15. Assume now $\theta^{\prime}_{z}$ has a unique lift. Let $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ be the unique lifts of $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ (first part of the lemma). By unicity of lifts, we have $\zeta=\hat{\zeta}^{\prime}\circ\hat{\zeta}^{\prime\prime}$ . We have $\hat{\zeta}^{\prime},\hat{\zeta}^{\prime\prime}\in L(\theta)$ , hence $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ have opposite orientations. It follows that $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations as well.

Consider now $\zeta^{\prime}:f(i_{1})\to f(i_{2})$ a smooth homotopy class of paths such that $f(\zeta)\circ\zeta^{\prime-1}$ and $\zeta^{\prime-1}\circ f(\zeta)$ are smooth and have the same orientation as $f(\zeta)$ and $\zeta^{\prime}$ . Let $\hat{\zeta}^{\prime}$ be the unique lift of $\zeta^{\prime}$ . Since $f(\zeta)\circ\zeta^{\prime-1}$ is smooth, it follows that $\hat{\zeta}^{\prime}(0)=i_{1}$ and $\zeta\circ\hat{\zeta}^{\prime-1}$ is smooth and has the same orientation as $\zeta$ . Similarly, $\hat{\zeta}^{\prime}(1)=i_{2}$ and $\hat{\zeta}^{\prime-1}\circ\zeta$ is smooth and has the same orientation as $\zeta$ . A similar statement holds for $\zeta$ replaced by $\bar{\zeta}$ . We deduce that $f(\zeta)\in D(\theta^{\prime})$ . ∎

0PBS

Remark 7.4.29. The picture below shows what would go wrong in Lemma 7.4.28 if we allowed unoriented points in $Z_{exc}$ . In the proof, we need $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ to be in $L(\theta)$ , which would not be true if this example were valid.

0PBT

Proposition 7.4.30. Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}(Z)}(I,J)$ . We have $i(\theta)=\sum_{\Omega\in\pi_{0}(Z)}|(L(\theta)\cap\Omega)/\mathrm{inv}|e_{\Omega}$ . In particular, $L(\theta)$ is finite.

0PBU

Proof. The statement is true for $Z=S^{1}$ unoriented by Lemmas 3.2.3, 7.4.19 and 7.4.20. It follows from Lemmas 7.4.28 and 7.3.22 that it holds for any connected non-singular $Z$ , by embedding it in $S^{1}$ . So, the lemma holds for any non-singular $Z$ . By realizing an arbitrary $Z$ as a quotient of its non-singular cover, we deduce from Lemmas 7.4.28 and 7.3.22 that the lemma holds for any $Z$ . ∎

Given $f:Z\to Z^{\prime}$ a morphism of curves, given $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}_{f}(Z)}(I,J)$ and given $\zeta\in L(\theta)$ , we have $f(\theta^{\zeta})=f(\theta)^{f(\zeta)}$ .

0PBV

Lemma 7.4.31. Given $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}(Z)}(I,J)$ and $\zeta\in L(\theta)$ , we have $\theta^{\zeta}\in\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}(Z)}(I,J)$ . We have $\zeta\in D(\theta)$ if and only if $\overline{\deg}_{D}(\theta^{\zeta})=\overline{\deg}_{D}(\theta)+1$ for some (or equivalently, any) finite subset $D$ of $T(Z)$ such that $D\cap\iota(D)=\emptyset$ .

0PBW

Proof. Let us show the first statement. We can assume $\theta_{\zeta(0)}^{\zeta}\neq\operatorname{id}\nolimits$ .

Assume $\theta_{\zeta(0)}^{\zeta}$ has the same orientation as $\zeta^{-1}$ . We have $\theta_{\zeta(1)}=\theta_{\zeta(0)}^{\zeta}\circ\zeta^{-1}$ . If $\gamma$ and $\gamma^{\prime}$ are minimal paths in $\theta_{\zeta(0)}^{\zeta}$ and $\zeta^{-1}$ , then $\gamma\circ\gamma^{\prime}$ is a minimal path in $\theta_{\zeta(1)}$ . Since $\gamma\circ\gamma^{\prime}$ is admissible, it follows that $\gamma$ is admissible, hence $\theta_{\zeta(0)}^{\zeta}$ is admissible.

