diff -urN -X stacks-0.2/src/documentation/dontdiff stacks-0.2.orig/src/categories.tex stacks-0.2/src/categories.tex
--- stacks-0.2.orig/src/categories.tex	2006-02-03 08:59:14.000000000 -0500
+++ stacks-0.2/src/categories.tex	2006-02-08 09:12:07.000000000 -0500
@@ -101,7 +101,6 @@
 \thispagestyle{fancy}
 
 \tableofcontents
-
 \section{Introduction}
 \label{section-introduction}
 
@@ -150,9 +149,15 @@
 A groupoid is a category where every morphism is an isomorphism.
 \end{definition}
 
-\smallskip\noindent
-A functor $F : \mathcal{A} \to \mathcal{B}$ between two categories
-$\mathcal{A}, \mathcal{B}$ is given by the following data:
+\begin{example}\label{example-group}
+A group $G$ can be thought of as a groupoid with a single object $x$
+and morphisms $\text{Mor}(x,x)=G$, with the composition rule
+given by the group law in $G$.
+\end{example}
+
+\smallskip\noindent A functor $F : \mathcal{A} \to \mathcal{B}$
+between two categories $\mathcal{A}, \mathcal{B}$ is given by the
+following data:
 \begin{enumerate}
 \item A map $F : \text{Ob}(\mathcal{A}) \to \text{Ob}(\mathcal{B})$.
 \item For every $x,y \in \text{Ob}(\mathcal{A})$ a map
@@ -181,6 +186,14 @@
 $x \in \text{Ob}(\mathcal{A})$ such that $F(x)$ is isomorphic to $z$ in
 $\mathcal{B}$.
 
+\begin{example}\label{example-group-homorphism-functor}
+A homomorphism $p\colon G\to H$ of groups gives rise to a functor
+between the associated groupoids in Example \ref{example-group}.  It is
+faithful (resp.\ fully faithful) if and only if $p$ is injective (resp.\ an
+isomorphism).
+\end{example}
+
+
 \smallskip\noindent
 A transformation of functors $t : F \to G$ (or simply a morphism of funtors)
 between functors $F, G : \mathcal{A} \to \mathcal{B}$ is given by the following
@@ -214,9 +227,37 @@
 
 \subsubsection{Additional notions}
 \label{subsubsection-categories-additional}
+\begin{definition}\label{definition-fibre-products}
+Let $x,y\in \text{Ob}(\mathcal{C})$ and $f\in \text{Mor}_{\mathcal{C}}(x,z)$
+and $g\in \text{Mor}_{\mathcal C}(y,z)$.  The fibre product of $f$ and $g$ is
+an object $x\times_z y\in \text{Ob}(\mathcal{C})$ together with morphisms 
+$p\in \text{Mor}_{\mathcal C}(x\times_z y,x)$ and 
+$q\in\text{Mor}_{\mathcal C}(x\times_z y,y)$ making the diagram
+$$
+\xymatrix{
+&x\times_y z \ar[r]^{p} \ar[d]_{q} & x \ar[d]^{f} \\
+&y \ar[r]^{g} & z }
+$$
+commute, and such that the following universal property holds: for
+any $w\in \text{Ob}(\mathcal{C})$ and morphisms 
+$\alpha \in \text{Mor}_{\mathcal C}(w,x)$ and 
+$\beta \in \text{Mor}_{\mathcal{C}}(w,y)$ with $f \circ \alpha= g\circ \beta$
+there is a unique $\gamma\in \text{Mor}_{\mathcal C}(w,x\times_z y)$ making
+the diagram
+$$
+\xymatrix{
+w\ar[rrrd]^\alpha \ar@{-->}[rrd]_\gamma \ar[rrdd]_\beta &&\\
+&&x\times_y z \ar[r]_{p} \ar[d]_{q} & x \ar[d]^{f} \\
+&&y \ar[r]^{g} & z }
+$$
+commute.  If a fibre product exists it is unique up to unique
+isomorphism.  We say the category $\mathcal{C}$ has fibre products if
+the fibre product exists for any $f\in \text{Mor}_{\mathcal C}(x,z)$
+and $g\in \text{Mor}_{\mathcal C}(y,z)$.
+\end{definition}
 
 \noindent
-FIXME: Fibre products, contravariant functors, representable functors,
+FIXME: Contravariant functors, representable functors,
 representable morphisms (namely morphisms $x\to y$ such that for every
 $w \to y$ the fibre product $w\times_y x$ exists), etc.
 
@@ -374,6 +415,7 @@
 \text{Mor}_{\mathcal{S}_U}(x,y) = \{ \phi \in \text{Mor}_\mathcal{S}(x,y) :
 p(\phi) = \text{id}_U\}.
 $$
+
 In order to discuss the notion of ``category fibred in groupoids'' we
 temporarily introduce the notion of lifting.
 A {\it lift} of an object $U \in \text{Ob}(\mathcal{C})$ is an object 
@@ -429,18 +471,28 @@
 $z' \to z$. The uniqueness implies that the morphisms $z' \to z$ and
 $z\to z'$ are mutually inverse, in other words isomorphisms.
 
