"the standard two-fold branched cover of $CP^2$" What could the following sentence mean: > $\iota : S^2\times S^2 \rightarrow \mathbb{C}P^2$ is the standard two-fold branched cover, branched along the diagonal. What I can think of is to consider $S^2 \times S^2$ as embedded in $\mathbb{R}^6$, then identify $\mathbb{R}^6$ with $\mathbb{C}^3$ and then take the standard projection. But then which identification makes sense? Since the map is branched along the diagonal, I probably want the permutation map $(X_1,X_2) \mapsto (X_2,X_1) \in S^2 \times S^2$ to be a covering transformation. The identification $(x_1,y_1,z_1,x_2,y_2,z_2) \mapsto (x_1+ix_2,y_1+iy_2,z_1+iz_2)$ doesn't seem to fit this description. I feel like the solution is something entirely different.

Identify $S^2 \cong \mathbb{C}P^1$. Then act on $S^2 \times \ldots \times S^2$ by $S_n$. The quotient space is $\mathbb{C}P^n$ and the projection map is the branched cover in question. In my case n=2 and the diagonal sphere is the the fixed set under the involution of $S_2$, so it is the branching locus.