Otherwise, $\theta_{\zeta(0)}^{\zeta}$ has the same orientation as $\bar{\zeta}^{-1}$ and $\theta_{\zeta(0)}=\bar{\zeta}^{-1}\circ\theta_{\zeta(0)}^{\zeta}$ , hence we deduce as above that $\theta_{\zeta(0)}^{\zeta}$ is admissible.

Similarly, $\theta_{\zeta(1)}^{\zeta}$ is admissible and we deduce that $\theta^{\zeta}$ is a braid.

Let us prove the second part of the lemma. When $Z=S^{1}$ unoriented, this holds by Lemmas 7.4.20, 6.2.9 and 7.4.19 and Proposition 7.4.18. We deduce that the lemma holds when $Z$ is a connected non-singular curve, by embedding $Z$ in $S^{1}$ . So, it holds when $Z$ is a non-singular curve (since $\operatorname{supp}\nolimits(\zeta)$ is contained in a connected component of $Z$ ).

Consider now a general $Z$ and the non-singular cover $q:\hat{Z}\to Z$ . There is a braid $\hat{\theta}$ in $\hat{Z}$ with $q(\hat{\theta})=\theta$ (Lemma 7.4.5) and there is $\hat{\zeta}\in L(\hat{\theta})$ such that $\zeta=q(\hat{\zeta})$ (Lemma 7.4.28). The considerations above show that $\hat{\theta}^{\hat{\zeta}}$ is a braid in $\hat{Z}$ , hence $\theta^{\zeta}=q(\hat{\theta}^{\hat{\zeta}})$ is a braid in $Z$ . The statement on degrees follows from Lemmas 7.4.28 and 7.4.12. ∎

Given $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(I,J)$ , we put

d(\theta)=\sum_{\zeta\in D(\theta)/\mathrm{inv}}\theta^{\zeta}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z)}(I,J).

Note that the set $D(\theta)$ is finite by Proposition 7.4.30.

0PBX

Theorem 7.4.32. The map $d$ equips ${\mathcal{S}}(Z)$ with a structure of differential $\bar{\Gamma}(Z,Z_{exc}^{+})$ -graded ${\mathbf{F}}_{2}$ -linear category and ${\mathcal{S}}^{\bullet}(Z)$ with a structure of differential $\bar{\Gamma}(Z,Z_{exc}^{+})$ -graded pointed category.

Let $f:Z\to Z^{\prime}$ be a morphism of curves.

$\bullet\$ The functor $f:{\mathcal{S}}_{f}^{\bullet}(Z)\to{\mathcal{S}}^{\bullet}(Z^{\prime})$ is a faithful pointed functor and its restriction to ${\mathcal{S}}_{\{z\in Z\ |\ |f^{-1}f(z)|=1\}}^{\bullet}(Z)$ is a differential $\bar{\Gamma}(Z^{\prime},Z_{exc}^{\prime+})$ -graded pointed functor.

$\bullet\$ If $f$ is strict, then $f^{\#}:\operatorname{add}\nolimits({\mathcal{S}}(Z^{\prime}))\to\operatorname{add}\nolimits({\mathcal{S}}_{f}(Z))$ is a differential $\bar{\Gamma}(Z^{\prime},Z_{exc}^{\prime+})$ -graded functor commuting with coproducts.

$\bullet\$ If $f$ is a quotient morphism, then $f^{\#}$ is faithful and every map in ${\mathcal{S}}^{\bullet}(Z^{\prime})$ is in the image by $f$ of a map of ${\mathcal{S}}^{\bullet}_{f}(Z)$ .

0PBY

Proof. Lemma 7.4.28 shows that $d(f^{\#}(\theta^{\prime}))=f^{\#}(d(\theta^{\prime}))$ for any $\theta^{\prime}$ and that $d(f(\theta))=f(d(\theta))$ if $|f^{-1}f(\theta)|=1$ .

