--- a/text/a_inf_blob.tex	Wed Aug 26 23:10:55 2009 +0000
+++ b/text/a_inf_blob.tex	Tue Sep 15 02:55:39 2009 +0000
@@ -91,7 +91,15 @@
 Then filtration degree 1 chains associated to the four anti-refinemnts
 $KL\to K$, $KL\to L$, $K'L\to L$ and $K'L\to K'$
 give the desired chain connecting $(a, K)$ and $(a, K')$
-(see Figure xxxx).
+(see Figure \ref{zzz4}).
+
+\begin{figure}[!ht]
+\begin{equation*}
+\mathfig{.63}{tempkw/zz4}
+\end{equation*}
+\caption{Connecting $K$ and $K'$ via $L$}
+\label{zzz4}
+\end{figure}
 
 Consider a different choice of decomposition $L'$ in place of $L$ above.
 This leads to a cycle consisting of filtration degree 1 stuff.
@@ -99,9 +107,17 @@
 Choose a decomposition $M$ which has common refinements with each of 
 $K$, $KL$, $L$, $K'L$, $K'$, $K'L'$, $L'$ and $KL'$.
 \nn{need to also require that $KLM$ antirefines to $KM$, etc.}
-Then we have a filtration degree 2 chain, as shown in Figure yyyy, which does the trick.
+Then we have a filtration degree 2 chain, as shown in Figure \ref{zzz5}, which does the trick.
+(Each small triangle in Figure \ref{zzz5} can be filled with a filtration degree 2 chain.)
 For example, ....
 
+\begin{figure}[!ht]
+\begin{equation*}
+\mathfig{1.0}{tempkw/zz5}
+\end{equation*}
+\caption{Filling in $K$-$KL$-$L$-$K'L$-$K'$-$K'L'$-$L'$-$KL'$-$K$}
+\label{zzz5}
+\end{figure}
 
 \end{proof}