Hi, I'm currently reading through "Sheaves in Geometry and logic" by Mclane-Moerdijk and this one issue has been bugging me for a long time, which I hope you could help me resolve.

It is known that for a general presheaf on a Grothendieck Topology, we must in general apply the plus construction twice to obtain a sheaf. The first application turns an arbitrary presheaf into a separated presheaf, and one more application gives a sheaf. So, exactly, intuitively, what obstructs us from getting a sheaf from just one application of the plus construction when the sheaf is non-separated? Let us take the example: What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf? given by Sherry. When we apply it once we get a separated presheaf, OK. But what exact component of that presheaf hindered us from getting the sheaf we wanted? I agree with what Sherry wrote in that case, namely that : "So in our example, 1 and 3, over ABC and BCD, in our original presheaf were compatible on a refinement of BC but not on BC" Can we generalize this notion to become rigorous in the case for arbitrary non-separated presheaves?

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    $\begingroup$ I suggest that you should look at the other answers besides Sherry's at that other question. $\endgroup$ Apr 30, 2012 at 15:28
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    $\begingroup$ The problem is exactly that: in $X^+$ there may be families that become compatible, hence ought to have gluings, but were not compatible in $X$, so that we did not already add gluings for them. By definition of $X^+$, this happens precisely when their restrictions to overlaps agree upon further passage to a cover, which in turn can happen precisely when $X$ is not separated. So I think the theorem that $X^+$ is a sheaf when $X$ is separated is a "generalization of this notion to become rigorous". $\endgroup$ Apr 30, 2012 at 16:03
  • $\begingroup$ Related interesting MO questions: mathoverflow.net/questions/90969 , mathoverflow.net/questions/54128 $\endgroup$ Apr 30, 2012 at 18:17
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    $\begingroup$ It is possible to construct the associated sheaf functor in only one step. This is due to Eduardo Dubuc. The key idea is to consider "locally compatible families" instead of "compatible families". You can read the details here: cms.dm.uba.ar/academico/carreras/licenciatura/tesis/… (spanish, sorry) Page 19, (3.2). $\endgroup$ Apr 30, 2012 at 19:29

3 Answers 3


If I understand you correctly you believe in the counter example and just want an intuitive reason. Here is how I think about it (I am open for remarks!):

Definition: Let $\mathscr A$ be a (small) category with arbitrary (small) limits. A presheaf $F$ on a (small) site $\mathscr C$ with values in $\mathscr A$ is a sheaf, if the diagrams $F(U)\rightarrow \prod_i F(U_i)\rightrightarrows \prod_{i,j}F(U_i\times_UU_j)$ are equalizers for all covering families $U_i\rightarrow U$ of all objects $U$ of $\mathscr C$.

Construction: Let $F$ be an $\mathscr A$-valued presheaf on a site $\mathscr C$ and $U\in \mathscr C$. Then for any covering family $\mathscr U=\{U_i\rightarrow U\}$, define $F_U(\mathscr U)$ to be the equalizer of the sequence $F_U(\mathscr U)\rightarrow \prod_i F(U_i)\rightrightarrows \prod_{i,j}F(U_i\times_UU_j).$ Further define $F^+(U)$ as the colimit $F^+(U)=colim_{\mathscr U}F_U(\mathscr U)$.

And here is what happens: We do the whole construction for each $U$ seperately. And this is precisely the dangerous part: Look at the set $U=ABC$ in Sherry's example. Here, the set $BC$ is part of a covering and thus has influence on $F^+(U)$. But at the same time we apply our construction to the set $BC$, which doesn't even see the element $A$. So while we are trying to adjust our sheaf on $ABC$, so that it fits with $BC$, we change the sheaf on $BC$ independently. In the end more sections are compatible than before, but we didn't include their glueings yet. Morally the reason for that phenomenon is (non-)separatedness. By Definition we just have precisely one glueing for every compatible family in $F^+$. That means on the one hand that $F^+$ is automatically separated, but it means also that if $F$ is not separated then the different glueings for the same compatible family become "identified". In our example we have the elements $1$ and $3$ in $F(BC)$, which give the same compatible family in $F(C)$ and $F(B)$. In $F^+(BC)$ we have just one element for every family in $F(C)$ and $F(B)$.So $F^+(BC)$ looks quite different than $F(BC)$.


"Why" questions can often be answered in multiple ways; I'll give an answer that's different from the other good comments and answers that you've already had.

