checking sheaf condition

Question

Let $X=\mathbb{C}$be a topological space. a Let $\mathcal{F}$ be a presheaf of abelian group on $X$ such that $\mathcal{F}(\mathbb{C}) = \mathbb{C}$ and $\mathcal{F}(U) = 0 $ for any open set $U$ which is not $\mathbb{C}$ and restriction maps are $res^{\mathbb{C}}_{\mathbb{C}}=id_{\mathbb{C}}$ and any other restriction maps $res^U_V$ are zero map . In my book,writer says that $\mathcal{F}$ satisfies only gluing condition. But my first attempt: Let me check that $Im\iota =ker\nu$ in $0 \to \mathcal{F}(U) \xrightarrow{\iota} \prod_{i \in I} \mathcal{F}(U_i) \xrightarrow{\nu} \prod_{i,j \in I}\mathcal{F}(U_i \cap U_j) $.Let me take an open cover $\{ \mathbb{C} \}$ of $\mathbb{C} $ we get $ 0 \to \mathcal{F}(\mathbb{C} ) \xrightarrow{\iota} \mathcal{F}(\mathbb{C} ) \xrightarrow{\nu} \mathcal{F}(\mathbb{C} ) $ but here $\iota$ and $\nu$ are isomorphisms $id_{\mathbb{C}}$ of $\mathbb{C}$ so it is not exact. Thus $\mathcal{F}$ does not satisfies gluing condition. Is that right? feel free to leave your comment. thank you.

Answer 1

pf) Claim : $\mathcal{F}$ satisfies gluing condition.

Let $\{U_i\}_{i \in I}$ be any open cover of $U$. Let $(s_i)_{i \in I}$ be in $ker\nu$ where $\nu$ is $0 \to \mathcal{F}(U) \xrightarrow{\iota} \prod_{i \in I} \mathcal{F}(U_i) \xrightarrow{\nu} \prod_{i,j \in I}\mathcal{F}(U_i \cap U_j)$ .

case i ) If $U$ is not $\mathbb{C}$ , then all section are zero all restriction map are zero map so $Im\iota =ker\nu$.

case ii ) We consider the case for $U$ = $\mathbb{C}$.

case ii -1) If $\{U_i\}_{i \in I}$ is $\{ \mathbb{C} \}$ , then $0 \to \mathcal{F}(\mathbb{C}) \xrightarrow{\iota} \mathcal{F}(\mathbb{C}) \xrightarrow{\nu}\mathcal{F}(\mathbb{C})$ where $\iota$ is an isomorphism $f \mapsto f$ and $\nu$ is a zero map $f \mapsto f-f=0$. So $ker\nu = Im \iota$.

case ii -2) Let $\{U_i\}_{i \in I}$ be any open cover of $\mathbb{C}$ other than $\{ \mathbb{C} \}$. then $ \mathcal{F}(U_i \cap U_j)$ is { $0$ } for any $i,j \in I$. so $\nu$ is a zero map. then to verify $Im\iota = ker\nu$, we need to show that $\iota$ is surjecive.

Given $(s_i)_{i \in I} \in \prod_{i \in I} \mathcal{F}(U_i)$, if the open cover does not contains $\mathbb{C}$. all $\mathcal{F}(U_i)= \{0\}$ so take $0 \in \mathcal{F}(\mathbb{C})$.

On the other hands, Suppose that the open cover $\{U_i\}_{i \in I}$ contains $\mathbb{C}$ and $s_0$ is in $\mathcal{F}(\mathbb{C})$, then we take $s_0$ as a global section mapping to $(s_i)_{i \in I} $ via $\iota$.