(Floer products involving the identity, continued) In the context of Example 6.1, the orientation isomorphism (69) for the holomorphic triangle is determined by whether the isotopy of the Lagrangian boundary conditions respects the relative spin structure. Since the relative spin structure on $\phi_{\epsilon H}(L)$ is induced from $L$, this problem is equivalent to the corresponding isotopy problem for the limiting holomophic strip. The holonomy factor of the local systems for the holomorphic triangle, is also reduced to that of the limiting strip.

In the simplest case when we are given closed elements $\alpha\in CF^{0}(L,L^{\prime}),\beta\in CF^{0}(L^{\prime},L)$ each involving only one intersection point, the local systems are trivial, and only one holomorphic curve contributes to the Floer product, then $\beta\circ\alpha=1_{L}\in HF^{0}(L,L)$ means that for the holomorphic strip from $\alpha$ to $\beta$ passing through a generically chosen point $r\in L$, the Lagrangian boundary condition on the disk obtained by gluing $TL,TL^{\prime}$ and the two Lagrangian paths at the two strip like ends, can be contracted to constant, respecting the prescribed relative spin structures on $L,L^{\prime}$ and the two ends. More generally, many intersections points and holomorphic strips may contribute to the Floer product, and $\beta\circ\alpha=1_{L}\in HF^{0}(L,L)$ means a weighted signed count of holomophic strips is equal to one.

Under sufficient transversality assumptions, we can form the $(n-1)$ dimensional moduli space of holomorphic strips from $\alpha$ to $\beta$, and thereby produce an $(n+1)$-dimensional universal family $\mathcal{C}$, as in the main text section 3.1. Using the relative spin structures on $L,L^{\prime}$ and the Lagrangian paths associated with the ends, we use (69)(70) and Remark 6.9 to induce a canonical orientation on the moduli space from $\beta\otimes\alpha\in CF^{0}(L^{\prime},L)\otimes CF^{0}(L,L^{\prime})$. Using the complex orientation on the holomorphic curve $\Sigma$, and inserting an extra minus sign, we obtain an orientation on $\mathcal{C}$. This tricky minus sign accounts for the difference between the counterclockwise orientation of $\partial\Sigma$ compatible with the complex orientation, and the clockwise orientation of $\partial\Sigma$ compatible on the $L$-boundary with the translation vector field $-\partial_{s}$. Putting everything together, $\beta\circ\alpha=1_{L}\in HF^{0}(L,L)$ means in the sense of weighted counts, that $\partial\mathcal{C}$ passes once through a generic point $r\in L$ in the same orientation as $\lambda(TL)$. In other words, the $L$-boundary evaluation of $\partial\mathcal{C}$ sweeps out the oriented cycle $L$.

The same argument says that if $\alpha\circ\beta=1_{L^{\prime}}\in HF^{0}(L^{\prime},L^{\prime})$, then the moduli space of holomorphic strips from $\beta$ to $\alpha$ produces a universal family $\mathcal{C}^{\prime}$, whose $L^{\prime}$-boundary evaluation map sweeps out the oriented cycle $L^{\prime}$. The subtle point is that due to the reversal of the $\mathbb{R}$-translation vector fields, $\mathcal{C}^{\prime}$ has the reverse orientation as $\mathcal{C}$. Therefore, the $L^{\prime}$-boundary evaluation of $\partial\mathcal{C}$ sweeps out the oriented cycle $-L^{\prime}$ instead of $L^{\prime}$. Here ends the example.