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Let $C$ be a site and $F$ an abelian presheaf on $C$. Suppose that for each object $U$ in $C$ there is a covering $\{ U_i\to U \}$ such that $F(U_i)=0$. Is it true that $F^{sh}=0$? This should be true, and is trivially if the presheaf is separated, but I don't see how to prove it in general.

edit: slight clarification

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    $\begingroup$ I am confused by the role of the $U_i$ in the question. As it is, it seems to me you are saying that the presheaf is zero on each object U of the site, in other words it's the zero presheaf therefore its sheafification is the zero sheaf (by preservation of limits if you want to use fancy stuff). $\endgroup$
    – pro
    Commented Nov 10, 2015 at 16:22
  • $\begingroup$ Yes I forgot the subscript for the covering. Thanks Edoardo for the edit. $\endgroup$
    – Rene Recktenwald
    Commented Nov 10, 2015 at 16:51
  • $\begingroup$ have sorted this out? also, is it a specific site you want it to work for? if the site has enough points (whatever that means) you should just be able to look at stalks and clearly all stalks of F are zero. $\endgroup$
    – pro
    Commented Nov 21, 2015 at 11:55
  • $\begingroup$ ah, or maybe, can you not say that the collection of the $U_i$ as U varies is a kind of basis for the site. Hence sheaves on the basis should be the same as sheaves on the site. F on the basis is zero, hence its sheafification is zero, hence sheafification of F on the actual site is also zero. was I convincing? $\endgroup$
    – pro
    Commented Nov 21, 2015 at 11:57
  • $\begingroup$ @pro Yes, see the answer I just posted. Please let me know if there is a mistake. $\endgroup$
    – Rene Recktenwald
    Commented Nov 22, 2015 at 12:58

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Indeed in the situation above

$$F^\nmid=\varinjlim_{\lbrace U_i\to U\rbrace} \ker \left(\prod_i F(U_i)\rightrightarrows \prod_{i,i'} F(U_i\times_U U_{i'})\right)=:H(\lbrace U_i\to U\rbrace, F)$$

is already zero. Fix a covering $\lbrace U_i\to U \rbrace$. By assumption we find $\lbrace V_{ij}\to U_i\rbrace$ such that $F(V_{ij})=0$ for all $i,j\in I\times J$. Also by the definition of a site $\lbrace V_{ij}\to U\rbrace$ is a covering. The canonical refinement $\lbrace V_{ij}\to U\rbrace\to \lbrace U_i\to U\rbrace$ induces

$$ H(\lbrace U_i\to U\rbrace, F)\to H(\lbrace V_{ij}\to U\rbrace, F)=0 $$

where the right side is zero because the product of $F(V_{ij})$ is zero and so the kernel can't be anything else but zero.

Hence we have a system in which every object has a map to zero IN the system, and the colimit of such a system is zero as well.

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    $\begingroup$ From the last line, it follows that the colimit of every system of vector spaces is zero. $\endgroup$
    – Vidit Nanda
    Commented Dec 22, 2015 at 14:08
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    $\begingroup$ Right, that last line can't be true. If that were the case then any colimit in a category with a zero object (e.g. pointed sets, abelian groups, spectra) would be trivial, which cannot be true. $\endgroup$
    – Jonathan Beardsley
    Commented Dec 22, 2015 at 15:25
  • $\begingroup$ @Jon Beardsley I think that Round the corner meant that the colimit of a direct system that has the property that every object has a "greater" object which is 0, must be 0. (For abelian groups this is clear by the standard construction). That is, he's talking about the diagram category, not the category of vector spaces/pointed sets. $\endgroup$
    – Phil Tosteson
    Commented Dec 22, 2015 at 18:10
  • $\begingroup$ Yes I meant what @Phil said. Treat the system as a system in the category of abelian groups, i.e. think of it as the image of the functor. Then every object maps to zero IN the direct system. Is the last line still wrong? $\endgroup$
    – Rene Recktenwald
    Commented Dec 29, 2015 at 20:34

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