TITLE: Minimal Degree of map $S^2\times S^2\mapsto \mathbb{CP}^2$
QUESTION [8 upvotes]: I am having troubles finding the minimal d such there is a map of such degree from $S^2\times S^2\mapsto \mathbb P^2$.
I know that the cohomology ring of $\mathbb P^2$ is thus I know that the degree will be given by $f^*(x^2)=(f^*(x))^2$ which will be the square of a class in $H^2(S^2\times S^2)$.
Thus $f^*(x)^2=(ha_1+ka_2)^2=2hk a_1a_2$ where $a_1$ and $a_2$ are the classes corresponding to $S^2\times pt$ and $pt\times S^2$.
How can I finish from here?
Thanks
Edit: It is complex projective space.

REPLY [2 votes]: $\mathbb{C}P^n$ 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.