added lec08, thanks Nils!

lec.tex

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 \input{lec05}
 \input{lec06}
 \input{lec07}
+\input{lec08}
 
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 \bibliography{refs}

lec08.tex

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+\documentclass{lecture}
+
+\begin{document}
+		\section*{§7 fppf-topology and algebraic spaces}
+		Let $S$ be a scheme and $Sch_S$ the category of $S$-Schemes. We have the following problem. Given an $S$-group scheme $G$ and an $S$-sub-group-scheme $H$ the quotient $G/H$ may not exist in the category of $S$-schemes. \\
+		We remedy this by forming the quotient in a larger category, namely the category of algberaic spaces or fppf-sheaves, and then study conditions under which the constructed quotient lives in $Sch_S$.
+		\begin{bem}
+			Given a scheme $S$ we have the following topologies on $Sch_S$
+			\[
+			\text{Zariski} \subset \text{étale} \subset \text{fppf} \subset \text{fpqc}.
+			\]
+		\end{bem}
+		\begin{theorem}[Grothendieck, 023Q Stacks]
+			Let $S$ be a scheme. Then the representable $\hom_{Sch}(-,S)$ is a fpqc-sheaf.
+		\end{theorem}
+		\begin{bem}
+			\begin{itemize}
+				\item If $G/H$ as above is a scheme its yoneda image must be an fpqc-sheaf. 
+				\item However sheafification does not work for fpqc-presheaves.
+			\end{itemize}
+		\end{bem}
+		\begin{definition}
+			We call a family of morphisms $\{f_i:X_i\to X\}$ of schemes an fppf-covering iff
+			\begin{enumerate}[1)]
+				\item $f_i$ are flat and of finite presentation $\forall i$ and
+				\item they are jointly surjective, i.e. $X = \bigcup_i f_i(X_i)$.
+			\end{enumerate}
+		\end{definition}
+		\begin{bem}
+			A preseheaf $F:Sch_S^{op} \to Grp$ is an fppf-sheaf iff the associated set-valued presheaf is an fppf-sheaf, since the forgetful functor from groups to sets commutes with limits (as it has an adjoint).
+		\end{bem}
+		Exactness of a sequence of sheaves can be checked just as for topological spaces
+		\begin{bem}
+			Let $\tau$ be a topology on the site $Sch_S$. And let $0\to F\to G\to H$ be a sequence of sheaves with values in abelian groups/modules/\ldots. It is exact iff 
+			\begin{enumerate}[1)]
+				\item for all $X\in Sch_S$ the sequence $0\to F(X)\to G(X) \to H(X)$ is exact and 
+				\item For all $X\in Sch_S$ and all $h\in H(X)$ there is a covering $\{X_i \to X\}$ s.t. $h|_{X_i}$ is in the image of $G(X_i)\to H(X_i)$. 
+			\end{enumerate}
+		\end{bem}
+		\begin{bsp}
+			Let $n\ge 1$ be an integer. The sequence $0\to \mu_{n,S} \to \mathbb{G}_{m,S} \overset{(-)^n}{\to} \mathbb{G}_{m,S}$
+			is an exact sequence of presheaves on $Sch_S$. If $n\in \mathcal{O}_S(S)^\times$, then $(-)^n$ is surjective w.r.t. to the étale topology on $Sch_S$. If $n\notin \mathcal{O}_S(S)^\times$ then $(-)^n$ is not surjective w.r.t. the étale topology but w.r.t. the fppf-topology on $Sch_S$.
+		\end{bsp}
+		Let $\mathcal{C}$ be any category. We denote by $y:\mathcal{C} \to PSh(\mathcal{C}), \ X\mapsto y(X)= \hom_\mathcal{C}(-,X)$ the  yoneda embedding.
+		\begin{definition}
+			\begin{enumerate}[(i)]
