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	2005-10-15 15:14:24.000000000 -0400
+++ stacks-0.2/src/categories.tex	2006-02-03 08:59:14.000000000 -0500
@@ -440,7 +440,7 @@
 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 garantees that $f^\ast$ is a
+commutes. Again uniqueness of this arrow guarantees that $f^\ast$ is a
 functor $ f^\ast : \mathcal{S}_U \to \mathcal{S}_V$. 
 
 \begin{lemma}
@@ -463,9 +463,61 @@
 $$
 \end{lemma}
 
-\begin{proof} FIXME.
+\begin{proof} 
+To show all fibre categories $\mathcal{S}_U$ for $U \in \text{Ob}(\mathcal{C})$
+are groupoids, we must exhibit for every $f : y \to x$ in $\mathcal{S}_U$ an
+inverse morphism.  The diagram on the left (in $\mathcal{S}_U$) is mapped by
+$p$ to the diagram on the right:
+$$
+\xymatrix{
+y \ar[r]^f & x & U \ar[r]^{id_U} & U \\
+x \ar@{-->}[u] \ar[ru]_{id_x} & & U \ar@{-->}[u]\ar[ru]_{id_U} & \\
+}
+$$
+Since only $id_U$ makes the diagram on the right commute, there is a unique
+$g : x \to y$ making the diagram on the left commute, so $fg = id_x$.  By a
+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.
 \end{proof}
 
+\noindent
+Conversely, given $p : \mathcal{S} \to \mathcal{C}$, we can ask: if the fibre
+category $\mathcal{S}_U$ is a groupoid for all $U \in \text{Ob}(\mathcal{C})$,
+must $\mathcal{S}$ be fibred in groupoids over $\mathcal{C}$? We can see the
+answer is no as follows. Start with a category fibred in groupoids
+$p : \mathcal{S} \to \mathcal{C}$. 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.
+
+\begin{example}
+Let $ \text{Ob}(\mathcal{C}) = \{A,B,T\}$ and 
+$\text{Mor}_\mathcal{C}(A,B) = \{f\}$, $\text{Mor}_\mathcal{C}(B,T) = \{g\}$,
+$\text{Mor}_\mathcal{C}(A,T) = \{h\} = \{gf\},$ plus the identity morphism for 
+each object. See the diagram below for a picture of this category. Now let 
+$\text{Ob}(\mathcal{S}) = \{A',B',T'\}$ and 
+$\text{Mor}_\mathcal{S}(A',B') = \emptyset$,  
+$\text{Mor}_\mathcal{S}(B',T') = \{g'\}$,  
+$\text{Mor}_\mathcal{S}(A',T') = \{h'\},$ plus the identity morphisms. The 
+functor $p : \mathcal{S} \to \mathcal{C}$ is obvious. Then for every 
+$U \in \text{Ob}(\mathcal{C})$, $\mathcal{S}_U$ is the category with one 
+object and the identity morphism on that object, so a groupoid, but the 
+morphism $f: A \to B$ cannot be lifted. Similarly, if we declare 
+$\text{Mor}_\mathcal{S}(A',B') = \{f'_1, f'_2\}$ and 
+$ \text{Mor}_\mathcal{S}(A',T') = \{h'\} = \{g'f'_1 \} = \{g'f'_2\}$, then 
+the fibre categories are the same and $f: A \to B$ in the diagram below has 
+two lifts. 
+$$
+\xymatrix{
+B' \ar[r]^{g'} & T' &  & B \ar[r]^g & T & \\
+A' \ar@{-->}[u]^{??} \ar[ru]_{h'} & & \ar@{}[u]^{above} & A \ar[u]^f \ar[ru]_{gf = h} & \\
+}
+$$ 
+\end{example}
+
 \begin{example}
 \label{example-functor-groupoids}
 Suppose that $F : \mathcal{C} \to \text{Groupoids}$ is a contravariant functor
@@ -474,15 +526,15 @@
 and \hyperref[functor-into-2-category]{Remark~\ref{functor-into-2-category}}). 
 For $f : V \to U$ in $\mathcal{C}$ we will
 suggestively write $F(f) = f^\ast$ for the functor from $F(U)$ to $F(V)$. 
-From this we can construct a category fibred
-in groupoids over $\mathcal{C}$ as follows. Define 
+From this we can construct a category fibred in groupoids over $\mathcal{C}$ 
+as follows. Define 
 $$
 \text{ob}(\mathcal{S}) =
-\{(U,x) \mid U\in \text{Ob}(\mathcal{C}), x\in \text{Ob}(F(U)).
+\{(U,x) \mid U\in \text{Ob}(\mathcal{C}), x\in \text{Ob}(F(U)\}.
 $$ 
 For $(U,x), (V,y) \in \text{Ob}(\mathcal{S})$ we define
 $$
-\text{Mor}_\mathcal{S}(y,x) = 
+\text{Mor}_\mathcal{S}((V,y),(U,x)) = 
 \{ (f, \phi) \mid f\in \text{Mor}_\mathcal{C}(V,U), 
 \phi \in \text{Mor}_{F(V)}(y, f^\ast x)\}.
 $$
@@ -492,10 +544,27 @@
 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 $U$ garantees that $\mathcal{S}$ is
-fibred in groupoids over $\mathcal{C}$. (Check this. FIXME?)
+that $F(U)$ is a groupoid for every $U$ guarantees 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 $f$. 
+Uniqueness means $h$ in the diagram on the left determines $(h,\nu)$ on 
+the right:
+$$
+\xymatrix{
+V \ar[r]^f & U & (V,y) \ar[r]^{(f, \phi)} & (U,x) \\
+W \ar@{-->}[u]^h \ar[ru]_g & &
+(W,z) \ar@{-->}[u]^{(h,\nu)} \ar[ru]_{(g, \psi)} & \\
+}
+$$
+Then $\nu = (h^\ast \phi)^{-1} \circ \psi $ and the uniqueness of inverses
+guarantees this is the only lift making the diagram commute.
+
+\noindent
 We will write $\mathcal{S}_F \to \mathcal{C}$ for the resulting functor
-if we want to indicate the dependence on $F$. 
+if we want to indicate the dependence on $F$. Because we can think of 
+objects of $\mathcal{S}_F$ as pairs $(U,x)$, we sometimes say $\mathcal{S}_F$ 
+is a {\it split} category fibed in groupoids.
 \end{example}
 
