Let $X$ be a topological space.
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 $s|_{U_i}=0$ $\forall i \in I$.
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!
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 $s \neq 0$.