I'm trying to follow the explanation given in Olsson's "Sheaves on Artin stacks" for the lack of functoriality for lisse-étale topology: Let $f:Y \to X$ be a morphism of algebraic stacks. The functor $f^{-1}$ sends $\def\liset{\text{lis-ét}}N \in X_\liset$ to the sheaf associated to the presheaf which to any $V \in \def\LisEt{\text{Lis-Et}}\LisEt(Y)$ associates the colimit $\lim_{V \rightarrow U} N(U)$ where the colimit is taken over all morphisms over $f$ from $V$ to objects $U \in \LisEt(X)$. Olsson says this functor $f^{-1}$ is not left exact, because the colimit is not filtering (it is connected but equalizers do not exists). And therefore $(f^{-1}, f_*)$ does not define a morphism of topoi $Y_\liset \to X_\liset$.

I have a couple of questions:

  1. When $f$ is a map of schemes (with the Zariski topology), I think $f^{-1}$ is exact, since $f^{-1}N_{y} = N_{f(y)}$ (correct me if I'm mistaken). My argument roughly is that given any open neighborhood $U$ of $f(y)$, $f^{-1}(U)$ is an open neighborhood of $y$. And so $f$ maps a basis of neighborhoods of $y$ to subsets contained in a basis around $f(y)$, giving the result. Why doesn't the analogous argument hold for the lisse-étale topology? I suspect the answer is in whatever it means that "the colimit is not filtering" but I don't know what this means. I see from Olsson's Example 3.3 (omitted here) that equalizers need not exists in $\LisEt(Y)$, but I'm not sure what this has to do direct limits.

  2. In response to the comments/answers received, here is a more concrete version of what I'm trying to ask in question 1 above. Presumably Olsson's Example 3.4 (described below) is crucially using the lisse-étale topology, but it appears to apply to the Zariski topology as well. Let $k$ be a field, let $X=\mathop{Spec} k[t]$, $Y=\mathop{Spec} k$ and $f:Y \to X$ the inclusion of the origin. Let $g:\mathcal{O}_X \to \mathcal{O}_X$ be given by multiplication by $t$. This map has kernel zero. But $f^{-1}\mathcal{O}_X = \mathcal{O}_Y$ as lisse-étale sheaves (because $Hom(f^{-1}\mathcal{O}, G)=Hom(\mathcal{O}, f_*G)=f_*G(\mathbb{A}^1_X)=G(\mathbb{A}^1_Y)$. We have that $f^{-1}g$ has non-zero kernel. Why doesn't this argument work in the Zariski topology? My guess is that $f^{-1}\mathcal{O}_X = \mathcal{O}_Y$ is not true in the Zariski topology (in fact I calculate $f^{-1}\mathcal O_X=k[t]_{(t)}$ but I don't see why the calculation given above doesn't apply.

  3. Why does $f^{-1}$ being not left exact imply we don't have a morphism of topoi? For a morphism of topoi, we need $f^{-1}$ to be left adjoint to $f_*$, and hence $f^{-1}$ needs to be right exact.

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    $\begingroup$ About question 2 : the left exactness of $f^{-1}$ is part of the definition of a morphism of topoi, see ncatlab.org/nlab/show/geometric+morphism $\endgroup$
    – Alex
    Jan 28, 2011 at 3:11
  • $\begingroup$ About question 1 : The limit is actually a colimit. Filtering colimit are nice because they are often left exact, and in particular they would be in your case, see mathoverflow.net/questions/2150/exactness-of-filtered-colimits. The existence of equalizers would guarantee that your colimit is filtering; it follows easily (I think, but it's late and I'm falling asleep) from the definitions of these two things, see en.wikipedia.org/wiki/Filtered_colimit and en.wikipedia.org/wiki/Equaliser_%28mathematics%29 $\endgroup$
    – Alex
    Jan 28, 2011 at 3:44
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    $\begingroup$ okay, I understand your answer to question 2 - basically $f^{-1}$ has to preserve finite limits, and kernels are an example of finite limits (they are the equalizer when one of the morphisms is the zero morphism). thanks. $\endgroup$
    – ykm
    Jan 28, 2011 at 4:04
  • $\begingroup$ You still have two instances of 'limit' where it should probably say colimit. $\endgroup$
    – David Roberts
    Jan 28, 2011 at 4:34
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    $\begingroup$ Q1: It has nothing to do with stacks. Also in the case of schemes the lisse-étale topos (or the lisse-lisse topos) is not functorial. You can find an example of an equalizer which is not preserved by pullback in Example 4.4 of Anton Geraschenko's notes from a course on stacks taught by Martin Olsson, math.berkeley.edu/~anton/written/Stacks/Stacks.pdf $\endgroup$ Jan 28, 2011 at 8:30

2 Answers 2


As noted in the comments this has nothing to do with stacks.

Let F and G be a pair of adjoint functors between categories C and D (with F the left adjoint). Denote by AB(C) and AB(D) the categories of abelian group objects of C and D. Then it doesn't automatically follow that F induces a morphism from AB(C) to AB(D), unless F is left exact.

Now for C and D topoi (i.e., categories of sheaves of sets on some site), AB(C) and AB(D) are just categories of sheaves of abelian group, and the condition that F be left exact (F is the $f^{-1}$ of your example) is exactly the condition needed for F of a sheaf of abelian groups to be a sheaf of abelian groups. (This is very clear if you think of `abelian group objects of C' instead of sheaves of abelian groups; if you think of this for awhile and can't figure out what I mean I'll be happy to add more detail.)

Added: This doesn't address the first part of your question (I hope that the example from Anton's notes suggested in the comments helps). For your second question -- left exactness of $f^{-1}$ is part of the definition of a morphism of topoi, and above I explain why that is part of the definition.

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    $\begingroup$ Left exactness is about compatibility with finite limits. However, for an adjunction to be compatible with algebraic structures like (abelian) group objects, we just need compatibilities with finite products. Moreover, in the case of the lisse-étale site, the functors $f^{-1}$ preserve finite products, so that there is no problem there. The problem of the functoriality of sheaves on the lisse-étale site arises at the level of derived categories of sheaves, where we have to understand why the adjunction $(f^{-1},f_*)$ leads to an adjunction of derived functors $(Lf^{-1},Rf_*)$. $\endgroup$ Jan 29, 2011 at 1:54
  1. You are right, $f^{-1} \mathcal O_X \neq \mathcal O_Y$. in Zariski topology. Your Yoneda-type calculation doesn't apply in Zariski topology because sheaf $\mathcal O_X$ is not representable in Zariski topology. The object $\mathbb A^1_X$ is not an object of Zariski topology, however it is an object of Lisse-etale site.

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