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I have a question about the verification of remark 1.2 in James Milne's book Étale Cohomology stated on page 156:

Assume $X$ be a normal & connected scheme with generic point $g: \eta \to X$.
Then the claim is that an étale sheaf $\mathcal{F}$ on $X$ with finite stalks is locally constant (i.e., there exists an étale cover $(U_i \to X)_i$ such that each $\mathcal{F} \rvert_{U_i}$ is a constant sheaf, cf. §1, Prop. 1.1) if and only if the unit map $\mathcal{F} \to g_*g^*\mathcal{F}$ is an isomorphism and the Galois action by $\operatorname{Gal}(\overline{k(\eta)}/ k(\eta))$ on the stalk $\mathcal{F}_{\overline{\eta}}$ factors through the étale fundamental group $\pi_1(X, \overline{\eta})$.

Question: How to see that this holds?

Few remarks:

  • $\pi_1(X, \overline{\eta})$ equals the Galois group of maximal unramified extension of $k(\eta)$; see p. 41.

  • it seems that the author suggests that one can use here the equivalence of categories between locally constant sheaves and finite $\pi_1(X, \overline{x})$-sets given by $\mathcal{F} \mapsto \mathcal{F}_{\overline{x}}$, respectively $M \mapsto \mathcal{F}(U)= \operatorname{Hom}_{\pi_1} (\operatorname{Hom}_X(\overline{x}, U),M) $.

The implication that for $\mathcal{F}$ locally constant the $\operatorname{Gal}(\overline{k(\eta)}/ k(\eta))$-action on $\mathcal{F}_{\overline{\eta}}$ factors through $\pi_1(X, \overline{\eta})$ is standard, e.g., (tag/0DV4). Essentially, that's the equivalence above & independence of geometric(= base) point.

But I not understand how the part with the unit map $\mathcal{F} \to g_*g^*\mathcal{F}$ fits in the picture. Especially, why local constancy of $\mathcal{F}$ implies the unit map is an isomorphism and respectively why it's reversely necessary alongside the assumption on $\mathcal{F}_{\overline{\eta}}$. In light of a result discussed here which appears to be maybe useful here, the main problem with the calculation of the stalks of $g_*g^*\mathcal{F}$ is how to control $g^*\mathcal{F}$. In general it's even not locally constant again. What do we know about it in this context which might help here to establish this characterization of locally constant sheaves?
Can this problem be reduced to case $\mathcal{F}$ via descent techniques, ie after passing to an etale trivialization $U_i \to X$ of $\mathcal{F}$ to deal with $\mathcal{F} \vert _{U_i} \to g_*g^*\mathcal{F} \vert _{U_i}$ simplifying(?) the situation? But then what is the term on the rhside?

#UPDATE\progress: About that the unit $\mathcal{F} \to g_*g^*\mathcal{F}$ is an iso assuming $\mathcal{F}$ is locally constant.
Keeping in mind the philosophy @Balazs suggest that the involved locally constant sheaves are "rigid" enough that the "whole information" is concentrated more less in a single stalk (...please correct me if I misunderstood your point) - the stalk & the action datum - and it suffice to reconstruct the whole sheaf, it seems reasonable to me that we need only to show two things:

(1) That the unit map is an isomorphism in the stalk at generic point; still not know how to manage it. Maybe that's exactly the point where the normality assumption is going to be exploited.

UPDATE #2(on point (1)) I think one part where normality of $X$ is exploited is to show that the stalk of $g_*g^*\mathcal{F}$ and $\mathcal{F}$ at geometric points coinside abstractly, compare again with this already linked above. But nevertheless, at this stage it's a priori not clear me why this should neccessarily imply that the unit map have to be an iso, or not?

(2) That $g_*g^*\mathcal{F}$ is locally constant; a pullback of a locally constant sheaf retain this property, and to same is true for pushforward assuming it is performed along an etale map, but $g$ is not etale as noticed in the comments below.
How else one can firsly deduce that $g_*g^*\mathcal{F}$ is locally constant (motivation: in order to simplify to check that the unit map is iso it suffice to sheck it on a single stalk (+ action compatibility) - the generic point is the plausible candidate - assuming(!) we already know that both unvolved sheaves are locally constant...

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  • $\begingroup$ I suppose that $f=g$? $\endgroup$ Commented Jan 15 at 23:15
  • $\begingroup$ @TomGoodwillie: Yes sure, thanks! $\endgroup$
    – user267839
    Commented Jan 15 at 23:38
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    $\begingroup$ I am not an expert on the etale side of things, but I think this condition is trying to capture the notion that ''nothing extra happens at closed points'': topological behaviour is controlled by what is happening on the general fibre. You may wish to look at the notes virtualmath1.stanford.edu/~conrad/Weil2seminar/Notes/L3.pdf. Theorem 4.1 here gives you a condition for a etale sheaf to be locally constant in terms of specialisation maps being bijective, which I believe is equivalent to the more concise condition in terms of the unit map. $\endgroup$
    – Balazs
    Commented Jan 16 at 12:23
  • $\begingroup$ @Balazs: so far I understand the philosophy you suggesting correctly, the idea on the the establishing that the unit is an iso "indirectly", is based on showing two things: Firstly, that it is an iso in the generic point, and secondly, that $g_*g^*\mathcal{F} $ is locally constant. (the idea is so far I understand the philosophy correctly is that for locally constant essentially the whole story is told at only one stalk + action data; it seems natural to to work with stalk at generic point. $\endgroup$
    – user267839
    Commented Jan 16 at 20:45
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    $\begingroup$ @user267839 “Etale”, according to standard usage, includes “locally finitely presented” as one of the requirements. What you call “etale” is closer to “formally etale.” $\endgroup$ Commented Jan 17 at 14:46

1 Answer 1

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(1) $ \mathcal F \to g_* g^* \mathcal F$ is an injection on the generic point.

