Understanding the cartesian product of complex projective lines. I am trying to understand the space obtained by taking the cartesian product $\mathbb{C}\mathbb{P}^1\times \mathbb{C}\mathbb{P}^1$ and identifying some of its points by the rule $(x,y)\sim (y,x)$. Viewing $\mathbb{C}\mathbb{P}^1$ as a CW complex with one 0-cell and one 2-cell I computed the homology of $\mathbb{C}\mathbb{P}^1\times \mathbb{C}\mathbb{P}^1/\sim$ which matches that of $\mathbb{C}\mathbb{P}^2$ but I can't seem to visualize an "obvious" homeomorphism between the two spaces. My question is the following: is $\mathbb{C}\mathbb{P}^1\times \mathbb{C}\mathbb{P}^1/\sim$ homeomorphic to $\mathbb{C}\mathbb{P}^2$ and, if so, how?
 A: It turns out that even more is true: the $n-$fold symmetric product of $\mathbb{C}\mathbb{P}^1$ is homeomorphic to $\mathbb{C}\mathbb{P}^n$!
To see this in the $2-$fold case: consider homogeneous polynomials of degree two $\mathbb{C}[x,y]^{(2)}$ whose elements are of the form $ax^2+bxy+cy^2$ and notice that for $\lambda\in\mathbb{C}^\times$, $$\lambda[ax_0^2+bx_0y_0+cy_0^2]=0\iff ax_0^2+bx_0y_0+cy_0^2=0.$$ This allows us to identify points of $\mathbb{C}\mathbb{P}^2$ with elements of $\mathbb{C}[x,y]^{(2)}/\sim$, where $\sim$ identifies polynomials having the the same roots. The map from $\mathbb{C}\mathbb{P}^2$ to the symmetric product of two copies of $\mathbb{C}\mathbb{P}^1$ is then given by
$$(a:b:c)\mapsto ax^2+bxy+cy^2=(\alpha x+\beta y)(\alpha'x+\beta'y)\mapsto [(\alpha:\beta),(\alpha':\beta')]$$ where the equality comes from the fundamental theorem of algebra.
A: I believe you are on the right track, and a homeomorphism from $\mathbb{CP}^1 \times \mathbb{CP}^1/\sim$ to $\mathbb{CP}^2$ is given by $$[((z_1:z_2),(w_1:w_2))] \mapsto (z_1 w_1: z_2 w_2: z_1 w_2 + z_2 w_1)$$
Note that elements of the form $[(1:z),(1:w)]$ map to $(1:zw:z+w)$, i.e., the coordinates are given by the elementary symmetric functions of $z$ and $w$, so the map is a homeomorphism restricted to this subspace onto the subspace of $\mathbb{CP}^2$ given by points with non-zero first coordinate. I have not worked out all the details, but I am pretty sure that this argument can be promoted to show that the map is actually a homeomorphism between your spaces.
