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?
[Math] Understanding the cartesian product of complex projective lines.
algebraic-topologygeneral-topologyhomology-cohomology
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Many thanks to Steve D, user17786, and Dave Hartman for their helpful corrections.
First, I put a cell structure on the twice-punctured disk with 3 0-cells, 5 1-cells, and 1 2-cell:
Note that the boundary of the 2-cell $D$ is $$d_2D=\alpha+\beta+\gamma-\beta+\delta+\epsilon-\delta=\alpha+\gamma+\epsilon,$$ and that the boundaries of the 1-cells are $$\begin{align} d_1\alpha&=0\\ d_1\beta&=y-x\\ d_1\gamma&=0\\ d_1\delta&=z-x\\ d_1\epsilon&=0 \end{align} $$ Now, we identify $y$ with $z$, and $\gamma$ with $\epsilon$, to produce a cell structure on $X$:
For $X$, the chain groups are $$\begin{align} C_0(X)&=\langle x,y\rangle\\ C_1(X)&=\langle \alpha,\beta,\gamma,\delta\rangle\\ C_2(X)&=\langle D\rangle \end{align}$$ where $D$ is our 2-cell, and we have $$\begin{align} d_1\alpha&=0\\ d_1\beta&=y-x\\ d_1\gamma&=0\\ d_1\delta&=y-x \end{align} $$
$$d_2D=\alpha+\beta+\gamma-\beta+\delta+\gamma-\delta=\alpha+2\gamma.$$
Thus, $$H_0(X)=\ker(d_0)/\mathrm{im}(d_1)=\langle x,y\rangle/\langle y-x\rangle=\left\langle\overline{x}\right\rangle\cong\mathbb{Z}$$ $$H_1(X)=\ker(d_1)/\mathrm{im}(d_2)=\langle \alpha,\gamma,\beta-\delta\rangle/\langle \alpha+2\gamma\rangle=\left\langle\overline{\gamma},\overline{\beta-\delta}\right\rangle\cong\mathbb{Z}^2$$ $$H_2(X)=\ker(d_2)/\mathrm{im}(d_3)=0/0\cong 0.$$
You're quite right. We need not restrict ourselves to metric spaces or countable or finite products:
Let $A_i, B_i, i \in I$ be a family of topological spaces such that $h_i: A_i \rightarrow B_i$ is a homeomorphism for every $i$. Then $h: A = \prod_{i \in I} A_i \rightarrow B = \prod_{i \in I} B_i$ defined by $h( (x_i) ) = h(x_i)_i $ is a homeomorphism as well.
The fact that $h$ is a bijection is simple set theory. Any map $f: X \rightarrow \prod_{i \in I} X_i$ is continuous iff for every $i$, $\pi_i \circ f$ is continuous between $X$ and $X_i$, where $\pi_i$ is the projection onto the $i$'th coordinate. This is the universal property for the product topology.
From the universal property $h$ is continuous as by construction $\pi_i \circ h = h_i \circ \pi_i$, which is continuous, as a composition of continuous functions. The product of the inverses of the $h_i$ is the required continuous inverse, using the universal property for the $\prod_i A_i$ instead.
Best Answer
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.