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\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}