 \noindent
@@ -514,8 +583,8 @@
 some set, see Sets, \autoref{sets-section-reflection-principle}). Its 
 $1$-morphisms will be functors $F : \mathcal{S} \to \mathcal{S}'$
 such that $p' \circ F = p$, and its $2$-morphisms $t : F \to G$
-will be morphisms of functors such that $p'(t_x) = \text{id}_x$
-for all $x \in \text{Ob}(\mathcal{C})$.
+will be morphisms of functors such that $p'(t_x) = \text{id}_{p(x)}$
+for all $x \in \text{Ob}(\mathcal{S})$.
 \end{definition}
 
 \noindent
@@ -525,10 +594,10 @@
 
 \begin{lemma} 
 \label{lemma-fibred-strict}
-Let $ p : \mathcal{S} \to \mathcal{C}$ be a category
-fibred in groupoids. There exists a functor 
-$F : \mathcal{C} \to \text{Groupoids}$ such that $\mathcal{S}$ is
-equivalent to $\mathcal{S}_F$ over $\mathcal{C}$.
+Let $ p : \mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids.
+There exists a functor $F : \mathcal{C} \to \text{Groupoids}$ such that 
+$\mathcal{S}$ is equivalent to $\mathcal{S}_F$ over $\mathcal{C}$. In other 
+words, every category fibred in groupoids is equivalent to a split one.
 \end{lemma}
 
 \begin{proof} 
@@ -554,3 +623,4 @@
 \bibliographystyle{alpha}
 
 \end{document}
+