Minimal Degree of map $S^2\times S^2\mapsto \mathbb{CP}^2$

$\mathbb{C}P^n$ is the $n$th symmetric product of $\mathbb{C}P^1$.

The map from the $n$th product of $\mathbb{C}P^1$ to $\mathbb{C}P^n$, which is the quotient map by the symmetric group, has $n!$ pre-images. All pre-images have degree $1$, since permutations are holomorphic maps, and holomorphic maps are always orientation preserving. So the quotient map has degree $n!$.

In this case, $\mathbb{C}P^1 \times \mathbb{C}P^1$ to $\mathbb{C}P^2$ by quotient map has degree $2$.

To see why $\mathbb{C}P^n$ is the nth symmetric product of $\mathbb{C}P^1$, note that a point in $\mathbb{C}P^1$ is the unique solution of a linear homogenous polynomial. e.g. $[a:b]$ is the solution to the homogenous polynomial $bx - ay$. Given $n$ points in $\mathbb{C}P^1$, multiply the corresponding $n$ linear homogenous polynomials together to get a degree $n$ homogenous polynomial. The $n+1$ coefficients of this polynomial become a single point in $\mathbb{C}P^n$, since global multiplication of a homoegenous polynomial doesn't change its roots. Finally, each degree $n$ homogenous polynomial has $n$ unordered roots, which is where the action of the symmetric group comes from.