# Weighted projective plane as a quotient of $\Bbb CP^2$

For positive integers $$a_0,\dots,a_n$$, consider the weighted projective space $$\Bbb C\Bbb P(a_0,a_1,\dots,a_n)$$, which is the quotient of $$\Bbb C^{n+1}-\{0\}$$ by the action of $$\Bbb C^*=\Bbb C-\{0\}$$ defined by $$t\cdot (z_0,\dots,z_n)=(t^{a_0}z_0,\dots,t^{a_n}z_n)$$.

According to https://en.wikipedia.org/wiki/Weighted_projective_space, the weighted projective space $$\Bbb C\Bbb P(a_0,a_1,\dots,a_n)$$ is isomorphic to the quotient of projective space $$\Bbb {CP}^n$$ by the group that is the product of the groups of roots of unity of orders $$a_0,a_1,\dots,a_n$$ acting diagonally. I am trying to verify this in the simple case $$n=2$$. For $$m\geq1$$ let $$\mu_m$$ be the group of roots of unity of order $$m$$, and for $$a,b,c\geq 1$$ let $$G=\mu_a\times \mu_b\times \mu_c$$. There is a well-defined diagonal action of $$G$$ on $$\Bbb {CP}^2$$. I have to show that $$\Bbb {CP}^2/G=\Bbb {CP}(a,b,c)$$. Maybe one way to do this is to define a map $$\Bbb {CP}^2\to \Bbb {CP}(a,b,c)$$ and descend to the quotient. But the map $$\Bbb {CP}^2\to \Bbb {CP}(a,b,c)$$, $$[x,y,z]\to [x,y,z]$$ is not well-defined and I got stuck. Any hints?

Suppose $$\rho_a$$ is a primtive $$a$$-th root of unity, and likeiwse for $$\rho_b$$ and $$\rho_c$$. Define an action of $$\mu_a\times \mu_b\times \mu_c$$ on $$\mathbb{C}P^2$$ as follows. For an element $$(x,y,z)\in \mu_a\times \mu_b\times \mu_c$$ and an element $$[z_0:z_1:z_2]\in \mathbb{C}P^2$$, we set $$(x,y,z)\ast [z_0:z_1:z_2] = [\rho_a^x z_0: \rho_b^y z_1: \rho_c^z z_2].$$
Define a map $$\mathbb{C}P^2\rightarrow X:=\mathbb{C}P(a,b,c)$$ by $$f([z_0:z_1:z_2])\rightarrow [z_0^a: z_1^b: z_2^c]$$.
I claim this is well defined. Indeed, for $$\lambda \in \mathbb{C}\setminus\{0\}$$, we have $$f([\lambda z_1: \lambda z_2:\lambda z_3]) = [\lambda^a z_0^a: \lambda^b z_1^b: \lambda^c z_2^c] = [z_0^a:z_1^b:z_2^c] = f([z_0:z_1:z_2]).$$
Moreover, it's not too hard to see that $$f$$ is invariant under the action of $$\mu_a\times \mu_b\times \mu_c$$, so we get an induced map $$\mathbb{C}P^2/(\mu_a\times \mu_b\times \mu_c)\rightarrow \mathbb{C}P(a,b,c)$$. Can you prove that this map is a homeomorphism?