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In the end of the Voevodsky’s lectures on cross functors, P. Deligne considers a couple of axioms which define (using the vocabulary of Ayoub's thesis) a stable homotopical 2-functor. Among them, we have that

  1. (Homotopy invariance) If $p$ is the projection $\mathbb{A}^1_X\to X$, the adjunction morphism $\operatorname{id}\to p_*p^*$ is an isomorphism;
  2. (Stability) If $s$ is the zero-section of $p$, then $p_\#\circ s_+$ is an equivalence of categories. (Where $p_\#$ is the left adjoint of $p^*$, which exists since $p$ is smooth.)

He then affirms that the two axioms above are well known in the $\ell$-adic setting. My first question then is: how are they proven? (I think a description of the proof would be nice for the MO community, but I would also be happy with a reference.)

The axiom of homotopy invariance surely axiomatises what its name describes: in Sheaves and Manifolds, M. Kashiwara and P. Shapira deduce the homotopy invariance of sheaf cohomology from the fact that the projection $X\times [0,1]\to X$ satisfies the axiom above. The axiom 1 then refers to this.

Now, I don't really understand whats the role of the axiom 2 (of "stability"). (Perhaps because I don't really have much of an intuition for $p_\#$.) So my second question is: how should one think about this axiom? Perhaps it is more intuitive in the $\ell$-adic context?

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    $\begingroup$ About axiom 2 : cohomological purity implies that $s^!p^*\Lambda \simeq \Lambda(-1)[-2]$ (Milne's étale cohomology , theorem 6.1). We even have (by remark 5.2 in loc. cit.) that for any locally constant étale $\Lambda$-sheaf $F$ on $S$, $s^!p^*F \simeq F\otimes\Lambda(-1)$, therefore the functor $s^!p^*$ coincides with twisting by (-1). It is natural to ask that twisting is invertible, and as $p_\sharp s_*$ is left adjoint to $s^!p^*$ it also equal its inverse, thus condition 2 is ''twisting by 1 is an invertible functor''. $\endgroup$
    – Swann
    May 9 at 15:48

1 Answer 1

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Proof of homotopy invariance: This follows from a base change/Kunneth type statement and the calculation of the cohomology of $\mathbb A^1$.

Specifically, Lemma 7.6.7 of Lei Fu's etale cohomology theory, specialized to $S$ a point, $f$ the map from $X$ to a point, $g$ the map from $\mathbb A^1$ to a point, so that $g' = p$, $K$ an arbitrary complex on $X$, and $L$ the constant sheaf, implies that $p_* p^* K = K \otimes f^* g_* \mathbb Z_\ell$ as soon as $K$ is strongly locally acyclic relative to $f$, which it is by Lemma 9.3.4.

Then $g_* \mathbb Z_\ell$ is the cohomology of the affine line, which is simply $\mathbb Z_\ell$, so $K \otimes f^* g_* \mathbb Z_\ell = K \otimes f^* \mathbb Z_\ell = K \otimes \mathbb Z_\ell= K$.

Proof of stability: As Deligne notes in the very last paragraph, in the etale setting $p_\#$ is $p_!$ up to a shift and twist: For a smooth morphism, $p^*$ and $p^!$ agree up to a shift and twist, and $p^!$ has a left adjoint $p_!$, which thus agrees with $p_\#$ up to the dual shift and twist.

Since shifting and twisting are equivalences of categories, it suffices to check that $p_! s_*$ is an equivalence of categories. Now $s_*$ of any sheaf is compactly supported over the base $X$ (since a section is compact), which means $p_! s_* = p_* s_*$, and $p_* s_*$ is the identity, and thus an equivalence, by the Leray spectral sequence.

Not sure on the motivation.

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  • $\begingroup$ Dear @WillSawin, the Lemma 7.6.7 of Lei Fu's book needs $g$ to be an open immersion, but the Theorem 7.6.9 works in the needed generality. Thank you for your answer :) $\endgroup$
    – Gabriel
    May 9 at 15:48
  • $\begingroup$ It would probably be better to consider $\mathbb{Z}_l$-linear (co?)homology in your answer. I believe that your arguments work in this setting without difficulty, whereas $\mathbb{Z}_l$-cohomology contains more information than $\mathbb{Q}_l$-one and it factors through motives with integer coefficients as well. $\endgroup$ May 11 at 19:37
  • $\begingroup$ @MikhailBondarko Sure! Interestingly I looked to see which is used in the document linked in the question and as far as I can see it does not specify. $\endgroup$
    – Will Sawin
    May 11 at 20:08

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