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In Proposition 1.14, page 25 in the book "3264 and all that Intersection Theory in Algebraic Geometry" the authors define a right exact sequence: $$ Z(\mathbb{P}^1 \times X) \rightarrow Z(X) \rightarrow A(X) \rightarrow 0 $$ where the left-hand map takes any subvariety $\Phi \subset \mathbb{P}^1 \times X $ to $0$ if $\Phi$ is contained in a fiber of the projection $\mathbb{P}^1 \times X \rightarrow \mathbb{P}^1$ and otherwise to $$ \langle \Phi \cap ( \lbrace t_0 \rbrace \times X ) \rangle - \langle \Phi \cap ( \lbrace t_1 \rbrace \times X) \rangle. $$

I see the idea of the exact sequence but I can not figure out what $t_0$ and $t_1$ are for a given $\Phi$. Can someone help me?

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    $\begingroup$ You can choose them to be any two distinct rational points: the automorphism group of the projective line is doubly transitive (even triply transitive). $\endgroup$
    – Jason Starr
    Commented Dec 30, 2023 at 14:24
  • $\begingroup$ Could you elaborate on this in an answer, please? I don't get why this transitivity implies that you can choose any two distinct rational points. $\endgroup$
    – Andarrkor
    Commented Dec 30, 2023 at 15:59
  • $\begingroup$ The group $\textbf{PGL}_2$ of automorphisms of $\mathbb{P}^1$ is a dense open subset of $\mathbb{P}^3$. Thus, the images under the homomorphism are rationally equivalent, independent of the choice of rational point $(t_0,t_1)$ of $\mathbb{P}^1\times \mathbb{P}^1 \setminus \Delta(\mathbb{P}^1)$. $\endgroup$
    – Jason Starr
    Commented Dec 30, 2023 at 21:48
  • $\begingroup$ But in $Z(X)$ we do not take the quotient yet. $Z(X)$ is the free abelian group of the set of subvarieties, so the map is not well defined, right? We need the image to be exactly the same, not rationally equivalent. $\endgroup$
    – Andarrkor
    Commented Dec 31, 2023 at 16:41
  • $\begingroup$ The image is exactly the same. The homomorphisms differ by rational equivalence, but the images are equal. $\endgroup$
    – Jason Starr
    Commented Dec 31, 2023 at 18:51

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