%! TEX root = AG2.tex % vim: tw=50 % 16/10/2024 10AM \begin{remark*} \phantom{} \begin{enumerate}[(1)] \item If $\mathcal{F}$ is a \gls{sheaf}, then $\mathcal{F}(\emptyset) = 0$ since the empty cover is a cover of $\emptyset$. \item Properties (1) and (2) of the definition of a \gls{sheaf} together can be stated by saying \[ 0 \to F(U) \stackrel{\alpha}{\to} \bigoplus_{i \in I} \mathcal{F}(U_i) \twofuncs{\beta_1}{\beta_2} \bigoplus_{i, j \in I} \mathcal{F}(U_i \cap U_j) \] is \emph{exact}, for all $U \subseteq X$ open and open covers $\{U_i\}$ of $U$. Here, \begin{align*} \alpha(S) &= (S|_{U_i})_{i \in I} \\ \beta_1((S_i)_{i \in I}) &= (S_i|_{U_i \cap U_r})_{i, j \in I} \\ \beta_2((S_i)_{i \in I}) &= (S_j|_{U_i \cap U_r})_{i, j \in I} \end{align*} Exactness means:w \begin{enumerate}[(1)] \item $\alpha$ is injective (this is property (1) of a \gls{sheaf}) \item $\beta_1 \circ \alpha = \beta_2 \circ \alpha$ (obvious) \item For any $(S_i) \in \bigoplus_{i \in I} \mathcal{F}(U_i)$ with $\beta_1((S_i)) = \beta_2((S_i))$, there exists an $s \in \mathcal{F}(U)$ with $\alpha(S) = (S_i)$. (this is property (2) of a \gls{sheaf}) \end{enumerate} \end{enumerate} \end{remark*} \begin{remark*} $\alpha$ is the \emph{equalizer} of $\beta_1, \beta_2$ (from category theory). \end{remark*} This definition works, even if e.g. we a set rather than abelian group. \begin{example*} \phantom{} \begin{enumerate}[(1)] \item $X$ any topological space \[ \mathcal{F}(U) = \{\text{continuous functions $f : U \to \Rbb$}\} \] is a \gls{sheaf}. \item $X = \Cbb$ with the Euclidean topology. Then for \[ \mathcal{F}(U) = \{f : U \to \Cbb \mid \text{$f$ bounded and holomorphic}\} \] gluing fails because one may not preserve boundedness. \item Let $G$ be a group, and set $\mathcal{F}(U) = G$ for all $U \subseteq X$. $\rho_{UV} = \id$. This is a \gls{pres}, known as the \emph{constant presheaf}. If we give $G$ the discrete topology, set \[ \mathcal{F}'(U) = \{f : U \to G \text{continuous}\} .\] These are locally constant functions. Obviously a \gls{sheaf}, called the \emph{constant sheaf}. \item If $X$ is a variety, denote by $\mathcal{O}_X(U)$ the set of regular functions $f : U \to k$. $\mathcal{O}_X$ is a \gls{sheaf}, called the \emph{structure sheaf} of $X$. See Part II \courseref{AG} for definitions of these. \end{enumerate} \end{example*} \begin{fcdefn}[Stalk / germ] Let $\mathcal{F}$ be a \gls{pres} on $X$, $p \in X$. Then the \emph{stalk} of $f$ at $p$ is \[ \mathcal{F}_p \defeq \{(U, s) \st \text{$U$ an open neighbourhood of $p$, $s \in \mathcal{F}(U)$}\} / \equiv \] where $(U, s) \equiv (V, t)$ if there exists $W \subseteq U \cap V$ a neighbourhood of $p$ such that $s|_W = t|_W$. The equivalence class of $(U, s) \in \mathcal{F}_p$ is written as $s_p$, and is the \emph{germ} of $s$ at $p$. \end{fcdefn} Note that given a morphism $f : \mathcal{F} \to \mathcal{G}$, we obtain $f_p : \mathcal{F}_p \to \mathcal{G}_p$ via $f_p(U, s) = (U, f_U(s))$. Note