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Assume we have two standard conic bundles $\pi:C \rightarrow X$ and $\pi': C'\rightarrow X'$. That is $\pi$ and $\pi'$ are flat morphims of smooth varieties over $\mathbb{C}$ and both maps are relatively minimal, i.e. the preimage of an irreducible divisor in $X$ resp. $X'$ is an irreducible divisor in $C$ resp. $C'$.

According to [1, 5.3.] these conic bundles correspond to rank 4 sheaves of maximal orders $\mathcal{A}$ and $\mathcal{A}'$ over $X$ resp $X'$ such that the degeneration divisors of the conic bundles coincide with the branch (ramification) divisors of the orders and satify $sing(C)=\bigcup\limits_{i,j} (C_i\cap C_j)$ where $C_i$ and $C_j$ are nonsingular transversally intersecting components of $C$.

Looking at the generic points in $X$ and $X'$ we get two quaternion algebras $A$ and $A'$ over the function fields $\mathbb{C}(X)$ and $\mathbb{C}(X')$, with classes $a$ and $a'$ in the Brauer group of $Br(\mathbb{C}(X))$ resp $Br(\mathbb{C}(X')$.

$\underline{\text{Question}:}$ If we have a birational map $\phi: X \dashrightarrow X'$, that ist $\mathbb{C}(X)\cong \mathbb{C}(X')$, which maps the class $a$ to the class of $a'$, can we somehow construct a birational map $\psi: C\dashrightarrow C'$ such that we have a commutative diagramm $\phi\circ\pi = \pi'\circ\psi$ (that is where theses maps are defined)?

[1] On conic bundle structures - V.G.Sarkisov

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    $\begingroup$ The title of the question is misleading (as it is written, the answer is surely no). $\endgroup$ Jul 6, 2016 at 13:47
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    $\begingroup$ If the birational transformation $X\dashrightarrow X'$ pulls back $a'$ to $a$, then the fibers of $C$, resp. $C'$, over the generic points of $X$, resp. $X'$ , are birational. $\endgroup$ Jul 6, 2016 at 17:24
  • $\begingroup$ @Francesco: Thanks. I shortened the title a bit, because I did not want it to be too long. Maybe it got shortened too much. I hope now it is not misleading. $\endgroup$
    – Bernie
    Jul 6, 2016 at 21:09
  • $\begingroup$ @Jason: Thanks! This seems good. Since the birational map between the generic curves is induced by $\phi$, this birational map between the generic curves gives rise to the desired birational map between $C$ and $C'$. $\endgroup$
    – Bernie
    Jul 6, 2016 at 21:10

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