$c_0$) around the perimeter. We denote this action by $$\rho^{S^1}: S^1 \times \Cact(n) \rightarrow \Cact(n)$$ With this action we can define the twisted gluing $$\begin{aligned}
\label{circtheta}
\circ_i^{S^1}:\Cact(n) \times S^1(n) \times \Cact(m) &\rightarrow& \Cact(n+m-1)\nn\\
(C,\theta,C')&\mapsto& C \circ \rho^{S^1}(\theta_i,C') =: C
\circ_i^{\theta_i}C'\end{aligned}$$

Given a cactus without spines $C\in \Cact(n)$ the orientation reversed perimeter (i.e. going around the outer circle [*clockwise*]{} i.e. reversing the orientation of the source of $\phi_0$) gives a map $\Delta_C: S^1 \rightarrow (S^1)^n$.

As one goes around the perimeter the map goes around each circle once and thus the map $\Delta_C$ is homotopic to the diagonal $ \Delta_C (S^1) \sim \Delta(S^1)$.

We can use the map $\Delta_C$ to give an action of $S^1$ and $(S^1)^{\times n}$. $$\rho^C: S^1 \times(S^1)^{\times n}\stackrel{\Delta_C}
{\rightarrow} (S^1)^{\times n} \times (S^1)^{\times n}
\stackrel{\mu^n}{\rightarrow}(S^1)^{\times n}$$ here $\mu_n$ is the diagonal multiplication in $(S^1)^{\times n}$ and $\bar \circ_i$ is the operation which forgets the $i$-th factor and shuffles the last $m$ factors to the $i$-th, …, $i+m-1$st places. Set $$\begin{gathered}
\label{perturbdef} \circ_i^C:(S^1)^{\times n} \times (S^1)^{\times
m} \stackrel{(id \times \pi_i)(\Delta) \times id}
{\longrightarrow} (S^1)^{\times n} \times
S^1\times (S^1)^{\times m}\\
\stackrel{id \times \rho^C}{\longrightarrow} (S^1)^{\times n}
\times (S^1)^{\times m}
\stackrel{\bar\circ_i}{\longrightarrow}(S^1)^{\times n+m-1}\end{gathered}$$ These maps are to be understood as perturbations of 