Consider the $\mathbb Z_2$-action $g:\mathbb C^2\to \mathbb C^2, z\mapsto-z$ on $\mathbb C^2$ and its quotient $X:=\mathbb C^2/{\mathbb Z_2}$.
This is a singular surface with singular point the image of $0$ (there is a bijection to the cone $\mathbb V(x^2-yz)\subset \mathbb C^3$).
I was told that the singularity could be removed by blowing up $\mathbb C^2$ at $0$ and then taking the quotient $Y:=Bl_0\mathbb C^2/{\mathbb Z_2}$.
We have $Bl_0\mathbb C^2=\mathcal O_{\mathbb CP^1}(-1)\subset \mathbb C^2\times \mathbb CP^1$ and the $\mathbb Z_2$-action on $\mathbb C^2$ extends to a $\mathbb Z_2$-action on $Bl_0\mathbb C^2$:
If $z\neq 0\in \mathbb C^2$, then $(z,[z])\mapsto (-z,[-z])=(-z,[z])$, where $[z]$ denotes the line through $z$.
If $z=0$, then $(0,l)\mapsto (0,l)$.
But why is the quotient $Y$ a manifold?
The idea would be to calculate charts for $Y$. We already have charts for $Bl_0\mathbb C^2$ given by the local trivializations of the line bundle $\mathcal O_{\mathbb CP^1}(-1)$. So the charts are of the form $U\times \mathbb C$ and the $\mathbb Z_2$ action fixes $U$ and flips $\mathbb C$.
How do I proceed from here?
Also: Is it true that first blowing up and then taking the quotient is the same as first taking the quotient and then blowing up?