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Let $P_1=(1:0:0),P_2=(0:1:0),P_3=(0:0:1) \in \mathbb{P}_2$ (over an algebraic closed field). Denote $U=\mathbb{P}^2 \setminus \{P_1,P_2,P_3 \}$ and consider the map

$$ f:U \rightarrow \mathbb{P}^2, (a_0:a_1:a_2) \mapsto (a_1a_2:a_0a_2:a_0a_1) $$

Let $\tilde{\mathbb{P}}^2$ be the bow up at $\{P_1,P_2,P_3 \}$.

According to the Andreas Gathmann notes, there is an isomorphism $\tilde{f}:\tilde{\mathbb{P}}^2 \rightarrow \tilde{\mathbb{P}}^2$ that extends $f$.

Can someone explain me, or give me a hint on how can we proove that?

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  • $\begingroup$ Look at the closure of the graph of $f$ in $\mathbb{P}^2\times\mathbb{P}^2$ and show that the projections are both blowing up the 3 points. $\endgroup$
    – Mohan
    Apr 16, 2016 at 17:36
  • $\begingroup$ Sorry, I am fairly new in algebraic geometry... What do you mean by "the brojections are both blwoing up the 3 points"? Btw, thank you ;) $\endgroup$
    – J.L
    Apr 16, 2016 at 20:39
  • $\begingroup$ You used the word `blowing-up'. What exactly do you mean by that? I assume you have no problems with projections to the two factors. $\endgroup$
    – Mohan
    Apr 16, 2016 at 21:10
  • $\begingroup$ The blow up of $\mathbb{P}^2$ at $P_1,P_2,P_3$ in this case is precisely the closure of $f$ in $\mathbb{P}^2 \times \mathbb{P}^2$. What I dont get is what does it means the projections are blowing up the 3 points... $\endgroup$
    – J.L
    Apr 17, 2016 at 9:19

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