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Suppose we have following commutative diagram (not a square i.e not a base change) of schemes:

$X\xrightarrow{p_1} Y$, $Y\xrightarrow{\pi_2} Z$, $X\xrightarrow{\pi_1} W$, $W\xrightarrow{p_2} Z$.

Let $E$ be a coherent sheaf on $X$.

Is there a natural morphism $R^ip_{1*}E\rightarrow \pi_2^*R^ip_{2*}(\pi_{1*}E)$ for all i?

or

Is there a natural morphism $\pi_2^*R^ip_{2*}(\pi_{1*}E)\rightarrow R^ip_{1*}E$ for all i?

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    $\begingroup$ How is that not a square? $\endgroup$
    – Jeremy Rickard
    Commented Oct 8, 2017 at 12:48
  • $\begingroup$ square means a base change or fibre product!! $\endgroup$
    – user111251
    Commented Oct 8, 2017 at 12:49
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    $\begingroup$ Thanks for clarifying, but I think you mean "Cartesian square". I don't think I've ever seen just "square" used to mean that. $\endgroup$
    – Jeremy Rickard
    Commented Oct 8, 2017 at 13:09
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    $\begingroup$ If you replace all functors by their derived functors, the second morphism takes place. $\endgroup$
    – Sasha
    Commented Oct 8, 2017 at 19:03

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