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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.

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  • $\begingroup$ $f$ 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. $\endgroup$ – Tabes Bridges Aug 11 at 22:09
  • $\begingroup$ Yes, that's true. Am I correct for $f^{-1}$? $\endgroup$ – cip Aug 12 at 8:07

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