Assume $Z=S^{1}$ (unoriented) and consider a finite subset $M$ of $Z$ as in §7.4.3. We use the notations of that section. It follows from Lemma 7.4.19 that the isomorphism $F$ of Proposition 7.4.18 induces an isomorphism of ${\mathbf{F}}_{2}$ -linear categories $F:{\mathbf{F}}_{2}[{\mathcal{H}}_{n}]\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}_{M}(Z)$ . It follows now from Lemma 7.4.20 that this isomorphism commutes with $d$ . In particular, $d$ is a differential on ${\mathcal{S}}_{M}(Z)$ . Since this holds for any finite subset $M$ of $Z$ , we deduce that $d$ is a differential on ${\mathcal{S}}(Z)$ .

Consider now a non-singular connected $Z$ and an injective morphism of curves $f:Z\hookrightarrow S^{1}$ . Since $f$ induces a faithful ${\mathbf{F}}_{2}$ -linear functor ${\mathcal{S}}(Z)\to{\mathcal{S}}(S^{1})$ commuting with $d$ , we deduce that $d$ is a differential on ${\mathcal{S}}(Z)$ .

The decomposition (7.4.3) is compatible with $d$ , hence $d$ is a differential on ${\mathcal{S}}(Z)$ for any non-singular $Z$ .

Consider now a general $Z$ and $q:\hat{Z}\to Z$ its non-singular cover. Since the additive ${\mathbf{F}}_{2}$ -linear functor $q^{\#}$ commutes with $d$ , it follows that $d$ is a differential on ${\mathcal{S}}(Z)$ .

The last statement of the theorem follows from Lemma 7.3.17. ∎

There is an isomorphism of differential pointed categories

(7.4.4)

{\mathcal{S}}^{\bullet}(Z^{\operatorname{opp}\nolimits})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}(Z)^{\operatorname{opp}\nolimits},\ I\mapsto I,\ \theta\mapsto(\theta_{s}^{-1})_{s}.

Note that the construction $Z\mapsto\operatorname{add}\nolimits({\mathcal{S}}(Z))$ and $f\mapsto f^{\#}$ defines a contravariant functor from the category of curves with strict morphisms to the category of differential categories.

7.4.8. Strands on non-singular curves

We consider as in §7.4.3 a family $M=\{a_{1},\ldots,a_{n}\}$ of points on $S^{1}$ and $z\in S^{1}-M$ such that $a_{1},\ldots,a_{n},z$ is cyclically ordered.

The next proposition follows immediately from Proposition 7.4.18 and Lemmas 7.4.19 and 7.4.20.

0PBZ

Proposition 7.4.33. The functor $F$ induces an isomorphism of differential pointed categories ${\mathcal{H}}_{n}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}_{M}(S^{1})$ . It restricts to isomorphisms of differential pointed categories

{\mathcal{H}}_{n}^{+}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}_{M}(\dot{S}^{1}),\ {\mathcal{H}}_{n}^{++}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}_{M}(\vec{S}^{1}),\ {\mathcal{H}}_{n}^{f}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}_{M}(I)\text{ and }{\mathcal{H}}_{n}^{f++}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}_{M}(\vec{I}).

The isomorphism $\Gamma_{[1,n]^{+}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\Gamma_{M}(\vec{S^{1}})$ of Lemma 7.4.19 restricts to an isomorphism of groups $\Gamma^{f}_{[1,n]^{+}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\Gamma_{M}(\vec{I})$ and the isomorphism ${\mathcal{H}}_{n}^{f++}\to{\mathcal{S}}^{\bullet}_{M}(\vec{I})$ of Proposition 7.4.33 is compatible with the grading by those groups.

Consider $Z={\mathbf{R}}_{>0}$ as an unoriented curve. We denote by ${\mathcal{S}}_{\otimes}^{\bullet}({\mathbf{R}}_{>0})$ the full subcategory of ${\mathcal{S}}^{\bullet}({\mathbf{R}}_{>0})$ with objects the subsets of the form $\{1,\ldots,n\}$ for some $n\in{\mathbf{Z}}_{\geq 0}$ . We define a monoidal structure on the differential pointed category ${\mathcal{S}}_{\otimes}^{\bullet}({\mathbf{R}}_{>0})$ by $\{1,\ldots,n\}\otimes\{1,\ldots,m\}=\{1,\ldots,n+m\}$ and $\theta^{\prime\prime}=\theta\otimes\theta^{\prime}$ is defined by $\theta^{\prime\prime}_{i}=\theta_{i}$ if $i\leq n$ and $\theta^{\prime\prime}_{i}=\theta^{\prime}_{i-n}$ otherwise.

