# Proving that a map is birational of $\mathbb{P}^2$ to itself

Prove that the map $$y_0 = x_1x_2$$, $$y_1= x_0x_2$$, $$y_2 = x_0x_1$$ defines a birational map of $$\mathbb{P}^2$$ to itself. At which points are $$f$$ and $$f^{-1}$$ not regular? What are the open sets mapped isomorphically by $$f$$?

My attempt I tried to invert the equations and express $$x_i$$ in terms of $$y_i$$. So I got $$y_1/y_0 = x_0/x_1$$ and $$y_2/y_1 = x_1/x_2$$ and it looks like $$f$$ is always regular, while $$f^{-1}$$ is not regular for $$x_1 =0$$ and $$x_2 =0$$. Is this correct? As for the last question, I don’t really know how to address it: could the open sets be $$x_1 \neq 0$$ and $$x_2 \neq 0$$? I am sorry if I made any gross mistakes, but I am an absolute beginner at the subject. Thank you.

• $f$ is regular except at 3 points, $(0:0:1), (0:1:0)$, and $(1:0:0)$ since all three $y_i$ vanish at those points. – Tabes Bridges Aug 11 at 22:09
• Yes, that's true. Am I correct for $f^{-1}$? – cip Aug 12 at 8:07