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Showing the support of a sheaf may not be closed (Liu 2.5)

This is question 2.5 of Qing Liu.

I am new in algebraic geometry and really stuck on it and can’t do anything to solve it.

The question:
Let $F$ be a sheaf on $X$. Let $operatorname{Supp} F={xin X:F_xneq 0}$. We want to show that in general, $operatorname{Supp} F$ is not a closed subset of $X$. Let us fix a sheaf $G$ on $X$ and a closed point $x_0in X$. Let us define a pre-sheaf $F$ by $F(U)=G(U)$ if $x_0notin U$ and $F(U)= {sin G(U):s_{x_0}=0}$ otherwise. Show that $F$ is a sheaf and that $operatorname{Supp} F = operatorname{Supp}Gsetminus {x_0}$.

I don’t know how to solve this question:
To show a pre-sheaf is a sheaf I need to check the "uniqueness" and "gluing local sections".

For the uniqueness: Let $U$ be an open subset of $X$ , $sin F(U)$, if $x_0notin U$ , then since $G$ is a sheaf, I don’t see a problem for $F$ to be a sheaf.

If $sin F(U)$ and $x_0 in U$ and ${U_i}_i$ be an open covering of $U$,then there exists an $i_0$ such that $x_0in U_{i_0}$. the image of $s$ in the stalk $F_{x_0}$ is $s_{x_0}$. $F(U_{i_0})={sin G(U):s_{x_0}=0}$ by definition. I don’t know what to do now? (so sorry and I know this is an easy question…)

Mathematics Asked by user468730 on November 14, 2021

3 Answers

3 Answers

Side note: a presheaf $F$ has the same stalks as its sheafification $F^#$. Thus, $operatorname{supp}(F) = operatorname{supp}(F^#)$ and it suffices to find a presheaf $F$ such that $operatorname{supp}(F)$ is not closed, if all you want to know is that the support of a sheaf may or may not be closed. This doesn't matter here because the problem tells you specifically to prove $F$ is a sheaf, but it's good to know this trick.


$DeclareMathOperator{res}{res}$ The first thing we need to do when checking that $F$ is a sheaf is to keep in mind the entire datum of the presheaf $F$; not just the objects $F(U)$ for $U$ an open subset of $X$, but also the restriction morphisms $res_{U, V} : F(U) to F(V)$ when $V subseteq U$. In this case, $res_{U,V}$ is just the restriction (haha) of the restriction morphism $G(U) to G(V)$ (which we check is well-defined). In this situation (where $F(U)$ is always a subset of $G(U)$ and the restriction morphisms of $F$ are induced by those of $G$), $F$ is called a subpresheaf of $G$. Nothing crazy going on here, but it's important to fully understand the object you're working with. For convenience, I will write $sigma|_V$ to mean $res_{U,V}(sigma)$, since $U$ can always be determined from context.

Next, as you said, we should check "uniqueness". Remember, the uniqueness axioms says that for all open covers ${U_i}_{i in I}$ of an open set $U subseteq X$ and all $sigma in F(U)$, if $sigma|_{U_i} = 0$ for all $i in I$, then $sigma = 0$. We should try to prove this in the most straightforward possible way:

Let ${U_i}_{i in I}$ be an open cover of an open set $U subseteq X$. Let $sigma in F(U)$ be arbitrary. Suppose that $sigma|_{U_i} = 0$ for all $i in I$. Since $F$ is a subpresheaf of $G$, we have in particular that $sigma in G(U)$ and $sigma|_{U_i} = 0 in G(U_i)$ for all $i in I$. Since $G$ is assumed to be a sheaf, we must have $sigma = 0$, as desired.

Indeed, this very simple argument shows that any subpresheaf of a sheaf is separated (a.k.a. satisfies the uniqueness axiom): no need to mention anything about $x_0$!

Now you just need to check gluing, which I'll omit because this proof appears in other answers.

Answered by diracdeltafunk on November 14, 2021

This is just definition-pushing, and you've already made a good start.

To see uniqueness, let $Usubset X$ be an open subset and ${U_i}$ an open cover of $U$. Let $s,tin F(U)$ and let $s_i,t_iin F(U_i)$ be their restrictions. Then the condition that $s_i=t_i$ in $F(U_i)$ means that $s_i=t_iin G(U_i)$, which means that $s=t$ in $G(U)$ and as $F(U)subset G(U)$, we have that $s=t$ in $F(U)$.

To check gluing, let $s_i$ be a collection of sections of $F(U_i)$ so that $s_i|_{U_icap U_j}=s_j|_{U_icap U_j}$ as elements of $F(U_icap U_j)$. Then this equality is also true in $G(U_icap U_j)$, and by the assumption that $G$ is a sheaf, there is a section $sin G(U)$ so that $s|_{U_i}=s_i$. This implies that $s_{x_0}=0$ (if $x_0in U$ - if not, we have nothing to worry about), as the maps $G(U)to G(U_i)to G_{x_0}$ commute, so $sin F(U)$ as well and thus $F$ satisfies gluing.

Answered by KReiser on November 14, 2021

This kind of a statement follows from "unpacking definitions". You unpack the sheaf properties of $G$ together with their connection with $F$ to get that $F$ is a sheaf. In principle this is easy and there is nothing happening. In practice you can get lost, in particular if you are not familiar with the general concepts at play.

Since $G$ is a sheaf and $F(U)subseteq G(U)$ for any $U$ you get uniqueness of any gluings automatically, since they are also gluings in $G$. But lets be explicit, if $U_alpha$ is a collection of open sets and $s,s' in F(bigcup_alpha U_alpha)$ with $slvert_alpha = s'lvert_alpha$ for all $alpha$ we want to see that $s=s'$ must follow. Since $F(bigcup U_alpha)subseteq G(bigcup U_alpha)$ you get that $s,s'$ are both in $G(bigcup U_alpha)$ with $slvert_alpha = s'lvert_alpha$, since $G$ is a sheaf you get $s=s'$.

So whats left is to show is the existence of a gluing.

So suppose $U_alpha$ is a collection of open sets and $s_alphain F(U_alpha)$ for all $alpha$ with $s_alphalvert_{U_alphacap U_beta} = s_betalvert_{U_alphacap U_beta}$ for any $alpha,beta$, ie the compatibility conditions of gluing are satisfied. We want to show that there is an $sin F(bigcup_alpha U_alpha)$ with $slvert_alpha = s_alpha$ for every $alpha$. Now since $G$ is a sheaf and $s_alphain G(U_alpha)$ you can glue them to in $G$ get an $sin G(bigcup_alpha U_alpha)$ for which $slvert_alpha =s_alpha$, we just need to check that $sin F(bigcup U_alpha)$, ie that $s_{x_0}=0$ if $x_0inbigcup_alpha U_alpha$ (if $x_0notin bigcup_alpha U_alpha$ there is nothing to check). Suppose $x_0in U_gamma$ for a fixed $gamma$, since $s_gamma in F(U_gamma)$ this means that $(s_gamma)_{x_0}=0$, ie there is some open set $Usubseteq U_gamma$ containing $x_0$ for which you have that $s_gamma lvert_U =0$. But then: $$slvert_U=(slvert_{U_gamma})lvert_{U}=s_gammalvert_U=0$$ implying that $s_{x_0}=0$, ie $sin F(bigcup_alpha U_alpha)$.

Answered by s.harp on November 14, 2021

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