-\smallskip\noindent
-Suppose that for every $f : V \to U$ and $x\in \text{Ob}(\mathcal{S}_U)$
-as in the first condition we choose a lift $f^\ast x \to x$ of $f$; this is 
-possible by the axiom of choice. For every morphism $\phi : x \to x'$ in
-$\mathcal{S}_U$ there is a unique morphism $f^\ast \phi : f^\ast x \to
-f^\ast x'$ in $\mathcal{S}_V$ such that
+\begin{example}\label{example-group-homomorphism-fibreedingroupoids}
+A homomorphism of groups $p : G \to H$ gives rise to a functor 
+$p\colon \mathcal{S}\to\mathcal{C}$ as in Example 
+\ref{example-group-homorphism-functor}. This functor
+$p\colon \mathcal{S}\to\mathcal{C}$ is fibred in groupoids if and only if 
+$p$ is surjective.  The fibre category $\mathcal{S}_{U}$ over the (unique)
+object $U\in \text{Ob}(\mathcal{C})$ is the category associated to the
+kernel of $p$ as in Example \ref{example-group}.
+\end{example}
+
+\smallskip\noindent Suppose that for every $f : V \to U$ and $x\in
+\text{Ob}(\mathcal{S}_U)$ as in the first condition we choose a lift
+$f^\ast x \to x$ of $f$; this is possible by the axiom of choice. For
+every morphism $\phi : x \to x'$ in $\mathcal{S}_U$ there is a unique
+morphism $f^\ast \phi : f^\ast x \to f^\ast x'$ in $\mathcal{S}_V$
+such that
 $$
 \xymatrix{
 f^\ast x \ar[r]^{f^\ast \phi} \ar[d] & f^\ast x' \ar[d] \\
 x \ar[r]^{\phi} & x' }
 $$
-commutes. Again uniqueness of this arrow guarantees that $f^\ast$ is a
+commutes. Again uniqueness of this arrow garantees that $f^\ast$ is a
 functor $ f^\ast : \mathcal{S}_U \to \mathcal{S}_V$. 
 
 \begin{lemma}
@@ -448,19 +500,20 @@
 If $p : \mathcal{S} \to \mathcal{C}$ is a category fibred in groupoids then
 all fibre categories are groupoids. Choose functors $f^\ast$ as above.
 Then for any pair of composable
-morphisms $f : V \to W$, $g : U \to V$ there is a unique isomorphism of 
-functors $\mathcal{S}_W \to \mathcal{S}_U$ 
+morphisms $f : V \to U$, $g : U\to W$ there is a unique isomorphism of 
+functors $\mathcal{S}_W \to \mathcal{S}_V$ 
 $$
-t : g^\ast f^\ast \to (f \circ g)^\ast 
+t : g^\ast f^\ast \to (g \circ f)^\ast 
 $$ 
-such that for every $x\in \text{Ob}(\mathcal{S}_W)$ the following
+such that for every $y\in \text{Ob}(\mathcal{S}_W)$ the following
 diagram commutes
-$$
+
+\begin{equation}
 \xymatrix{
-g^\ast f^\ast x \ar[r] \ar[d]_{t_x} & f^\ast x \ar[d] \\
-(f\circ g)^\ast x \ar[r] & x
-}
-$$
+f^\ast g^\ast y \ar[r] \ar[d]_{t_y} & g^\ast y \ar[d] \\
+(f\circ g)^\ast y \ar[r] & y
+}\label{eq:lemma-fibred-groupoids-commutes}
+\end{equation}
 \end{lemma}
 
 \begin{proof} 
@@ -479,8 +532,27 @@
 similar argument there is a unique $h : y \to x$ so that $gh = id_y$.  Then
 $fgh = f : y \to x$.  We have $fg = id_x$, so $h=f$.
 
-\noindent
-FIXME.
+\smallskip\noindent
+Now let $y\in \text{Ob}(\mathcal S_W)$ and consider the diagram
+\begin{equation}\label{eq:lemma-fibred-groupoids-commutes2}
+\xymatrix{
+f^\ast g^\ast y \ar@{-->}[d]_{t_y} \ar[r] & g^\ast y \ar[r] & y \\
+(g\circ f)^\ast y \ar[rru] & &
+}
+\xymatrix{
+V\ar@{-->}[d]_{\text{id}_V} \ar[r]^f & U \ar[r]^g & W \\
+V \ar[rru]_{g\circ f} & &
+}
+\end{equation}
+The morphism $t_y \colon f^\ast g^\ast y \to (g\circ f)^\ast y$ is the
+unique lift of of $\text{id}_V$ making
+\ref{eq:lemma-fibred-groupoids-commutes2} (resp.\ 
+\ref{eq:lemma-fibred-groupoids-commutes}) commute.  If $\phi\colon
+y'\to y$ is a morphism in $\mathcal S_W$ the compositions $(f^\ast
+g^\ast \phi) \circ t_y$ and $((g\circ f)^\ast \phi)\circ t_{y'}$ are
+both lifts of $\text{id}_V$, so are equal making $t$ is a
+transformation of functors.  Essentially the same construction applies
+to give the inverse transformation $t^{-1}$, so $t$ is an isomorphism.
 \end{proof}
 
