=  \delta_0\circ\tau_n$,

-   $\sigma_i = F({\mathfrak{a}}^{n+1,1}_{i+1,1,n-i})\circ F(\Id^{\ot i+1}\ot u\ot \Id^{\ot n-i})\circ F({\mathfrak{a}}_{n,1}^{i+1,0,n-i})$, for $0 \leq i \leq n$.

Note that for $0 \leq i \leq n-1$ we have $$\delta_i=F(\delta_i^{(n)})$$ and for $0 \leq i \leq n-2$ we have $$\delta_i=F(\delta_i^{(n-1)}\ot Id).$$ Similarly, for $0 \leq i \leq n$ we have $$\sigma_i=F(\sigma_i^{(n)})=F({\mathfrak{a}}_{1,n+1}^{n+1,1})F(Id\ot \sigma_{i-1}^{(n-1)})F({\mathfrak{a}}_{n,1}^{1,n}),$$ and observe that $\sigma_{-1}$ makes sense and is useful. These observations become relevant in the following Proposition.

\[trace1\] For any $A\in Alg({\mathcal{C}})$ and any symmetric 2-trace $F: {\mathcal{C}}\to {\text{Vec}}$ we have a cyclic object $C_{\bullet}(A)=F(A^{\ot \bullet +1})$ in ${\text{Vec}}$.

To see the simplicial relations we apply the functor $F$ to the simplicial relations that are classically satisfied by $\delta_i^{(n)}$’s and $\sigma_i^{(n)}$’s, with the exception of those involving the special $\delta_n$. One can check that the latter all follow formally from the former simplicity relations and the cyclicity relations below.

Here we check the cyclicity relations. First for any $1\leq i \leq n$, we show that $\delta_{i}\tau_n= \tau_{n-1} \delta_{i-1}$. We begin with the case $1\leq i \leq n-1$: $$\begin{aligned}
  \tau_{n-1} \delta_{i-1} &= F({\mathfrak{a}}_{1,n-1}^{n-1,1})\iota_A(A^{\ot n-1})F(\delta_{i-1}^{(n-1)}\ot Id)\\
   &= F({\mathfrak{a}}_{1,n-1}^{n-1,1}) F(Id\ot \delta_{i-1}^{(n-1)})\iota_A(A^{\ot n})\\
   &= F(\delta_i^{(n)})F({\mathfrak{a}}_{1,n}^{n,1})\iota_A(A^{\ot n})\\
   &= \delta_i \tau_n.\end{aligned}$$ For $i=n$ we observe that $\delta_n\tau_n=\tau_{n-1}\delta_{n-1}$ iff $\delta_0\tau^2_n=\tau_{n-1}\delta_{n-1}$, since $\delta_n=\delta_0\tau_n$ by definition, and so: 