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Suppose we are given a six-functor formalism and a cartesian diagram $$\require{AMScd} \begin{CD} X @>\tilde{g}>> Z \\ @V \tilde{f} V V @V Vf V \\ Y @>g>> W\end{CD} \,.$$ There are two ways of defining a natural transformation $\tilde{g}_* \to \tilde{g}_* \tilde{f}^! \tilde{f}_!$. First, one can take the transformation $\operatorname{id} \to \tilde{f}^! \tilde{f}_!$ adjoint to the identity $\tilde{f}_! \to \tilde{f}_!$ and compose on the left with $\tilde{g}_*$ to get a transformation $\tilde{g}_* \to \tilde{g}_* \tilde{f}^! \tilde{f}_!$. Second, one can compose $\operatorname{id} \to f^! f_!$ on the right with $\tilde{g}_*$ to get a morphism $\tilde{g}_* \to f^! f_! \tilde{g}_*$, and then compose with the morphisms $f^! f_! \tilde{g}_*\to f^! g_* \tilde{f}_! \xrightarrow{\sim} \tilde{g}_* \tilde{f}^! \tilde{f}_!$ induced by base change.

Are these two natural transformations $\tilde{g}_* \to \tilde{g}_* \tilde{f}^! \tilde{f}_!$ necessarily equivalent?

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    $\begingroup$ It’s confusing that $f$ appears twice. Do you secretly have in mind the case that $h$ is an isomorphism? $\endgroup$
    – Dan Petersen
    Commented Oct 7 at 7:05
  • $\begingroup$ I guess there is no reason why the vertical arrows need to be the same - the question still makes sense in general. I will edit the question. $\endgroup$
    – dgulotta
    Commented Oct 7 at 21:59
  • $\begingroup$ What's the map $f_!\tilde g_\ast \to g_\ast \tilde f_!$ ? There is an obvious one if $g$ is proper... $\endgroup$
    – Dan Petersen
    Commented Oct 8 at 4:13
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    $\begingroup$ These compositions agree in general, though the proof is reasonably long. The idea is to expand the $*-!$ base change into a factorization, and prove it in the proper/etale cases separately. This is still formal, just using the triangle identities and unravelling base change. I'm writing up a general method for these kinds of problems, should be on arxiv in the new year. Let me know if this proof sketch doesn't suffice. $\endgroup$
    – Chris H
    Commented Oct 8 at 12:48
  • $\begingroup$ @DanPetersen The map $f_! \tilde{g}_* \to g_* \tilde{f}_!$ is the one adjoint to $g^* f_! \tilde{g}_* \xrightarrow{\sim} \tilde{f}_! \tilde{g}^* \tilde{g}_* \to \tilde{f}_!$. $\endgroup$
    – dgulotta
    Commented Oct 8 at 23:05

1 Answer 1

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Yes, this is the diagram (5HU) of my daily six-functor.

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