0
$\begingroup$

Let $f: X \to Y$ be a morphism of schemes and $\mathcal{F}$ sheaf of sets/Abelian groups on the small étale site $Y_{ét}$. Assume we manage somehow to show thatat every geometric point $\overline{y} \to Y$ the stalks $(f_*f^* \mathcal{F})_{\overline{y}} $ and $\mathcal{F}_{\overline{y}} $ are isomorphic.

Does from this automatically follow that the unit map $u:\mathcal{F} \to f_*f^* \mathcal{F}$ is an iso?

Motivation #1: In several arguments from eg Milne's online notes on Etale Cohomology and Lei Fu's book is argued by showing only that the stalks $(f_*f^* \mathcal{F})_{\overline{y}} $ and $\mathcal{F}_{\overline{y}} $ are isomorphic, it follows that then the unit map $u$ must be an iso.
Motivation #2: If that's true, this would solve also this problem I posted some days ago.

Drawing the analogy to dual situation where we may ask if the counit map $c:f^*f_* \mathcal{G} \to \mathcal{G}$ is an iso, assuming we know that the stalks of both sheaves coincide, the answer should be positive,
because by explicit construction of the counit map (see eg Görtz', Wedhorn's AlGeom, Proposition 2.27 for Zariski site, which can be imitated for etale case), if we know that that the inductive system over which the limit is taken to calculate the stalk $f^{\dagger} f_* \mathcal{G}$ (where $f^{\dagger} \mathcal{H} $ is the inverse presheaf (see somewhere in Milne's notes) which sheafified gives $f^*\mathcal{H} $ )
is cofinal in the inductive system over which the limit is taken to calculate the stalk $\mathcal{G}_x$, then by construction of the counit map we know that it must be an iso.

The crucial point is here in case of counit that the sheafification of $f^{\dagger} f_* \mathcal{G}$ is exactly $f^* f_* \mathcal{G}$, so these have the same stalks.

In contrast, for unit this argument would break down, since $f_*f^* \mathcal{F}$ is not the sheafification of $f_*f^{\dagger} \mathcal{F}$, so even if we manage to show that the inductive system over which the limit is taken to calculate the stalk $\mathcal{F}_y$ is cofinal in the inductive limit over which the limit is taken to calculate the stalk $(f_*f^{\dagger} \mathcal{F})_y$, we in general don't know something about what the inductive system calculating $\mathcal{F}_y$ has to do with inductive system over which the limit is taken to calculate the stalk $(f_*f^*\mathcal{F})_y$.

Therefore the unit case is seemingly more complicated. So if what I have wrote so far make sense the key difference between the argumentation for counit and unit is that for counit we can relate inductive systems used to calculate the stalks $\mathcal{G}_x$ and $(f^{\dagger} f_* \mathcal{G})_x$ and since $f^* f_* \mathcal{G}$ is sheafification of $f^{\dagger} f_* \mathcal{G}$ we have control over the stalks of the latter, since canonical sheafification map is identity on stalks.

On the other hand for unit we can relate inductive systems used to calculate the stalks $\mathcal{F}_y$ and $(f_*f^{\dagger} \mathcal{F})_y$ , but unfortunately $f_*f^* \mathcal{F}$ is not the sheafification of $f_*f^{\dagger} \mathcal{F}$, so we haven't control over the inductive system over which the limit is taken in the stalk calculation and cannot especially relate it to the inductive system over which the stalk $\mathcal{F}_y$ is calculated.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ A morphism of sheaves is an isomorphism if induces isomorphisms on the stalks. This is what Milne and Fu are using. It is not enough to know simply that the stalks are isomorphic. $\endgroup$
    – anon
    Commented Jan 19 at 6:24
  • $\begingroup$ @anon: Yes, that's exactly the strategy showing that a map between sheaves is an iso I have in mind too. But see eg Cor. 5.3.10 in Fu's book linked above: There the autor seemingly finishes the argument abruptly after having shown that the stalks are isomorphic (due to 5.3.7). But that's exactly the thing, as you said to show that the map is iso it not enough to know that the stalks are isomorphic! One must show that the map on stalks must be iso, but seemingly Fu - if I'm not missing something - shows only the former statement, which as you said is in general not enough. $\endgroup$
    – user267839
    Commented Jan 19 at 9:46
  • $\begingroup$ So the question is essentially if thats a "special feature" of the unit map to be of "such canonical nature" that it suffice already to see that the stalks are isomorphic, which seems rather strange for me in light of your plausible explanations to argue that way since seemingly such "argument" would not the shape of the map itself take into account $\endgroup$
    – user267839
    Commented Jan 19 at 10:03
  • $\begingroup$ I do not have a copy of Fu's book and cannot access it via the link. In 5.3.7, is there any information about the isomorphism on stalks? In particular, is there a specific isomorphism being described? $\endgroup$
    – Brian Shin
    Commented Jan 19 at 16:02
  • $\begingroup$ @BrianShin: In 5.3.7 the statement is that for finite map $f:X \to Y$ and $F$ sheaf on $X$ for the stalk at geometric pt $y$ (I skip the bar notation) of pushforward of $F$ on etale site holds $(f_*F)_y \cong \oplus_{x \in X_y} F_x$. The proof firstly reduces to the case that there lies only one geom point $x$ over $y$, and then constructs the iso $(f_*F)_y \cong F_x$ naturally by using that under assumption of finiteness for $f$ the inductive system of $U \times_Y X$ - where $U$ run over connected etale neighborhoods of $y=f(x)$ (="symbolical" notation) - is cofinal in the inductive $\endgroup$
    – user267839
    Commented Jan 20 at 11:36

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .