On Ozsváth and Szabó's branched covering description of holomorphic disks in symmetric products

Question

On page 25 of Holomorphic Disks and Topological Invariants for 3-manifolds (https://arxiv.org/pdf/math/0101206.pdf), the following lemma appears.

Given any holomorphic disk $u \in M(x,y)$, there is a g-fold branched covering space $p: \hat{\mathbb{D}} \rightarrow \mathbb{D}$ and a holomorphic map $\hat{u}: \hat{\mathbb{D}} \rightarrow \Sigma$, with the property that for each $z \in \mathbb{D}$, $u(z) = \hat{u}(p^{-1}(z))$.

What am I missing here? Obviously $u: \mathbb{D} \rightarrow \text{Sym}_g(\Sigma)$, and $\hat{u}: \hat{\mathbb{D}} \rightarrow \Sigma$, have different codomains. The only way I know to fit $\Sigma$ in $\text{Sym}_g(\Sigma)$ is via the diagonal embedding, but in the proof of the lemma, they exclude the case that the image of $u$ is contained in the diagonal.

What's really going on with the statement that $u(z)$ and $\hat{u}(p^{-1}(z))$ agree?

Answer 1

They really mean to evaluate $\hat u$ on the $g$ points (with multiplicity) in $p^{-1}(z)$, so $u(z)=[\hat u(z_1),\ldots,\hat u(z_g)]$ where $p^{-1}(z)=\lbrace z_1,\ldots,z_g\rbrace$ (with possible repetitions) and the ordering doesn't matter since we passed to the quotient $\Sigma^{\times g}\to Sym^g(\Sigma)$. In particular, wherever $p$ is branched $u$ hits the big diagonal.