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:

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.

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.

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.