\documentclass[10pt]{amsart} \usepackage[leqno]{amsmath} \usepackage{amssymb,mystyle,amscd,diagrams} \topmargin=-0.2in \oddsidemargin=-0.2in \evensidemargin=-0.2in \textwidth=6.9in \textheight=8.7in \DeclareMathOperator{\Mor}{Mor} \DeclareMathOperator{\Rad}{Rad} \DeclareMathOperator{\Frac}{Frac} \DeclareMathOperator{\supp}{Supp} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\height}{ht} \DeclareMathOperator{\Char}{char} \renewcommand{\Im}{{\rm Im}} \newcommand{\B}{\mathscr{B}} \newcommand{\calC}{\mathscr{C}} \newcommand{\calV}{\mathscr{V}} \begin{document} \smallskip \centerline{\sc Math 632, Lecture 5\\ January 16, 2004} \section{Sheafification} \begin{theorem}\label{she} Let $\calF$ be a presheaf of sets on a topological space $X$. Then there exists a pair $(\calF^+,\theta:\calF\rightarrow\calF^+)$ with $\calF^+$ a sheaf, such that for any sheaf $\calG$ on $X$ and a map $\calF\rightarrow\calG$, there exists a unique map $\calF^+\rightarrow \calG$ making the diagram \begin{diagram} \calF & & \rTo & & \calG \\ &\relax\rdTo_{\theta} & & \relax\ruTo & \\ & &\calF^+ & & \\ \end{diagram} commute, i.e. we have a bijection $\Hom_X(\calF^+,\calG)\stackrel{\circ\theta}{\longleftrightarrow}\Hom_X(\calF,\calG)$. Moreover, $\calF^+$ is unique up to unique isomorphism and for all $x\in X$ we have an isomorphism $\calF_x\simeq \calF^+_x$. \end{theorem} We call $(\calF^+,\theta)$ (or by abuse of language, $\calF^+$) the {\em sheafification} of $\calF$. \begin{enumerate} \item Let $\calF$ be the constant presheaf on $X$ associated to the set $\Sigma$. Then $\calF^+=\underline{\Sigma}$ is the constant sheaf associated to $\Sigma$ (i.e. the sheaf of locally constant functions with values in $\Sigma$). We claim that $\Hom_X(\calF,\calG)=\{\Sigma\rightarrow \calG(X)\}$. Indeed, since $\calF(U)=\Sigma$ for all $U\neq\emptyset$ with restriction maps the identity, to give maps $\varphi_U:\calF(U)\rightarrow \calG(U)$ for all open $U\subseteq$ such that the diagram $$ \begin{CD} \calF(X)=\Sigma@>>> \calG(X)\\ @V\id VV @VV\rho_{X,U}V\\ \calF(U)=\Sigma @>>> \calG(U) \end{CD} $$ commutes is equivalent to giving a map $\psi:\Sigma\rightarrow \calG(X)$ since commutativity forces all maps $\calF(U)\rightarrow\calG(U)$ to be induced by $\psi$. \item Let $M$ be a $C^{\infty}$ manifold and $\calF$ the presheaf on $M$ given by $U\mapsto \wedge^k_{\O_M(U)}(\Omega^1_M(U))$. Then we have a canonical map $$\varphi_U: \wedge^k_{\O_M(U)}(\Omega^1_M(U))\longrightarrow \Omega^k_M(U)$$ and we claim that the sheaf $U\mapsto \Omega^k(U)$ {\em is} $\calF^+$. Indeed, by the universal property of sheafification, we have a unique map $\calF^+\rightarrow \Omega^k_M$ making the diagram \begin{diagram} \calF & \rTo^{\theta} & \calF^+\\ & \rdTo& \dTo\\ & & \Omega^k_M \end{diagram} commute. But the map $\theta_x:\calF_x\rightarrow\calF^+_x$ is an isomorphism on stalks, and it is not