Proof: In other words, the map $g^* \mathcal F \to g^* g_* g^* \mathcal F$ arising from the unit is an isomorphism. By a basic property of adjunctions, composing this with the map $g^* g_* g^* \mathcal F \to g^* \mathcal F$ coming from the counit gives the identity, so this map must be injective on stalks and the counit must be surjective on stalks so that their composition can be isomorphic on stalks.

(2) Let $\mathcal F$ be locally constant - in particular assume it is locally the constant sheaf associated to the set $\Lambda$. Then $g_* g^* \mathcal F$ is locally the constant sheaf associated to the set $\Lambda$.

Proof: Both $g_* $ and $g^*$ are local constructions so we may work étale-locally, and thus we may assume $\mathcal F$ is actually isomorphic to $\Lambda$. (Here we use that every étale open set in a normal variety is again normal.) Thus $g^* \mathcal F$ is $\Lambda$. So we must check that $g_* \Lambda$ is isomorphic to $\Lambda$. Recall that $\Lambda$ is the sheaf that sends an open set $U$ to the group $\Lambda$ raised to the set of connected components of $U$. There is a natural map $\Lambda \to g_* \Lambda$ arising from the natural map from the set of connnected components of $g^{-1}(U)$ to the set of connected components of $U$. This map is an isomorphism since $X$ is normal: $g^{-1}(U)$ is just the set of generic points of components of $U$ and each component has exactly one generic point.

Now if $\mathcal F$ is locally constant with finite stalks, this immediately implies $\mathcal F \to g_* g^* \mathcal F$ is an isomorphism, since at the generic point it's a map $\Lambda \to \Lambda$ which is injective and hence surjective, and then as you note since it's an map of locally constant sheaves which is an isomorphism at one point it's an isomorphism everywhere. If you want this for infinite sheaves it should be possible to do this by writing them as a limit of finite sheaves but I'm not sure if it's easier to make the limit play nice with the adjunction than to do this a different way.

This gives the "only if" direction (combined with what you already noted). For if, just note that if the Galois action on the stalk of $\mathcal F$ at $\eta$ factors through the étale fundamental group then $g^* \mathcal F$ is isomorphic to $g^* \mathcal L$ where $L$ is the locally constant sheaf corresponding to that representation of the étale fundamental group. So $\mathcal F \cong g_* g^* \mathcal F \cong g_* g^* \mathcal L$ is locally constant by the above.

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  • $\begingroup$ In (1) you used that the counit $g^*g_* \mathcal{F} \to \mathcal{F}$ is an iso. Do I understand it correctly that in etale world this holds for all immersions? Also, do you know a "standard way" how to see it? I'm (almost) familar with one eg from Sheaves in Geometry and Logic, p 368 where is presented an argument strictly speaking on level of topological spaces, but it should be easy to mimic it nearly literally to obtain a proof for etale topos, up to one thing which I not fully understood: $\endgroup$
    – user267839
    Commented Jan 19 at 14:36
  • $\begingroup$ The proof there showed only that the stalks of $g^*g_*F$ and $F$ at any point $y$ coincide, but strictly speaking the question is about if the counit map is an iso. Do I understand this crucial point here correctly that this argument from the quoted book (showing only that the stalks are isomorphic; compare also with this problem where I tried to eloborate this problem) is only "legal" here because of $\endgroup$
    – user267839
    Commented Jan 19 at 14:39
  • $\begingroup$ the explicit construction of the counit map as more or less a "simultaneous" restriction map? More concretely, going through the explicit construction of counit, firstly it is declared explicitly on level of $f^{\dagger}f_* F \to F$ (where $f^{\dagger}G$ is the usual inverse image presheaf) mapping naturally a section $s \in F(f^{-1}(V) \subset f^{\dagger}f_xF(U)$ (for $U \subset f^{-1}(V) $) to $\text{res}^{f^{-1}(V)}_U(s)$; ie beeing given "intuitively" more less as the "simultaneous" restriction map, $\endgroup$
    – user267839
    Commented Jan 19 at 14:41
  • $\begingroup$ and then extending it to $g^*g_* F$ using that the latter is the sheafification of $f^{\dagger}f_* F$. And then - but only because of the observation in last sentence & invariance of stalks under sheafification - the question just becomes if these $f^{-1}(V)$ for with $f(x) =y \in V$ become cofinal in the inductive system of etale neighborhoods of $x$, and this is answered just by showing that the stalks are isomorphic, that's the trick right? $\endgroup$
    – user267839
    Commented Jan 19 at 14:43
  • $\begingroup$ And - if what I wrote so far is correct - this argument is based on really a special feature of counit map, which cannot mimicked for unit map, exactly due to the reason - so far I understood your explanation correctly in 5th comment under this answer - that in contrast to the issue for counit that $g^*g_* F$ beeing sheafification of $f^{\dagger}f_* F$, the $f_*f^* G$ is not sheafification of $f_* f^{\dagger}G$, and that's the crucial difference where the argument cannot be "dualized", right? $\endgroup$
    – user267839
    Commented Jan 19 at 14:48

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