$\newcommand{\card}[1]{\lvert{#1}\rvert}$Introduction and outline
I define $\mathbb{P}^{n}$ as the quotient of $\mathbb{C}^{n+1}\setminus\{(0,0,\ldots,0)\}$ by the scaling action of $\mathbb{C}^{\times}=\mathbb{C}\setminus\{0\},$ and give $\mathbb{P}^{n}$ the quotient topology induced by the canonical map $\pi\colon\mathbb{C}^{n+1}\setminus\{(0,0,\ldots,0)\}\to\mathbb{P}^{n}.$ I call this the strong topology on $\mathbb{P}^{n}.$ I am only interested in the complex case and do not care about projective spaces over other fields.
Consider the unit sphere in $\mathbb{C}^{n+1},$ and quotient that by the scaling action of $S^{1},$ the group of complex numbers of modulus $1$. That is, the group $S^{1}$ acts on $$S^{2n+1}=\{(a_{0},\ldots,a_{n})\in\mathbb{C}^{n+1}:\card{a_{0}}^{2}+\cdots+\card{a_{n}}^{2}=1\};$$ we look at $\Sigma^{n}=S^{2n+1}/S^{1},$ and give it the quotient topology induced by the canonical map $q\colon S^{2n+1}\to\Sigma^{n}.$ I want to prove that this is homeomorphic to $\mathbb{P}^{n}$ with the above definition.
My difficulty boils down to proving that a certain map (which I expect to be a homeomorphism) is open.
Method
The homeomorphism should be given by $\varphi\colon\Sigma^{n}\to\mathbb{P}^{n},$ defined so that $\varphi\circ q=\pi{\vert_{S^{2n+1}}}.$ Indeed, this is the method suggested by Mumford in Algebraic Geometry I, and is the method given here on nLab. I can prove that $\varphi$ is bijective (I just write down the inverse) and that it is continuous (this amounts to diagram chasing). Again, my difficulty is in seeing that the map $\varphi$ is open.
Details
Let $U\subseteq\Sigma^{n}$ be an open set. This means precisely that $q^{-1}(U)$ is open in $S^{2n+1},$ and we want to prove that $\varphi(U)=f(q^{-1}(U))=\pi(q^{-1}(U))$ is open in $\mathbb{P}^{n}.$ By the definition of the strong topology, this is true if and only if $\pi^{-1}(\pi(q^{-1}(U)))$ is open in $\mathbb{C}^{n+1}\setminus\{(0,0,\ldots,0)\}.$
Problem: How to prove that $\pi^{-1}(\pi(q^{-1}(U)))$ is open in $\mathbb{C}^{n+1}?$
On nLab, this final step is dismissed as being "clear" because a certain union of orbits is a "cylinder". My problems with this are as follows:
- I can't see how the mathematics outlined at the "cylinder" link relate to what is being discussed (and moreover, I don't see how some abstract notion of topological cylinder is helpful in proving that a specific subset of a Euclidean space is open);
- if I imagine the situation in real space, for example in $\mathbb{R}^{2},$ then I have a very clear picture: I have some segment of the circle (the "sphere" $S^{2\cdot0+1}$), and the set I want to show is open is then the unions of all the lines through the origin and those points, minus the origin itself. This is clearly open in this concrete picture, but I wouldn't want to try to prove it from scratch!
Consider $g:\mathbb{C}^{n+1}-\{0\}\rightarrow S^{2n+1}$ defined by $g(x)={x\over{\|x\|}}$,$\pi^{-1}(\pi(q^{-1}(U))=g^{-1}(q^{-1}(U))$.
To show that $\mathbb{P}^n$ is compact, it is enough to remark that $q$ is continuous since the image of a compact set by a continuous map is compact.