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The following is a very elementary question but I can't find the error:

Denote $D$ the diagonal $\{[x_0:x_1],[x_0:x_1]\} \subset \mathbb P^1$x $\mathbb P^1$. Let $\phi: \mathbb P^1 \longrightarrow \mathbb P^2$ defined by $\phi([t_0:t_1])=(t_0^2:2t_0 t_1:t_1^2)$ and let P be its image. Similar to the Veronese map, this is an isomorphism and we have $P = \{[u_0: u_1: u_2] | u_0 u_2 = \frac {u_1^2} 4\}.$

Let $i$ be the natural isomorphism $\mathbb P^1 \longrightarrow D \subset\mathbb P^2$.

Let $\psi:\mathbb P^1$ x $\mathbb P^1 \longrightarrow \mathbb P^2$ be defined by $\psi([x_0:x_1],[y_0:y_1])= [x_0 y_0: x_0 y_1 + x_1 y_0 : x_1 y_1]$. One sees that $\psi(D) = P$.

We have $\psi \big |_{D}=\psi \big |_{im(i)}$ is an isomorphism to $P$, because $\phi$ is and $\psi \circ i = \phi$ (and it's easy to write down the inverse morphism explicitely).

But if I calculate it manually I get $\psi^{-1}(P) = V((x_0 y_1 - x_1 y_0)^2) \subset \mathbb P^1$x $\mathbb P^1$, which is 2 times the Diagonal $D$ (which must be true because we have a correspondence between biquadratic curves in $\mathbb P^1$x $\mathbb P^1$ and quadrics in $\mathbb P^2$). But this contradicts $\psi \big |_{D}$ being an isomorphism (and the latter in turn is used by an author to prove s.th., so I somehow hope it is not too far away from the truth / there is a way to fix his argument, but that's a different story).

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  • $\begingroup$ A nitpick: I would rather take $[t_0:t_1]$ as coordinates for the domain $\mathbb P^1$ of $\phi$ since $[x_0:x_1]$ already denotes the coordinates of the left-hand factor $\mathbb P^1$ of the product $ \mathbb P^1 \times \mathbb P^1 $. $\endgroup$ – Georges Elencwajg May 3 '13 at 23:33
  • $\begingroup$ At first I took [s:t], but I wanted to make sure that $\psi(D)= P$ is really obvious. But I see your point. I think $[t_0:t_1]$ is a good compromise. $\endgroup$ – JustSomeBungle May 4 '13 at 0:01
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Everything you write is correct and there is no contradiction!
An analogy may help:

Consider the morphism $ f: \mathbb C\to \mathbb C:z\mapsto z^2$.
It induces an isomorphism $\lbrace 0 \rbrace \to \lbrace 0 \rbrace$ but nevertheless the scheme-theoretic inverse image $f^{-1}(\lbrace 0 \rbrace )$ is twice ${\lbrace 0 \rbrace}$, since it has equation $z^2=0$.
Analogously, in your case the inverse image under $\psi$ of the cycle (actually a divisor) $P$ of $\mathbb P^2$ is the cycle (or divisor) $2D$ of $\mathbb P^1$x $\mathbb P^1$.
And this is perfectly compatible with $\psi|D:D\to P$ being an isomorphism : the point is to carefully distinguish the cycle $\psi^{-1}(P)$ of $\mathbb P^2$ from the image $(\psi|D)^{-1}(P)=D$ of the isomorphism of schemes $(\psi|D)^{-1}:P\to D$.

These considerations on cycles are basic in intersection theory: you might consult the first chapter of Fulton's Intersection Thory, the standard reference on the subject.

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  • $\begingroup$ Thanks, I am somewhat familiar with basic intersection theory and in particular with a part of Fulton's book, but I seemed to be blind / confused, somehow. Your words were very clarifying. (There is a minor typo in your answer: there is no $\psi^{-1}(D)$). $\endgroup$ – JustSomeBungle May 4 '13 at 0:18
  • $\begingroup$ Dear Laugerizor, I'm happy things are now clarified. Thanks for catching the typo: I have corrected it. $\endgroup$ – Georges Elencwajg May 4 '13 at 6:59

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