The next theorem follows immediately from Proposition 7.4.33.

0PC0

Theorem 7.4.34. There is an isomorphism of differential pointed monoidal categories ${\mathcal{U}}^{\bullet}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}_{\otimes}^{\bullet}({\mathbf{R}}_{>0})$ defined by $e\mapsto\{1\}$ and $\tau$ maps to the non-zero and non-identity element of $\operatorname{End}\nolimits_{{\mathcal{S}}^{\bullet}({\mathbf{R}}_{>0})}(\{1,2\})$ .

7.4.9. Products and divisibility

0PC1

Lemma 7.4.35. Consider braids $\theta^{\prime\prime}:I\to J$ and $\theta^{\prime}:J\to K$ and assume $\theta=\theta^{\prime}\cdot\theta^{\prime\prime}$ is non-zero. Let $\zeta\in D(\theta)\setminus(D(\theta)\cap D(\theta^{\prime\prime}))$ . Assume $\zeta$ and $\zeta^{-1}$ are oriented.

Define $\alpha^{\prime\prime}:I\to J$ by

\alpha^{\prime\prime}_{s}=\begin{cases}\theta^{\prime\prime}_{\zeta(1)}\circ\zeta&\text{ if }s=\zeta(0)\\ \theta^{\prime\prime}_{\zeta(0)}\circ\zeta^{-1}&\text{ if }s=\zeta(1)\\ \theta^{\prime\prime}_{s}&\text{ otherwise.}\end{cases}

Let $\zeta^{\prime}=\theta^{\prime\prime}_{\zeta(1)}\circ\zeta\circ(\theta^{\prime\prime}_{\zeta(0)})^{-1}$ and $\alpha^{\prime}=(\theta^{\prime})^{\zeta^{\prime}}$ . Then $\alpha^{\prime\prime}$ and $\alpha^{\prime}$ are braids and $\theta=\alpha^{\prime}\cdot\alpha^{\prime\prime}$ .

0PC2

Proof. Since $\zeta$ and $\zeta^{-1}$ are oriented, it follows that $\alpha^{\prime\prime}_{s}$ is oriented for all $s$ . Also, it follows from Lemma 7.4.31 that $\alpha^{\prime}$ is a braid.

Consider first the case where $Z=S^{1}$ unoriented. In that case, the lemma follows from Proposition 7.4.33 and Lemmas 7.4.20 and 6.2.10.

Assume now $Z$ is smooth and connected. There is an injective morphism of curves $f:Z\to S^{1}$ , where $S^{1}$ is unoriented. Since the lemma holds for $S^{1}$ , we deduce that it holds for $Z$ .

When $Z$ is only assumed to be smooth, the lemma follows from the case of the connected component containing $\zeta$ .

Consider now the general case. Let $f:\tilde{Z}\to Z$ be a smooth cover. Let $\tilde{\theta}$ be a braid lifting $\theta$ . There are unique braids $\tilde{\theta}^{\prime}$ and $\tilde{\theta}^{\prime\prime}$ in $\tilde{Z}$ with $\tilde{\theta}=\tilde{\theta}^{\prime}\cdot\tilde{\theta}^{\prime\prime}$ and $f(\tilde{\theta}^{\prime})=\theta^{\prime}$ , $f(\tilde{\theta}^{\prime\prime})=\theta^{\prime\prime}$ . There is a unique $\tilde{\zeta}\in D(\tilde{\theta})$ with $f(\tilde{\zeta})=\zeta$ (Lemma 7.4.28). We have $\tilde{\zeta}{\not\in}D(\tilde{\theta}^{\prime\prime})$ (Lemma 7.4.28). Since the lemma holds for $\tilde{Z}$ , we deduce it holds for $Z$ . ∎

7.4.10. Subcurves

Let $Z$ be a curve.

