I have a very quick question. Is there an easy example of a representable presheaf on a site that is not a sheaf? This certainly can't happen on a small FPPF site so I would expect a counterexample to arise on the big Zariski site.
Thanks!
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7$\begingroup$ By results of Grothendieck (descent theory), any representable functor on the category Schemes/S (where $S$ is a given base-scheme) is a sheaf for the fpqc topology. A site such that any representable functor is a sheaf is called subcanonical. For an example of a non-subcanonical topology, see sec. 2.3, Remark 2.56, p. 35 in A. Vistoli, "Grothendieck topologies, fibered categories..." in the book "FGA explained" edited by B. Fantechi, L. Illusie and others. $\endgroup$– Damian RösslerCommented Aug 9, 2012 at 14:31
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$\begingroup$ (addendum) in particular, there is no counterexample in the big Zariski site. $\endgroup$– Damian RösslerCommented Aug 9, 2012 at 14:35
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2$\begingroup$ The word to look up is subcanonical. A Grothendieck topology is called subcanonical if every representable presheaf is a sheaf. (See, for example, section III.4 of Mac Lane and Moerdijk's topos theory book.) Thus, you're looking for an example of a topology that is not subcanonical. Googling "not subcanonical" brings up some examples. $\endgroup$– Tom LeinsterCommented Aug 9, 2012 at 15:24
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$\begingroup$ @Damian - Thanks, I didn't know the fpqc fact. $\endgroup$– Lalit JainCommented Aug 12, 2012 at 14:38
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3 Answers
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Since you ask specifically for an easy example (not a natural, useful, etc., one), I suggest taking any category in which there are two objects $A$ and $B$ for which Hom$(A,B)$ is not a one-element set (i.e., the category is equivalent to neither the initial nor the terminal category) and giving it the topology in which every sieve (even the empty sieve) is covering. Then the only sheaf is the functor assigning, to each object, a singleton set, and my assumption about $A$ and $B$ ensures that the presheaf represented by $A$ is not a sheaf.
answered Aug 9, 2012 at 15:16
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7$\begingroup$ Or similarly, take any nontrivial monoid and regard it as a one-object category. Give it the topology Andreas describes. The sheaf condition applied to the empty family tells us that the only sheaf is the presheaf assigning to our object a one-element set. In particular, the single representable is not a sheaf. $\endgroup$ Commented Aug 9, 2012 at 15:25
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1$\begingroup$ Maybe, instead of "two objects $A$ and $B$", I should have written "two (not necessarily distinct) objects $A$ and $B$". Then my answer would have subsumed Tom's comment. On the other hand, some people regard distinctness of objects as an "evil" notion, not to be mentioned in polite company. $\endgroup$ Commented Aug 12, 2012 at 5:35
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You want a topology that is not "subcanonical", the definition of which is precisely that there are representable functors that are not sheaves. You could of course take a very fine topology, for instance the discrete topology in which every sieve is a covering sieve, and then most schemes are not sheaves (if you are using the category of schemes). In fact, the only one would be the terminal object: since the empty sieve covers everything, for any sheaf object $X$ and any other object $S$, the empty map from the empty covering of $S$ extends uniquely to a map from $S$ to $X$, so $X$ is terminal.
For a less trivial answer, the canonical topology is characterized as being generated by the universal strict epimorphisms: maps $X \to Y$ such that $Y$ is the quotient of $X$ by the relations $X \times_Y X \Rightarrow X$ (two projections), and for which this is stable under base change (it's easy to see from the definition of a quotient that this makes every representable functor a sheaf). The big Zariski site won't help you because the coverings are the same as in the small site; only the objects are expanded. Any fppf covering is subcanonical, and that includes any one of the "normal" Grothendieck topologies.
I am told that Voevodsky's h-topology, or the similar cdh-topology, is not subcanonical, but I don't know anything about it.
answered Aug 9, 2012 at 21:28
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$\begingroup$ "...strict epimorphisms..." also known as universal regular epimorphisms. $\endgroup$– David Roberts ♦Commented Aug 10, 2012 at 0:08
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1$\begingroup$ @David: according to nLab (I can't believe I'm citing nLab now), a regular epimorphism is the coequalizer of some pair of maps, and a strict epimorphism is the colimit of the diagram of all pairs of maps, or equivalently, if there are fiber products, the coequalizer of the projections from the fiber product. So I think strict is strictly the correct term here. $\endgroup$ Commented Aug 10, 2012 at 1:51
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Here is an easy, but very specific, example: let $T$ be the topology on the category of topological spaces with continuous maps where {$f_i: U_i \to X$} is a covering iff $X = \bigcup f_i(U_i)$. One can show that the only sheaves on this topology are those that are representable by topological spaces with trivial (indiscrete) topology. So for example, the presheaf represented by the discrete space $X = ${$a, b$} is not a sheaf. If $Y$ is $X$ with trivial topology, then $\{id: Y \to X\}$ is a covering, but the sequence $$\hom(X, Y) \to \hom(X, X) \rightrightarrows \hom(X, X \times_Y X)$$ is not exact.
