
diff -r 91973e94a126 -r 615c8a719a28 text/a_inf_blob.tex
--- a/text/a_inf_blob.tex	Mon Apr 18 22:18:00 2011 -0700
+++ b/text/a_inf_blob.tex	Mon Apr 18 22:28:44 2011 -0700
@@ -192,7 +192,7 @@
 We have $\psi(r) = 0$ since $\psi$ is zero on $(\ge 1)$-simplices.
  
 Second, $\phi\circ\psi$ is the identity up to homotopy by another argument based on the method of acyclic models.
-To each generator $(b, \ol{K})$ of $G_*$ we associate the acyclic subcomplex $D(b)$ defined above.
+To each generator $(b, \ol{K})$ of $\cl{\cC_F}(Y)$ we associate the acyclic subcomplex $D(b)$ defined above.
 Both the identity map and $\phi\circ\psi$ are compatible with this
 collection of acyclic subcomplexes, so by the usual method of acyclic models argument these two maps
 are homotopic.
@@ -248,12 +248,12 @@
 A fiber bundle $F\to E\to Y$ gives an example of a $k$-category over $Y$:
 assign to $p:D\to Y$ the blob complex $\bc_*(p^*(E))$, if $\dim(D) = k$,
 or the fields $\cE(p^*(E))$, if $\dim(D) < k$.
-($p^*(E)$ denotes the pull-back bundle over $D$.)
+(Here $p^*(E)$ denotes the pull-back bundle over $D$.)
 Let $\cF_E$ denote this $k$-category over $Y$.
 We can adapt the homotopy colimit construction (based decompositions of $Y$ into balls) to
 get a chain complex $\cl{\cF_E}(Y)$.
 The proof of Theorem \ref{thm:product} goes through essentially unchanged 
-to show that
+to show the following result.
 \begin{thm}
 Let $F \to E \to Y$ be a fiber bundle and let $\cF_E$ be the $k$-category over $Y$ defined above.
 Then
@@ -270,7 +270,7 @@
 $D\widetilde{\times} M$ is a manifold of dimension $n-k+j$ with a collar structure along the boundary of $D$.
 (If $D\to Y$ is an embedding then $D\widetilde{\times} M$ is just the part of $M$
 lying above $D$.)
-We can define a $k$-category $\cF_M$ based on maps of balls into $Y$ which a re good with respect to $M$.
+We can define a $k$-category $\cF_M$ based on maps of balls into $Y$ which are good with respect to $M$.
 We can again adapt the homotopy colimit construction to
 get a chain complex $\cl{\cF_M}(Y)$.
 The proof of Theorem \ref{thm:product} again goes through essentially unchanged 