hard to see that the canonical map $\varphi:\calF\rightarrow \Omega^k_M$ is also an isomorphism on stalks (because every $k$-form is locally a $k$-wedge power of 1-forms). Thus, $\calF^+\rightarrow\Omega_M^k$ is an isomorphism on stalks; since $\calF^+$ and $\calG$ are {\em sheaves}, it follows that $\calF^+\rightarrow \calG$ is an isomorphism. \end{enumerate} \begin{definition} A presheaf $\calF$ on $X$ is {\em separated} if the map $$\calF(U)\longrightarrow \prod \calF(U_i)$$ is injective for all open $U\subseteq X$ and all open covers $\{U_i\}$ of $U$. \end{definition} \begin{proof}[Proof of Theorem \ref{she}] Let $\Sigma_U$ be the set of all indexed open covers $\calV=\{V_i\}$ of $U$. We put a partial ordering on $\Sigma_U$ by $\{V_i\}_{i\in I}=\calV\ge \calV'=\{V'_j\}_{j\in J}$ if there exists a map $\tau:I\rightarrow J$ such that $V'_{\tau(i)}\supseteq V_i$ for all $i\in I$. Let $\calF$ be a presheaf and define $\calF_0$ by $$\calF_0(U)=\varinjlim_{\{V_i\}_{i\in I}\in \Sigma_U}\left\{ (s_i)\in \prod_{i\in I} \calF(V_i)\ :\ {s_i}\big|_{V_i\cap V_j}={s_j}\big|_{V_i\cap V_j}\ \text{in}\ \calF(V_i\cap V_j)\ \text{for all}\ i,j\in I \right\},$$ where the direct limit is formed as follows: for any $\{V'_i\}\ge \{V_j\}$ and any $\tau:I\rightarrow J$ we have the map $\prod \calF(V_j)\rightarrow\prod \calF(V'_i)$ given by $(s_j)\mapsto (s_{\tau(i)}\big|_{V'_{i}})$. It is evident that $s_{\tau(i)}$ and $s_{tau(i')}$ agree on $V'_i\cap V'_{i'}$ because $V'_i\cap V'_{i'}\subseteq V_{\tau(i)}\cap V_{\tau(i')}$ and we know that $s_{\tau(i)}$ and $s_{\tau(i')}$ agree on $V_{\tau(i)}\cap V_{\tau(i')}$ already. We claim that our definition of $\calF_0$ is independent of the choices of maps $\tau:I\rightarrow J$ that are used in forming the direct limit as described above. To see this, we must show that for any $\sigma,\tau:I\rightarrow J$ the sections $s_{\sigma(i)}$ and $s_{\tau(i)}$ agree on $V'_{i}$, where $V'_i\subseteq V_{\sigma(i)}\cap V_{\tau(i)}$. But this is clear, as $s_{\sigma(i)}$ and $s_{\tau(i)}$ already agree on $V_{\sigma(i)}\cap V_{\tau(i)}$. We define transition maps $\rho_{U,W}:\calF_0(U)\rightarrow \calF_0(W)$ as follows: given $(s_i)\in\prod\calF(V_i)$ with $\{V_i\}$ a cover of $U$, we obtain a cover of $W$ as $\{W_i=V_i\cap W\}$ and an element $(s_i\big|_{V_i\cap W})\in\prod\calF(V_i\cap W)$ with the $s_i$ compatible on overlaps; hence we get an element of $\calF_0(W)$. Now we assert that: \begin{enumerate} \item $\calF_0$ is a separated presheaf. \item For any separated presheaf $\calG$ and any map $\calF\rightarrow \calG$ there exists a unique map $\calF_0\rightarrow\calG$ making the diagram \begin{diagram} \calF & \rTo^{\theta_0} & \calF_0\\ & \rdTo& \dTo\\ & & \calG \end{diagram} commute. \end{enumerate} We first prove (1). We need to show that given an open cover $\{U_{\alpha}\}$ of $U$ and sections $s,t\in\calF_0(U)$ with $s\big|_{U_{\alpha}}=t\big|_{U_{\alpha}}$ in $\calF_0(U_{\alpha})$ then $s=t$ in $\calF_0(U)$. Therefore, suppose we have such $s,t$ and pick an open cover $\{V_i\}$ of $U$ such that there exist $(s_i)\in\prod \calF(V_i)$ and $(t_i)\in\prod \calF(V_i)$ representing $s,t\in\calF_0(U)$. Now for each $\alpha$, we see that $\{V_i\cap U_{\alpha}\}_{i\in I}$ is a cover of $U_{\alpha}$. Since $s\big|_{U_{\alpha}}=t\big|_{U_{\alpha}}$ in $\calF_0(U_{\alpha})$, for each $\alpha$ there exists a refinement of $V_i\cap U_{\alpha}$ (covering $U_{\alpha}$) such that the $s_i$ and $t_i$ agree under restriction. Putting these refinements together across all $\alpha$ we obtain a cover of $\{W_j\}$ of $U$ together with ``refinements" $(s_j)\in\prod\calF(W_j)$ and $(t_j)\in\prod\calF(W_j)$ such that $s_j=t_j$ in $\calF(W_j)$. Therefore, $s=t$ as elements of $\calF_0(U)$ and $\calF_0$ is separated. We now dispense with (2). Since $\calG$ is a sheaf, we evidently have an isomorphism $\calG\tilde{\rightarrow}\calG_0$ and any map $\varphi:\calF\rightarrow\calG$ induces a natural map $\varphi_0:\calF_0\rightarrow\calG_0$ such that the diagram $$ \begin{CD} \calF@>\varphi>>\calG\\ @V\theta_0VV @VVV\\ \calF_0 @>>\varphi_0>\calG_0 \end{CD} $$ commutes. We need only show that $\varphi_0$ is unique. But since $\calG$ is a sheaf, it suffices to show that the $(\varphi_0)_x: (\calF_0)_x\rightarrow (\calG_0)_x$ are unique for all $x$. But from the definition of $\calF_0$, it is clear that $(\theta_0)_x:\calF_x\rightarrow(\calF_0)_x$ is an isomorphism for all $x$. Since the two vertical maps in the diagram $$ \begin{CD} \calF_x@>\varphi>>\calG_x\\ @V(\theta_0)_xVV @VVV\\ (\calF_0)_x @>>(\varphi_0)_x>(\calG_0)_x \end{CD} $$ are isomorphisms, we see that $(\varphi_0)_x$ is uniquely determined by $\varphi_x$; hence $\varphi_0$ is unique. Now given a map $\varphi:\calF\rightarrow \calG$ with $\calG$ a sheaf, consider the following diagram: \begin{diagram} \calF & \rTo & \calF_0 & \rTo & (\calF_0)_0\\ & \rdTo^{\varphi} & \dTo& \ldTo & \\ & &\calG& & \\ \end{diagram} We have seen that $\varphi$ induces a unique map $\calF_0\rightarrow \calG$, and applying this fact twice, we get a unique map $(\calF_0)_0\rightarrow\calG$. We claim that if $\calF$ is any separated presheaf, then $\calF_0$ is a sheaf. This essentially follows from the definition of $\calF_0$ as the space of ``solutions to glueing problems'' and the fact that when $\calF$ is separated, such solutions are {\em unique}. \end{proof} We end by recording one obvious property of sheafification: If $U\subseteq X$ is any open set and $\calF$ is a presheaf on $X$, then there is a unique map $\left(\calF\big|_{U}\right)^+\rightarrow \calF^+\big|_{U}$ making the diagram \begin{diagram} \calF\big|_{U} & \rTo & \calF^+ \big|_U \\ & \rdTo & \uTo \\ & & \left(\calF\big|_{U}\right)^+ \end{diagram} commute. \end{document}