Sheaves are the first rung on an infinite ladder of concepts. The next rung is "stack". A stack is something like a "sheaf of categories", but there are added complications. A typical example of a stack on a topological space $X$ is the assignment $U \mapsto Sh(U)$, sending each open subset $U \subseteq X$ to the category $Sh(U)$ of sheaves on $U$. Sheaves on subsets of $X$ can be patched together, not quite uniquely, but uniquely up to canonical isomorphism. From this, you can work out what the definition of stack must be. This appears as an exercise somewhere in Mac Lane and Moerdijk.

After stacks, there are 2-stacks (something like sheaves of 2-categories), 3-stacks, and so on. (According to this terminological scheme, stacks are 1-stacks and sheaves are 0-stacks.) And just as sheaves are presheaves satisfying certain conditions, $n$-stacks are "$n$-prestacks" satisfying certain conditions.

The point now is that "twice suffices" fits into a larger pattern. For stacks, you need to apply (a version of) the plus construction three times. For 2-stacks, you need to apply it four times. In general, for $n$-stacks, you need to apply it $n+2$ times.

(I confess that this is something I've only heard in conversation, I believe from Andrew Kresch. I don't know the details.)

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    $\begingroup$ The nLab page on the plus construction says basically the same thing. It would be nice if someone could chase down a proof/reference! $\endgroup$
    – Zhen Lin
    May 2, 2012 at 0:43
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    $\begingroup$ As it happens, you also have $0$-th and $-1$st rungs on this ladder, where "once suffices" and "tautological sheaf" hold, respectively. $\endgroup$
    – S. Carnahan
    Jun 29, 2012 at 5:56

From my answere on https://mathoverflow.net/posts/95655/edit

let $F$ a presheaf on a topological space $X$ (all is straight generalizable to a general site).

THen the separate pre-sheaf $L(F)$ associate is definited as follow:

$L(F)(U)$ is the colimit of the sets $C^>(R, F)$ where: $C$ is the category of the opens of $X$, $C^>$ category of presheaves on $C$, and $R\subset h_U$ is a $X$-cover (covering cribles) of $U$, and the colimts is about all $X$-cover of $U$ and its inclusion morphisms.

Then (by Yoneda lemma, and natural representation os a presheaf as the colimits of representable by the comma category on it ) we can represent the elements of $L(F)(U)$ as a class of equivalence of families $[(U_i, x_i)_{i\in I}]$ with $U_i$ form a open covering of $U$, $x:i\in U_i$ and identify two of these family: $(U_i, x_i)_{i\in I}$ and $(V_j, y_j)_{j\in J}$ if $\forall i, j\in I\times J: {x_i}_{|U_i\cap V_j}= {y_j}_{|U_i\cap V_j}$.

as in the usual theorem follow that $L(F)$ is a separate presheaf, is a sheaf if $F$ is separate, is isomorphic to $F$ is $F$ is a sheaf, and $LL(F)$ is the sheaf associate with the canonical universal property.

Give a example (ad hoc) of a preshaf $F$ such that $L(F)$ isnt a sheaf (neccessarly $F$ isnt separate).

Let the topolgy $\tau_X=${$X, U, V, A, B, A\cap B, \emptyset$} with $X=U\cup V$ and $U\cap V=A\cup B$.

Let $F(X)=\emptyset=F(\emptyset)$, $F(U)=${$a$}, $F(V)=${$b$} with $a_{|U\cap V}\neq b_{|U\cap V}$ but $a_A=b_A,\ a_B=b_B, $ then consider $\alpha:=[(U, a)]\in LF(U),\ \beta:=[(V, b)]\in LF(V)$ we have that

$\alpha_{U\cap V}= \beta_{U\cap V}=[${$(A, a_A), (B, b_B)$}$]$

but $\alpha$ and $\beta$ cannot come from a (global) element of $F(X)$.

THis example (I hope ) explain the the difficulty that prevents $ L (F) $ to be a sheaf.

for if $X=U\cup V$, in general gived $s\in L(F)(U)$, $s=[(U_i, x_i)_I]$, and $t\in L(F)(V)$, $t=[(V_j, y_j)_J]$,

with $s_{|U\cap V}= t_{|U\cap V}$ this last condiction could use a refiniment of $(U_i\cap V)_I$ and $(V_j\cap U)_J$ and could be that ${s_i}_{|U_i\cap V_j} \neq {t_j}_{|U_i\cap V_j}$ but these are equal on a more strink refiniment.


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