# The support of a sheaf is not necessarily closed

This is paraphrased from an exercise in Hartshorne.

Let $$\mathcal{F}$$ be a sheaf on $$X$$, and for a point $$P\in X$$ let $$\mathcal{F}_P$$ denote the stalk of $$\mathcal{F}$$ at $$P$$. We define the support of the sheaf $$\mathcal{F}$$, denoted $$\operatorname{Supp}\mathcal{F}$$, to be $$\{P\in X \mid \mathcal{F}_P \neq 0\}$$. Show that $$\operatorname{Supp}\mathcal{F}$$ is not necessarily closed.

The point of this exercise is to contrast with the fact that the support of a section defined over an open is closed. I imagine that there is a canonical example of such a sheaf $$\mathcal{F}$$ that I just don't know about.

The most common example is the sheaf $j_!\mathbb{Z}$. Here $j:U\rightarrow X$ is the inclusion of an open set, and $j_!:\operatorname{Sh}(U,\mathbb{Z})\rightarrow \operatorname{Sh}(X,\mathbb{Z})$ is the functor such that $j_!\mathcal{F}$ is the sheaf associated to the presheaf $$V\mapsto\left\{\begin{array}{ll} \mathcal{F}(V) & \text{if V\subset U}\\ 0 &\text{ otherwise}\end{array}\right.$$ The sheaf $j_!\mathcal{F}$ has the nice property that $(j_!\mathcal{F})_x=\mathcal{F}_x$ if $x\in U$ and $(j_!\mathcal{F})_x=0$ otherwise. From this it is obvious that the support of $j_!\mathbb{Z}$ is simply $U$ which is open.

This example also shows that there are sheaves with non closed support on any non discrete space !

• I am not sure that this works for "any nondiscrete space". The definition of lower shriek that I know only works for locally compact spaces. It might be possible to generalize to spaces for which all open subspaces are compactly generated, but not for arbitrary spaces. Dec 26, 2022 at 13:00
• @Yai0Phah Why not ? the definition I gave works for any space and has the property that $\operatorname{Supp}j_!\mathcal{F}=U$ for any space (recall that the support of the sheaf associated to a given presheaf is the same as the support of the presheaf...) Dec 30, 2022 at 21:48

Let $X$ be a space equipped with a topology of nested open sets $\{U_i\}_{i \in \mathbf{N}}$ telescoping down to a point $Q$. So we have $X = U_0 \supset U_1 \supset U_2 \supset \dotsb\;$ and $\bigcap U_i = \{Q\}$. Define the sheaf $\mathcal{F}$ on $X$ such that $\mathcal{F}(U_i) \cong \bigoplus_{\mathbf{N}}\mathbf{Z}$ for each $U_i$ and such that the restriction maps are given by projections that drop the first graded components of the direct sum. For example the restriction $\mathcal{F}(U_0) \to \mathcal{F}(U_{1})$ looks like

\begin{align} \mathbf{Z} \oplus \mathbf{Z} \oplus \mathbf{Z} \oplus \dotsb \; &\to \; \mathbf{Z} \oplus \mathbf{Z} \oplus \dotsb \\ (z_0, z_1, z_2, \dotsc) \; &\mapsto \; (z_1, z_2, \dotsc)\,. \end{align}

In general then, the restriction map $\mathcal{F}(U_i) \to \mathcal{F}(U_j)$ drops the first $j-i$ graded components of $\mathcal{F}(U_i)$.

So the idea is that since direct sums have only finitely many non-zero components, any section $s \in \mathcal{F}(U_i)$ must eventually map to $0 \in \mathcal{F}(U_j)$ for some $j>i$. So $\mathcal{F}_Q = 0$, but for any other point $P\neq Q$, $\mathcal{F}_P = \bigcap_{U_i \ni P} = U_j$ for some $j$ (it's the $j$ corresponding to the "smallest" $U_j$ that contains $P$), and $U_j \neq 0$. So $\operatorname{Supp}\mathcal{F} = X \setminus \{P\}$, which is not closed.

• I didn't understand the definition clearly. You mean $\mathcal{F}(U_i) = \oplus_{j \in \mathbb{N}} Z_j,$ where $Z_j=\mathbb{Z}$ for $j\geq i+1$ and $=(0)$, otherwise; and the restriction map $\mathcal{F}(U_i) \to \mathcal{F}(U_{i+1})$ is defined by identity of each component except on the $(i+1)$-th component, where it is zero, and define restriction map for any $i \geq j$ inductively? (I may have made mistake with the indexing $i$ and/or $i+1$) Sep 5, 2017 at 18:14
• @Krish Yeah, you've got the right idea. I've edited my post to hopefully clear up any confusion. I phrased the restriction maps a little differently, where you can think of the maps as shifting over the components rather than zeroing them out, but the definitions are the same. Sep 5, 2017 at 18:46

Let $$X$$ be a topological space where any two non-empty open set has non-empty intersection. Fix a point $$p\in X$$ and define the following for any open set $$U$$

$$\mathcal{F}(U)=\left\{\begin{array}{ll} 0 & \text{if U=\emptyset or p\in U}\\ A &\text{ otherwise}\end{array}\right.$$

where A is any non-trivial abelian group. Restriction maps are either identity or zero maps. This is kind of opposite of Skyscraper and gives a sheaf (strangely).

The stalks of this sheaf are given by

$$\mathcal{F}_q=\left\{\begin{array}{ll} 0 & \text{if q\in \overline{\{p\}}}\\ A &\text{ otherwise}\end{array}\right.$$

Then we have $$Supp\mathcal{F}=X\setminus\overline{\{p\}}$$ which is not necessarily closed and it is not for nice choice of $$X$$. For example, if you take $$X=\mathbb{A}^1$$ equipped with Zariski Topology and $$p=0$$ then $$Supp\mathcal{F}=\mathbb{A}^1\setminus\{0\}$$ which is not closed in Zariski topology.

It's worth mentioning that if $$\mathcal{F}$$ (sheaf of $$\mathcal{R}$$-$$\mathscr{Mod}$$) is of finite type (locally generated by finite number of sections) then the support is indeed closed.

Proof. let $$x\in X\setminus \text{supp}(\mathcal{F})$$ (assume it's nonempty), therefore $$\mathcal{F}|_{U} = \sum\mathcal{R}|_U \cdot s_i$$, therefore $$s_i(x) = 0$$ therefore locally exist some $$V\subset U$$(needs to use finite type condition) such that $$s_i(y) = 0$$ on $$y\in V\subset U$$ therefore $$\mathcal{F}|_V = 0$$.