that a morphism of \glspl{sheaf} is just a morphism of \glspl{pres}. \begin{proposition*} Let $f : \mathcal{F} \to \mathcal{G}$ be a morphism of \glspl{sheaf}. Then $f$ is an isomorphism if and only if $f_p$ is an isomorphism for all $p \in X$. \end{proposition*} \begin{proof} \phantom{} \begin{enumerate}[$\Rightarrow$] \item[$\Rightarrow$] Obvious. \item[$\Leftarrow$] Assume $f_p$ is an isomorphism for all $p$. We will show $f_U : \mathcal{F}(U) \to \mathcal{G}(U)$ is an isomorphism for all $U$, and can then define the inverse to $f$ by $(f^{-1})_U = (f_U)^{-1}$. $f_U$ is injective: Suppose $s \in \mathcal{F}(U)$, $f_U(s) = 0$. Then for all $p \in U$, $f_p((U, s)) = (U, f_u(s)) = (U, 0) = 0 \in \mathcal{G}_p$. Thus $s_p = 0$ since $f_p$ is injective. Thus there exists an open neighbourhoord $V_p \subseteq U$ of $p$ such that $s|_{V_p} = 0$. But $\{V_p\}_{p \in U}$ covers $U$, so by (1) in the definition of a \gls{sheaf}, we get $s = 0$. $f_U$ is surjective: Let $p \in \mathcal{G}(U)$. Then for all $p \in U$, there exists $s_p \in \mathcal{F}_p$ such that $f_p(s_p) = t_p$, i.e. there exists an open neighbourhood of $p \in U$ and a germ $(V_p, \tilde{s}_p)$ representating $s_p$ such that $(V_p, f_{V_p}(\tilde{s}_p)) = (U, t) = t_p$. Shrinking $V_p$ if necessary, we can assume $f_{V_p}(\tilde{s}_p) = t|_{V_p}$ on $V_p \cap V_q$. Then \[ f_{V_p \cap V_q}(\tilde{s}_p |_{V_p \cap V_q} - \tilde{s}_q|_{V_p \cap V_q}) = t|_{V_p \cap V_q} - t|_{V_p \cap V_q} = 0 .\] Since we have shown $f_{V_p \cap V_q}$ is injective, we get \[ \tilde{s}_p|_{V_p \cap V_q} = \tilde{s}_q|_{V_p \cap V_q} ,\] and by (2) in the definition of a \gls{sheaf}, there exists $s \in \mathcal{F}(U)$ such that $s|_{V_p} = \tilde{s}_p$ for all $p$. Now \[ f_U(s)|_{V_p} = f_{V_p}(s|_{V_p}) = f_{V_p}(\tilde{s}_p) = t|_{V_p} .\] Thus $f_U(s) - t = 0$ by (1) in the definition of a \gls{sheaf}, i.e. $f_U(s) = t$. Thus $f_U$ is surjective. \qedhere \end{enumerate} \end{proof} \begin{remark*} If instead $f_p : \mathcal{F}_p \to \mathcal{G}_p$ is injective for all $p$, then $f_U : \mathcal{F}(U) \to \mathcal{G}(U)$ is still injective. If instead $f_p : \mathcal{F}_p \to \mathcal{G}_p$ is surjective for all $p$, we need not have $f_U : \mathcal{F}(U) \to \mathcal{G}(U)$ surjective (we will see examples later). \end{remark*} \textbf{Sheafification:} Given a \gls{pres} $\mathcal{F}$, there exists a \gls{sheaf} $\mathcal{F}^+$ and a morphism $\theta : \mathcal{F} \to \mathcal{F}^+$ satisfying the following universal property: For any sheaf $\mathcal{G}$ and a morphism $\varphi : \mathcal{F} \to \mathcal{G}$, there exists a unique morphism $\varphi^+ : \mathcal{F}^+ \to \mathcal{G}$ such that \[ \begin{tikzcd} F \ar[r, "\theta"] \ar[rd, "\varphi"] & F^+ \ar[d, "\varphi^+"] \\ & \mathcal{G} \end{tikzcd} \] commutes. The pair $(\mathcal{F}^+, \theta)$ is unique up to unique isomorphism. Also, $\mathcal{F}_p \cong \mathcal{F}_p^+$ via $\theta_p$ for all $p \in X$.