+				\item Let $F,G \in PSh(Sch_S)$. We call a morphism $F\to G$ \textit{representable} iff for all $X\in Sch_S$ and all morphisms $y(X)\to G$ the fibre product $F \times_G y(X)$ is representable.
+				\item Let furthermore $\mathbb{P}$ be a property of morphisms of schemes which is closed under pre- and postcomoposition with isomorphisms. We say that a representable morphism $F\to G$ \textit{has $\mathbb{P}$} iff for all $X\in Sch_S$ and all $y(X)\to G$ the morphism of schemes corresponding to 
+				\[
+				F\times_G y(X) \to y(X)
+				\]
+				has the property $\mathbb{P}$.
+			\end{enumerate}  
+		\end{definition}
+		\begin{bem}
+			Note that in the second part of the above definition the object $F\times_G y(X)$ is representable s.t. the definition becomes meaningful.
+		\end{bem}
+		\begin{bem}
+			Representable morphisms are closed under composition and base change. A representable morphism with representable target has representable source, because: Let $F\to G$ be representable, $G=y(X)$. Then for $id:y(X)\to y(X)$ the object $F\times_G G=F$ is representable.
+		\end{bem}
+		\begin{lemma}
+			Let $F\in PSh(Sch_S)$. Then we have 
+			\[
+			\Delta : F\to F \times_S F \text{ representable} \iff \forall X\in Sch_S: \text{every morphism } y(X) \to F \text{ is representable.} 
+			\]
+		\end{lemma}
+		\begin{proof}
+			$\Rightarrow$: Let $y(X) \to F$ and $y(Y)\to F$ be given as in the definition. The diagram 
+			\[
+			\begin{tikzcd}
+				y(X) \times_F y(Y)\arrow{r} \arrow{d} & y(X) \times_S y(Y) = y(X\times_S Y) \arrow{d} \\
+				F \arrow{r}& F\times_S F
+			\end{tikzcd}
+			\]
+			is cartesian. So the upper map is representable with representable target. So the term $y(X)\times_F y(Y)$ is representable.\medskip\\ 
+			$\Leftarrow:$ Let $y(X) \to F\times_S F$. We claim 
+			\[
+			F\times_{F\times_S F} y(X) = y(X) \times_{y(X)\times_S y(X)} (y(X)\times_F y(X)).
+			\]
+			By assumption the term $y(X)\times_F y(X)$ is representable and since then every term on the RHS is representable so is the left term. So it remains to show the claim. For this consider the diagram (we omit the yoneda embedding from the notation)
+			\[
+			\begin{tikzcd}
+				X\times_{X\times_S X} (X\times_F X)\arrow{r}\arrow{d} & X\times_F X \arrow{r}\arrow{d}& F\arrow{d} \\
+				X\arrow{r} & X\times_S X\arrow{r} & F\times_S F.
+			\end{tikzcd}
+			\]
+			Both little square are cartesian. Thus, the outer square is cartesian which shows the claim.
+		\end{proof}
+		\begin{definition}
+			An \textit{algebraic space (over $S)$} is a sheaf $X\in PSh(Sch_S)$ with respect to the fppf-topology s.t. 
+			\begin{enumerate}[i)]
+				\item $X\to X \times_S X$ is representable and 
+				\item There exists an $S$-scheme $U$ and a morphism $y(U) \to X$ which is surjective in the étale topology.
+			\end{enumerate}
+			From this we obtain the full subcategory $AlgSpc_S \subset Sh_{fppf}(Sch_S)$.
+		\end{definition}
+		\begin{bem}
+			The category of algebraice spaces is closed under fibre products in $PSh(Sch_S)$ (what does this mean?)
+		\end{bem}
+		\begin{lemma}
+			Let $Y\in AlgSpc_S$ and $X\to Y$ representable. Then $X\in AlgSpc_S$. 
+		\end{lemma}
+		\begin{proof}
+			The proof of this was not given completely.
+		\end{proof}
+\end{document}