I'm reading Osváth-Szabó's notes on Heegard-Floer homology, in particular about the surgery exact triangle.

On page 14 (numbered 42 on the document), they describe an isomorphism between the space of homology classes of Whitney triangles $\pi_2(x,y,z)$ in $\mbox{sym}^g(\Sigma)$ and $\mathbb{Z} \times \mathcal{P} $, where $\mathcal{P}$ denotes the group of periodic domains in $\Sigma$.

I'm not sure if I understand this isomorphism correctly, or if there are some typos, or both. Here is how I understand the isomorphism works:

Given two elements $\psi, \psi_0 \in \pi_2(x,y,z)$, we can associate domains $\mathcal{D}(\psi), \mathcal{D}(\psi_0)$, by taking $$\mathcal{D}(\psi) = \sum n_{z_i}(\psi) D_i,$$

where $D_i$ are the components of $\Sigma - \{x \cup y \cup z\}$, and $z_i \in D_i$; and $n_{z_i}(\psi)$ is the algebraic intersection of $\psi$ with $z_i \times \mbox{sym}^{g-1}(\Sigma)$.

It follows that if $n_z(\psi) = n_z(\psi_0)$, the domain $E = \mathcal{D}(\psi) - \mathcal{D}(\psi_0)$ is periodic: ie it satisfies $n_z(E) = 0$ (here, $n_z(E)$ denotes the coefficient of the component of $\Sigma - \{x \cup y \cup z\}$ containing $z$ in $E$).

So all we have to do to define the isomorphism is pick a $\psi_0$ such that $n_z(\psi_0) = 1$: then we can subtract off $n \mathcal{D}(\psi_0)$ from $\mathcal{D}(\psi)$ to get a periodic domain. Given a fixed $\psi_0$, the isomorphism is then given by

$$\psi \mapsto (n, \mathcal{D}(\psi) - n\mathcal{D}(\psi_0)).$$

The inverse of the isomorphism should then be given by sending $(n, E)$ to something like $n \psi_0 + \phi$, where $\mathcal{D}(\phi) = E$; but this seems dependent on choice and I'm not entirely sure how to make sense of $n$ times a holomorphic triangle.

Is this right? How do I fix the last part?

Thanks very much.