4
$\begingroup$

Let $X = Y \times_Z W$, where $X,Y,Z,W$ are Noetherian schemes, and consider the pullback diagram associated to $X, Y, Z, W$. We have a diagram $$ \require{AMScd} \begin{CD} D(Z) @>>> D(Y)\\ @VVV @VVV\\ D(W) @>>> D(X) \end{CD}, $$ where $D(-)$ means either the bounded derived category of coherent sheaves, or the unbounded category of quasi coherent sheaves, and the maps are given by pullback of modules. I was wondering whether this diagram is a push out diagram. I don’t expect this to be true in the world of standard algebraic geometry, i.e. if this holds I would expect one has to take the derived fiber product and consider the categories as $\infty$ or DG categories. However, not having much knowledge of derived algebraic geometry, I don’t know whether this result holds.

$\endgroup$
4
  • 2
    $\begingroup$ No, this is false. Consider what happens when the intersection is empty, for instance two distinct points in the line. If you reverse the roles of pushout and pullback, then the statement is often true-- though I'm not sure what the correct level of generality is. $\endgroup$ – Phil Tosteson Jan 21 '20 at 17:06
  • $\begingroup$ @Phil Tosteson I see, thanks! $\endgroup$ – Federico Barbacovi Jan 21 '20 at 17:15
  • 2
    $\begingroup$ Empty intersection seems to work fine. For the case of dg categories of unbounded complexes of quasi-coherent sheaves, see Theorem 4.7 in arxiv.org/abs/0805.0157. The case of dg categories of coherent complexes is more complicated, see arxiv.org/abs/1312.7164. $\endgroup$ – Pavel Safronov Jan 22 '20 at 9:22
  • $\begingroup$ @PavelSafronov yeah, actually I gave it another thought yesterday and I wasn’t able to spot why empty intersection should not work. Thank you very much for the references! $\endgroup$ – Federico Barbacovi Jan 22 '20 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.