About a 5-fold cyclic branched cover of $\Bbb CP^1$ along 3 points Suppose $F$ is a smooth connected complex curve, and we have a cyclic 5-fold branched cover $\pi:F\to \Bbb CP^1$ along 3 points. Let $\gamma$ denote the corresponding $\Bbb Z_5$-action on $F$. We have $\chi(F)=5(\chi(\Bbb CP^1)-3)+3=-2$, so the genus of $F$ must be $2$. Let $D$ be the diagonal in $\Bbb CP^1\times \Bbb CP^1$ and let $D'=(\pi\times \pi)^{-1}(D)\subset F\times F$. Then we can decompose $D'$ into union of $5$-curves: $D_i:=\{(x,y)\in D': y=\gamma^i x\}$ $(i=1,\dots,5)$. On the other hand, note that $H_2(\Bbb CP^1\times \Bbb CP^1;\Bbb Z)=\Bbb Z^2$ with basis $\alpha:=[\Bbb CP^1\times x], \beta:=[y\times \Bbb CP^1]$. Since $[D]\cdot \alpha=1=[D]\cdot \beta
$, we see that $[D]=\alpha+\beta \in H_2(\Bbb CP^1\times \Bbb CP^1)$.
In this situation, how can we show that

*

*$[D']=5([F\times x]+[y\times F]) \in H_2(F\times F;\Bbb Z)$, and that

*$[D_i]^2= [D_j]^2$ (self-intersection) for $i,j=1,\dots,5$?

 A: Throughout I use Poincare duality without much explanation.
1.
The important fact here is that the map $\pi: F \rightarrow \mathbb{CP}^1$ is degree $5$. I.e the map $\pi^*: H^{2}(\mathbb{CP}^1,\mathbb{Z}) \rightarrow  H^{2}(F,\mathbb{Z})$ is $x \mapsto 5x$, with respect to the positive generators of these groups (both are $\mathbb{Z}$). The degree can be computed by taking the number of preimages of a generic point which is 5, which can be made rigorous with a little bit of work.
Then it follows that $(\pi \times \pi)^*: H^{2}(\mathbb{CP}^1  \times \mathbb{CP}^1 , \mathbb{Z}) \rightarrow H^{2}(F \times F , \mathbb{Z}) $ is the map $(x,y) \mapsto (5x,5y)$ in the obvious bases (i.e. the Poincare dual of the two diagonal factors). To prove this just restrict the map to the Poincare dual of the two factors.  Now 1. can be proved using the fact that the class of the diagonal in $H^{2}(\mathbb{CP}^1 \times \mathbb{CP}^1,\mathbb{Z})$ is $(1,1)$ (i.e. $\alpha + \beta$ in your notation, I am taking these as an integral basis of $H^2$).
2.
For this, note that $D'$ consists precisely of points $(x,y)$ such that $\pi(x) = \pi(y)$. In other words, this may be restated as: $D'$ consists of exactly the points $(x,y)$ such that $\exists g \in \mathbb{Z}_5 $ such that $gx=y$.
It is not hard to see that for any $g$, the subset $U_{g} = \{(x,y): gx=y\} \subset F \times F$ is a holomorphic curve, isomorphic to $F$. To see this note that the map $F \rightarrow U_g$ , $p \mapsto (p,gp)$ is an isomorphism of Riemann surfaces.
Moreover, there is a global biholomorphism of $F \times F$ taking $U_{g}$ to $U_{g'}$ for any $g,g'$, namely the map $(p,q) \mapsto (p, g' g^{-1} q)$. So in particular these holomorphic curves have the same self-intersection.
