# Presheaf that do not satisfy: If $\{U_i\}$ is an open cover of $U \subseteq X$ and $s \in \mathcal{F}(U)$, then $s=0$ iff $s|_{U_i}=0$ $\forall i$

Let $X$ be a topological space.

1. Find an example of a presheaf $\mathcal{F}$ that do not satisfy: If $\{U_i\}_{i \in I}$ is an open cover of $U \subseteq X$ and $s \in \mathcal{F}(U)$, then $s=0$ if and only if $s|_{U_i}=0$ $\forall i \in I$.

2. Find an example of a presheaf $\mathcal{F}$ that do not satisfy: If $\{U_i\}_{i \in I}$ is an open cover of $U \subseteq X$ and $\{s_i\}_{i \in I}$ is a colection of sections $s_i \in \mathcal{F}(U_i)$ such that $s_i|_{U_i \cap U_j}=s_j|_{U_i \cap U_j}$, then there exists $s \in \mathcal{F}(U)$ such that $s|_{U_i}=s_i$ $\forall i \in I$.

I know that if $X=\mathbb{R}^n$, then $\mathcal{F}(U)=\{\varphi:U \rightarrow \mathbb{R} \quad \text{constant function}\}$ is a presheaf but not a sheaf. I think that this example for 2. is the simpliest.

However I can't find a simple example for 1.

Some help here would be appreciated. Thanks!

• As you mention $0$, I assume you only consider (pre-)sheaves that are at least abelian-group-valued (i.e., not set-valued)? – Hagen von Eitzen Jun 22 '15 at 17:16
• When you encounter questions like this I think it's easiest to start off with the simplest thing possible. Take a space with two points, play around with the handful of a topologies available and groups like $\mathbb{Z}$, $\mathbb{Z}/2$, $0$. Then you can get fancy with examples that appear to mean something. [I must admit that a few of the 'counterexamples' for sheafifying seem to require three points. Anyway.] – Hoot Jun 22 '15 at 17:45

Define a presheaf on $\mathbb R$ with $F(\mathbb R) = \mathbb Z$, $F(U) = \{0\}$ for every proper open $U$, and the restriction morphisms are the zero morphism : $r_U^V : F(V) \to F(U), s \mapsto 0$ (or the identity if $U=V$).

Now pick a non-zero section $s \in F(\mathbb R)$, and take any proper cover $\{ U_i\}_{i \in I}$ of $\mathbb R$ (here proper means for all $i$, $U_i \neq \mathbb R$).

Then, $r_{U_i}^{\mathbb R}(s) = 0$ for all $i$ but $s \neq 0$.

• The point of this exemple is : sheaf is basically a way of stocking local data. Here my presheaf is not a sheaf because from local data $\{ s_{| U_i} \}$ it's impossible to know if a section $s \in F(\mathbb R)$ was the zero section or not. – user171326 Jun 22 '15 at 16:58
• Thanks. Seems very simple and it's what I need. – Leafar Jun 22 '15 at 17:22
• You're welcome ! For be honnest, 2 month ago I asked exactly the same question and this was the example that my professor gave to me. – user171326 Jun 22 '15 at 17:48

Let $X$ be a topological space and let $\mathcal F(U)$ be the space of maps $U\to\mathbb R$ modulo constant maps.

1. Let $X=\{1,2\}$, $U_1=\{1\}$, $U_2=\{2\}$. Then $\mathcal F(X)\cong \mathbb R$ wheras $\mathcal F(U_1)\cong \mathcal F(U_2)\cong 0$ so that the restrictions of a nonzero glocbal section are zero. You can do the same with any other not connected space $X$

2. Let $X=S^1$ and cover it by at least three open arcs, e.g., $U_1=\{\,(x,y)\in S^1\mid x>0\,\}$, $U_2=\{\,(x,y)\in S^1\mid y>0\,\}$, $U_3=\{\,(x,y)\in S^1\mid x+y<0\,\}$. For each arc, we can define a continuous branch of $\arg(z)$; these conincide (up to constant, which we modded out) on the intersections, but there is no global continuous $\arg$ function.

There is no assumption on $X,$ so we can take discrete topology on the set $X=\{a,b\}.$ Now let $$\mathcal{F}(X)=\mathcal{F}(\{a\})=\mathcal{F}(\{b\})=\{0,1\}.$$ and set $$\begin{array}{c|lcr} r_{\{c\}}^X(s) & s=0 & s=1\\ \hline c=a & 0 & 1\\ c=b & 1 & 0\\ \end{array}$$ It is an example when not $\mathcal{F}(U_i)=\{0\}.$