MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Before I start, please note that it is not going to be sufficient to assume that the topology is subcanonical or that the site $C$ has finite limits, since the application I have in mind does not satisfy either of these two criteria. Therefore, we take our definition of a topology to be the one in terms of covering sieves (a function $J$ assigning to each object $U$ a family of subfunctors $S\subseteq h_U$ satisfying conditions). Each sieve has a canonical "cover" attached to it, $\{U_i\to U\}_{i\in I}$ (although because our case is somewhat pathological, the resulting system of covers does not meet the requirements for being a Grothendieck pretopology (the requirements are not meaningful in this setting because our underlying category does not have pullbacks)). We will make use of these covers in a construction, but it is to be understood that these arise from an actual Grothendieck topology.

Let $(C,\tau)$ be a Grothendieck site. It's well-known that the double-plus construction $(-)^{++}$, which is actually the composite $((-)^+)^+$ is the left adjoint to the inclusion functor $Sh(C,\tau)\hookrightarrow Psh(C)$. Recall that $(-)^+$ replaces the functor $F(-)$ by the functor $H_\tau^0(-,F)$, or the $0^{th}$ sheaf of Čech cohomology, which can be constructed as the colimit $H^0_{\tau}(U,F)=L_\tau F(U):=Colim_{S\in J(U)^{op}} Hom(S,F)$.

Up until yesterday or so, I believed that the functor $(-)^+:Psh(C)\to SepPsh(C)$ is left adjoint to the inclusion $\iota:SepPsh(C,\tau)\hookrightarrow Psh(C)$, but I feel very inclined to doubt it now, since by a computation I did with a toy example, it doesn't seem to have the right universal property (I'm unsure, because the one place that I've found that addresses the issue says that it is).

However, when I was looking back through Vistoli's notes in the section on sheafification, I noticed that his construction of the sheafification is actually as $((-)^{sep})^+$, where $(-)^{sep}$ is defined by quotienting out by the following equivalence relation: $R(U)\subset F(U)\times F(U)$ where a pair $(a,b)\in F(U)\times F(U)$ is in $R(U)$ (that is to say, $a\sim b$) if there exists a cover $\{U_i\to U\}_{i\in I}$ such that $a|_{U_i}=b|_{U_i}$ for each $i\in I$.

First thing: It's clear that the relation is reflexive and symmetric, but is it also clear that it's transitive? If it's not, does Vistoli's construction work if we saturate the relation to an equivalence relation?

Second thing: Is the $(-)^+$ functor actually the left-adjoint of the inclusion $\iota$? If not, is $(-)^{sep}$? If the left adjoint is indeed $(-)^{sep}$, does it preserve monomorphisms (this is not to show that it's left-exact, since that will definitely fail (since separated presheaves do not form a topos). I have something totally different in mind unrelated to the rest of this question).

The toy example I mentioned above is the one to keep in mind, so I'll mention it here:

Consider the category $[1]$, which is the category with two objects $0$ and $1$ with one nonidentity morphism $0\to 1$. Let $S=h_0\subseteq h_1$ be the only nonidentity covering sieve of $1$. Then separated presheaves on this site are exactly the functors $F:[1]^{op}\to Set$ such that the restriction map $res_{0}:F(1)\to F(0)$ is injective. Given a presheaf $F$, $F^{sep}$ is given by the quotient where $F^{sep}(0)=F(0)$ and $F^{sep}(1)$ is the quotient of $F(1)$ by the equivalence relation noted above.

Notice also that sheaves for this topology are the presheaves such that $F(1)\to F(0)$ is a bijection.

We can easily verify the universal property of Vistoli's construction, since given a morphism $F\to G$ where $G$ is separated, we see that it obviously factors uniquely through the quotient. The $(-)^+$ construction does not seem to have this property in any obvious way.

share|cite|improve this question
I think for $(-)^+$ to be left-adjoint to $\iota$, we would need $(-)^+\circ\iota = id.$ – Kevin Ventullo Dec 15 '10 at 6:04
@Kevin: Yes, I think so too. I just wanted to make sure, since the nLab appears to disagree with us on this point. – Harry Gindi Dec 15 '10 at 6:20
Actually, yeah, you're completely right. I just found the result you're talking about in Gabriel-Zisman, so if you want to post that as an answer, I will be happy to accept it. – Harry Gindi Dec 15 '10 at 6:24
up vote 10 down vote accepted

If $(-)^+$ were left adjoint to $\iota$, we would have $(-)^+\circ\iota=id.$, but this is impossible since $(-)^+$ applied to a separated presheaf gives its sheafification.

share|cite|improve this answer

About your first question: the relation is transitive. If $a$ is equivalent to $b$ using a covering, and $b$ is equivalent to $c$ using another covering, it is easy to see that $a$ is equivalent to $c$ using a common refinement. And yes, $(-)^{\rm sep}$ is left adjoint to the inclusion of separated presheaves into presheaves, that's easy to check.

share|cite|improve this answer
Thanks! I've got a quick question for you, then! Why are your notes the only place that I can find the construction $(-)^{sep}$? The double-plus construction seems pretty unintuitive, and the fact that it works seems more like a coincidence than anything. – Harry Gindi Dec 15 '10 at 8:03
Well, it seems to me that I am the wrong person to ask. Anyway, I share your feelings about the plus construction. – Angelo Dec 15 '10 at 9:12
The plus construction seems very intuitive to me: to make F into a sheaf, you want the sections of F over U to be determined by matching families over covers of U, so you simply replace F(U) by the set of all such matching families, modulo the obvious equivalence relation. It's maybe a little naive, but it's natural. It converges after 2 steps because sets are discrete; for stacks you need 3 steps, and for ∞-stacks you need transfinite steps. You can also think of it as a version of the small object argument, with sheaves as the fibrant objects. – Mike Shulman Dec 22 '10 at 5:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.