Let $S$ and $T$ be two finite subsets of $Z$ . Let $S_{1}$ be a subset of $S$ and $S_{2}=S\setminus S_{1}$ . Let $T_{1}$ be a subset of $T$ and $T_{2}=T\setminus T_{1}$ . Let $\Phi_{i}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{i},T_{i})$ . We define $\Phi=\Phi_{1}\boxtimes\Phi_{2}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S,T)$ by $\Phi_{s}=(\Phi_{i})_{s}$ when $s\in S_{i}$ . This gives an injective map of pointed sets

\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{1},T_{1})\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{2},T_{2})\hookrightarrow\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S,T).

Note that this is not compatible with composition in general. We obtain an isomorphism of pointed sets

\bigvee_{\begin{subarray}{c}T^{\prime}_{1}\subset T\\ |T^{\prime}_{1}|=|S_{1}|\end{subarray}}\bigl(\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{1},T^{\prime}_{1})\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{2},T\setminus T^{\prime}_{1})\bigr)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S,T).

We have corresponding morphisms of ${\mathbf{F}}_{2}$ -modules between $\operatorname{Hom}\nolimits$ -spaces in ${\mathcal{S}}(Z)$ . Note these are not compatible with the differential.

Assume $S_{2}=T_{2}$ . The map $\Phi_{1}\mapsto\Phi_{1}\boxtimes\operatorname{id}\nolimits_{S_{2}}$ defines a canonical embedding of pointed sets (not compatible with the differential nor the multiplication in general)

\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{1},T_{1})\hookrightarrow\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S_{1}\sqcup S_{2},T_{1}\sqcup S_{2}).

Given $Z_{1}$ and $Z_{2}$ two disjoint closed subcurves of $Z$ , we obtain a faithful differential pointed functor

{\mathcal{S}}^{\bullet}(Z_{1})\wedge{\mathcal{S}}^{\bullet}(Z_{2})\to{\mathcal{S}}^{\bullet}(Z),\ (S_{1},S_{2})\mapsto S_{1}\sqcup S_{2}.

Let $Z_{1},\ldots,Z_{r}$ be the connected components of $Z$ . The construction above induces an isomorphism of differential pointed categories (cf (7.4.2))

(7.4.5)

{\mathcal{S}}^{\bullet}(Z_{1})\wedge\cdots\wedge{\mathcal{S}}^{\bullet}(Z_{r})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}(Z),\ (S_{1},\ldots,S_{r})\mapsto S_{1}\sqcup\cdots\sqcup S_{r}.

Let us record a case where the tensor product construction $\boxtimes$ is compatible with composition and the differential in the following immediate lemma.

0PC3

Lemma 7.4.36. Let $M$ be a subset of $Z$ and let $Z^{\prime}$ be a subcurve of $Z$ . Assume that given an admissible homotopy class of paths $\zeta$ in $Z$ with endpoints in $M$ , there is an admissible path $\gamma$ in $\zeta$ contained in $Z-Z^{\prime}$ . There is a faithful functor of differential pointed categories

	$\displaystyle{\mathcal{S}}^{\bullet}_{M}(Z)\wedge{\mathcal{S}}^{\bullet}(Z^{\prime})$	$\displaystyle\to{\mathcal{S}}^{\bullet}_{M\cup Z^{\prime}}(Z)$
	$\displaystyle(S,T)$	$\displaystyle\mapsto S\sqcup T$
	$\displaystyle(\alpha,\beta)$	$\displaystyle\mapsto\alpha\boxtimes\beta=(\alpha\boxtimes\operatorname{id}\nolimits)\cdot(\operatorname{id}\nolimits\boxtimes\beta)=(\operatorname{id}\nolimits\boxtimes\beta)\cdot(\alpha\boxtimes\operatorname{id}\nolimits).$

7.4.11. Bordered Heegaard Floer algebras

We consider a chord diagram $({\mathcal{Z}},{\mathbf{a}})$ as in §7.2.4. Let $Z_{1},\ldots,Z_{l}$ be the connected components of ${\mathcal{Z}}$ . Let $\tilde{{\mathbf{a}}}=\bigcup_{\{z,z^{\prime}\}\in{\mathbf{a}}}\{z,z^{\prime}\}$ , $n_{i}=|\tilde{{\mathbf{a}}}\cap Z_{i}|$ and let $q:\tilde{Z}\to Z$ be the quotient map.

