# Cremona Transformation and isomorphism

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?

• 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. Apr 16, 2016 at 17:36
• 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 ;)
– J.L
Apr 16, 2016 at 20:39
• 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. Apr 16, 2016 at 21:10
• 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...
– J.L
Apr 17, 2016 at 9:19