 \noindent
@@ -491,7 +563,9 @@
 $p : \mathcal{S} \to \mathcal{C}$. Altering the morphisms in $\mathcal{S}$
 which do not map to the identity morphism on some object does not alter the 
 categories $\mathcal{S}_U$. Hence we can violate the existence and uniqueness
-conditions on lifts. Here are some examples.
+conditions on lifts. One example is the functor from Example 
+\ref{example-group-homomorphism-fibreedingroupoids} when $G \to H$ is not
+surjective. Here is another example.
 
 \begin{example}
 Let $ \text{Ob}(\mathcal{C}) = \{A,B,T\}$ and 
@@ -544,7 +618,7 @@
 Namely, we define the composition of $\psi : z \to g^\ast y$ and 
 $ \phi : y \to f^\ast x$ to be $ g^\ast(\phi) \circ \psi$. It is clear
 what the functor $p : \mathcal{S} \to \mathcal{C}$ is. The condition
-that $F(U)$ is a groupoid for every $U$ guarantees that $\mathcal{S}$ is
+that $F(U)$ is a groupoid for every $U$ garantees that $\mathcal{S}$ is
 fibred in groupoids over $\mathcal{C}$. Lifts of morphisms exist: given 
 $f: V \to U$ in $\mathcal{C}$ and $(U,x)$ a lift of $U$, then 
 $(f, id_{f^\ast x}): (V, {f^\ast x}) \to (U,x)$ is a lift of $f$. 
@@ -601,9 +675,44 @@
 \end{lemma}
 
 \begin{proof} 
-FIXME. 
+We construct a new category $\mathcal{S}'$ as follows. First we choose 
+pullback functors $g^\ast : \mathcal{S}_V \to \mathcal{S}_{V'}$ for any 
+morphism $g : V' \to V$ of $\mathcal{C}$. (We can do this since 
+$\mathcal{S}$, $\mathcal{C}$ are sets. FIXME: We can do this proof without
+choosing these as well.) The objects of $\mathcal{S}'$ 
+are pairs $(x,f)$ consisting of a morphism $f : V \to U$ of $\mathcal{C}$
+and an object $x$ of $\mathcal{S}$ over $U$, i.e., 
+$x\in \text{Ob}(\mathcal{S}_U)$. The functor 
+$p' : \mathcal{S}' \to \mathcal{C}$ will map the pair $(x,f)$ to the source 
+of the morphism $f$, in on other words $p'(x,f:V\to U) = V$. A morphism 
+$\varphi : (x_1,f_1: V_1 \to U_1) \to (x_2, f_2 : V_2 \to U_2)$ is given by a 
+pair $(\varphi,g)$ consisting of a morphism $g : V_1 \to V_2$ and a morphism 
+$\varphi : f_1^\ast x_1 \to f_2^\ast x_2$ with $p(\varphi) = g$. It is no 
+problem to define the composition law: $(\varphi,g) \circ (\psi,h) = 
+(\varphi \circ \psi, g\circ h)$ for any pair of composable morphisms. 
+There is a natural functor $\mathcal{S} \to \mathcal{S}'$ which simply maps
+$x$ over $U$ to the pair $(x, \text{id}_x)$.
+
+\smallskip\noindent
+FIXME. We need to check that $p'$ makes $\mathcal{S}'$ into a category
+fibred in groupoids over $\mathcal{C}$, and we need to check that 
+$\mathcal{S} \to \mathcal{S}'$ is an equivalence of categories over 
+$\mathcal{C}$. 
+
+\smallskip\noindent
+Finally, we can define pullback functors on $\mathcal{S}'$ 
+by setting $g^\ast(x,f) = (x, f \circ g)$ on objects if $g : V' \to V$ and
+$f : V \to U$. On morphisms $(\varphi,\text{id}_V) : (x_1, f_1) \to (x_2,f_2)$ between morphisms in $\mathcal{S}'_V$ we set $g^\ast(\varphi,\text{id}_V) =
+(g^\ast\varphi, \text{id}_{V'})$ where we use the unique identifications
+$g^\ast f_i^\ast x_i = (f_i \circ g)^\ast x_i$ from Lemma 
+\ref{lemma-fibred-groupoids} to think of $g^\ast\varphi$ as a morphism from
+$(f_1 \circ g)^\ast x_1$ to $(f_2 \circ g)^\ast x_2$. Clearly, these pullback
+functors $g^\ast$ have the property that
+$g_1^\ast \circ g_2^\ast = (g_2\circ g_1)^\ast$, in other words $\mathcal{S}'$
+is split as desired.
 \end{proof}
 
+
 \smallskip\noindent
 To continue reading, 
 \begin{enumerate}
@@ -618,9 +727,7 @@
 
 
 
-
 \bibliography{my}
 \bibliographystyle{alpha}
 
 \end{document}
-