The isomorphism (7.4.5) associated with the decomposition $\tilde{Z}=\mathring{Z}_{1}\coprod\cdots\coprod\mathring{Z}_{l}$ together with the strands algebra description of §6.3.2 and the isomorphism of Proposition 7.4.33 induce an isomorphism of differential algebras

{\mathcal{A}}(n_{1})\otimes\cdots\otimes{\mathcal{A}}(n_{l})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{End}\nolimits_{\operatorname{add}\nolimits({\mathcal{S}}(\tilde{Z}))}(\bigoplus_{I\subset\tilde{{\mathbf{a}}}}I).

It is compatible with the gradings, via the embedding $G^{\prime}(n_{1})\times\cdots\times G^{\prime}(n_{l})\hookrightarrow\Gamma_{{\mathbf{a}}}(\tilde{Z})$ given by §6.3.2 and §7.4.8.

The differential algebra ${\mathcal{A}}({\mathcal{Z}})$ associated to ${\mathcal{Z}}$ is a differential $(G^{\prime}(n_{1})\times\cdots\times G^{\prime}(n_{l}))$ -graded non-unital subalgebra of ${\mathcal{A}}(n_{1})\otimes\cdots\otimes{\mathcal{A}}(n_{l})$ (cf [Za, Definition 2.6] and [LiOzTh1, Definition 3.23] for the original setting where $l=1$ ). There is a unique isomorphism of differential algebras

{\mathcal{A}}({\mathcal{Z}})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{End}\nolimits_{\operatorname{add}\nolimits({\mathcal{S}}(Z))}\bigl(\bigoplus_{S\subset{\mathbf{a}}}S\bigr)

making the following diagram commutative

7.4.12. Fukaya categories from strand algebras

Consider an oriented singular curve $Z$ with $n_{z}\in\{2,4\}$ for all $z$ and its corresponding chord diagram $({\mathcal{Z}},{\mathbf{a}})$ (cf §7.2.4). Let $(F,\Lambda,S^{+},S^{-})$ be the associated sutured surface. We assume that every component of $\partial F$ intersects $S^{+}$ non-trivially (cf §7.2.5). Choose for each component $E$ of $S^{-}$ a point $e_{E}\in E$ and let $S=\{e_{E}\}_{E}$ . We have obtained a pair $(F,S)$ where $S$ is a finite subset of $\partial F$ .

Consider the arcs $\omega_{z}$ for $z\in Z_{exc}$ (cf §7.2.4). Note that $F\setminus\bigl(\bigcup_{z\in Z_{exc}}\omega_{z}\bigr)$ is a union of discs, each of which contains one point of $S$ .

Auroux [Au2, Definition 8] considers a partially wrapped Fukaya category ${\mathcal{F}}(\mathrm{Sym}^{n}F,S)$ of the symmetric power $\mathrm{Sym}^{n}(F)$ of $F$ with set of stops $S\times\mathrm{Sym}^{n-1}(F)$ . This is an (ungraded) $A_{\infty}$ -category over $k$ .

Let $s,t\in Z_{exc}$ . A path in $Z$ gives rise to a path in $F$ and this defines a bijection $f$ from the set of admissible homotopy classes of paths $s\to t$ in $Z$ to the set $\bar{\chi}_{t}^{s}$ of [Au2, Proposition 11] (recall the orientation reversal, cf Convention 7.2.12). When $s=t$ , the trivial path is sent to the element $\bf{1}_{i}$ of Auroux.

Auroux [Au2, Proposition 11] relates the $A_{\infty}$ -category ${\mathcal{F}}(\mathrm{Sym}^{n}F,S)$ to the strand algebra associated with $Z$ .

0PC4

Theorem 7.4.37 (Auroux). There is a fully faithful $A_{\infty}$ -functor

\Phi:{\mathcal{A}}(Z,n)\to{\mathcal{F}}(\mathrm{Sym}^{n}F,S),\ I\mapsto\prod_{i\in I}\omega_{i},\ \theta\mapsto(\chi(\theta),(f(\theta_{s}))_{s})

inducing an equivalence of derived categories.

7.4. Strands