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Jul 25, 2014 at 0:40 comment added user27920 @Adeel: Aha, OK. So then you can now see why I was skeptical that David's condition of $f^{\ast}$ preserving sheaves is reasonable: in "real life" settings with small sites it is an amazingly restrictive condition. But I agree with your parenthetical, so I'll sign off here too.
Jul 24, 2014 at 20:14 comment added AAK @user52824, I see the confusion: note that I am talking about presheaves on the site of all S-schemes. (Probably we should stop hijacking David's question now.)
Jul 24, 2014 at 19:44 comment added user27920 @Adeel: Let $f:X \rightarrow Y$ be a surjection of schemes with the Zariski topology. Write $f^{-1}$ for the adjoint to $f_{\ast}$ between categories of presheaves, and write $f^{\ast}$ for the adjoint to $f_{\ast}$ on between categories of sheaves. For any sheaf $F$ on $Y$, $f^{\ast}F$ is the sheafification of $f^{-1}F$. You claim for any sheaf $F$ on $Y$ that $f^{-1}F$ is a sheaf. Since $f$ is surjective, the explicit formula for $f^{-1}$ gives $(f^{-1}F)(X)=F(Y)$. But the natural map $F(Y)\rightarrow (f^{\ast}F)(X)$ is generally not bijective: consider $X=Y\coprod Y$ and the natural $f$.
Jul 24, 2014 at 18:02 comment added AAK @user52824, $f^*$ preserves sheaves iff $f$ preserves covering families. When $f$ is the base change functor induced by a morphism of schemes, this is obviously true by the stability under base change of covering families.
Jul 24, 2014 at 17:47 comment added user27920 @Adeel: The functor $f^{\ast}$ on categories of presheaves that is adjoint to $f_{\ast}$ on presheaves does not carry sheaves to sheaves in general. Just think about the need to sheafify when building sheaf-pullbacks even for something as basic as a constant sheaf (rather than constant presheaf). It is hardly ever satisfied in traditional algebro-geometric situations.
Jul 24, 2014 at 17:37 comment added AAK @DavidCarchedi, yes, I meant continuous in the sense of SGA 4.
Jul 24, 2014 at 17:17 comment added David Carchedi @Adeel: I'm confused, if $g$ is right adjoint, then it is continuous automatically, unless you mean something different by "continuous" than "limit preserving". Do you mean "preserves covers"?
Jul 24, 2014 at 16:47 comment added AAK Just an obvious remark: in case $f : C \to D$ admits a right adjoint $g$, then $f_!$ preserves sheaves iff $g$ is continuous.
Jul 24, 2014 at 13:11 comment added AAK @user52824, actually the base change functor $f : \mathrm{Sch}(T) \to \mathrm{Sch}(S)$ induced by a morphism of schemes $S \to T$ preserves covering families for any topology, and so $f^*$ always preserves sheaves. So this is really a reasonable condition (though at the same time I doubt it has anything to do with $f_!$ preserving sheaves).
Jul 8, 2014 at 14:46 comment added David Carchedi Maybe we should move this discussion to email, but I wouldn't mind seeing your concrete counterexample. My situation is a bit different, but, if I can find a counterexample for the examples I am thinking about, I'm also happy. That being said, I do agree that it's certainly very far from automatic. I suspect that $f_!$ will only preserve sheaves under very specific circumstances, and I was hoping to know some of these.
Jul 8, 2014 at 14:40 comment added user27920 For open embeddings of topological spaces, etale maps between schemes equipped with the etale topology, and your poset example, $f$ carries covers to covers but your desired conclusion is false. For every example I can think of (without $\infty$-categories) in which your hypothesis holds, generally $f_{!}$ doesn't take sheaves to sheaves. Think about "disconnected" objects in the source site and constant sheaves (as opposed to constant presheaves) to see the origin of such counterexamples.
Jul 8, 2014 at 14:32 comment added user27920 For many sites in algebraic geometry (etale, fppf, Nisnevich, pro-etale, etc.) it is rare that $f^{\ast}$ on presheaves preserves sheaves, except when $f$ is a fully faithful with "no new coverings" (e.g., open embedding of top. spaces, etale map of schemes with etale sites, your example with the poset), and in such cases $f_{!}$ almost never carries sheaves to sheaves (e.g., generally fails for your poset example). So it surprising that $f^{\ast}$ on presheaves preserving the sheaf condition is a "rather standard condition" and that intuition from the classical case is irrelevant.
Jul 8, 2014 at 14:27 comment added David Carchedi I didn't mean to seem dismissive. I think we're used to working in different contexts is all. There are certainly many situations where $f^*$ does not preserve sheaves, all I am saying is that there are also many situations where it does, and my examples are of this form. I think this is a side-discussion, and it doesn't seem so relevant to my question. In my situations, $f$ preserves covers, so $f^*$ sends sheaves to sheaves, and I want to know under what conditions $f_!$ will send sheaves to sheaves. It's probably only under very stringent conditions.
Jul 8, 2014 at 9:56 comment added David Carchedi I would say that it is a side issue to what I am saying. In the examples I care about, none of the sites are opens of a topological space, so the observation you are making for topological spaces is irrelevant. In the setting of more general Grothendieck sites, it is a very reasonable condition to put on $f^*$, e.g., see my last example.
Jul 8, 2014 at 0:27 comment added user27920 If $f$ is a continuous map between topological spaces then it is very rare that $f^{\ast}$ on presheaves carries sheaves to sheaves; generally one has to sheafify the presheaf pullback functor (unless the map $f$ is very special, such as an open embedding). Is that consistent with what you are saying?
Jul 7, 2014 at 21:43 comment added David Carchedi There's tons of examples of morphisms of sites $f$ which have $f^*$ map sheaves to sheaves. It's a rather standard condition. E.g., if $X$ is a topological space, and $\Omega(X)$ is its poset of open subsets, the canonical functor $\Omega(X) \to Top,$ to topological spaces satisfies this condition.
Jul 7, 2014 at 21:33 comment added user27920 I don't know anything about $\infty$-categories, but in more traditional settings the only cases which come to mind when $f^{\ast}$ satisfies your hypotheses is when $f$ is an open embedding (in which case the desired conclusion is readily seen to be false). So do you mean that the easy counterexamples for open embeddings don't readily adapt to your fancier situation? Anyway, I'll stop here since the context for this question is way over my head.
Jul 7, 2014 at 21:06 comment added David Carchedi Let $:\pi:\mathit{Mfd} \to \mathit{Mfd}[W^{-1}]_\infty$ be natural functor from the category of manifolds to the $\infty$-category of manifolds with homotopy equivalences weakly inverted (and endow the latter with the induced Grothendieck topology).
Jul 7, 2014 at 20:44 comment added user27920 I am skeptical that you have a situation where the hypothesis actually holds (just because in essentially all situations which I can think of, the hypothesis on $f^{\ast}$ fails). Can you give an interesting case where the hypothesis can actually be verified and the conclusion is not obviously false?
Jul 7, 2014 at 15:29 comment added David Carchedi @user52824: Not really, I have instead situations where I would like this to hold, and do not know if it does. If I can prove that it holds, then I'll have some examples. I was hoping for a formal answer.
Jul 7, 2014 at 15:14 comment added user27920 Can you give one nontrivial example where the hypothesis on $f^{\ast}$ (which is very restrictive in the classical setting of continuous maps of topological spaces) does hold and the answer is affirmative? The hypothesis holds for open embeddings of topological spaces yet the conclusion fails for that case. What is the motivation for this question?
Jul 7, 2014 at 13:47 history asked David Carchedi CC BY-SA 3.0