diff -r 62a402dd3e6e -r b138ee4a5938 text/evmap.tex
--- a/text/evmap.tex	Thu Sep 23 18:10:35 2010 -0700
+++ b/text/evmap.tex	Fri Sep 24 15:32:55 2010 -0700
@@ -223,7 +223,6 @@
 \item For balls $B$, the map $U(B) \to \BD_1(B)$, $u\mapsto (B, u, \emptyset)$, is continuous,
 where $U(B) \sub \bc_0(B)$ inherits its topology from $\bc_0(B)$ and the topology on
 $\bc_0(B)$ comes from the generating set $\BD_0(B)$. 
-\nn{don't we need to say more to specify a topology on an $\infty$-dimensional vector space}
 \end{itemize}
 
 We can summarize the above by saying that in the typical continuous family
@@ -277,7 +276,7 @@
 whose value at $p\in P$ is the blob $B^n$ with label $y(p) - r(y(p))$.
 Now define, for $y\in \btc_{0j}$,
 \[
-	h(y) = e(y - r(y)) + c(r(y)) .
+	h(y) = e(y - r(y)) - c(r(y)) .
 \]
 
 We must now verify that $h$ does the job it was intended to do.
@@ -290,22 +289,21 @@
 \end{align*}
 For $x\in \btc_{1j}$ we have
 \begin{align*}
-	\bd h(x) + h(\bd x) &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x - r(\bd_b x)) + c(r(\bd_b x)) - e(\bd_t x) && \\
+	\bd h(x) + h(\bd x) &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x - r(\bd_b x)) - c(r(\bd_b x)) - e(\bd_t x) && \\
 			&= \bd_b(e(x)) + e(\bd_b x) && \text{(since $r(\bd_b x) = 0$)} \\
 			&= x . &&
 \end{align*}
 For $x\in \btc_{0j}$ with $j\ge 1$ we have
 \begin{align*}
-	\bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) - \bd_t(e(x - r(x))) - \bd_t(c(r(x))) + 
-											e(\bd_t x - r(\bd_t x)) + c(r(\bd_t x)) \\
-			&= x - r(x) - \bd_t(c(r(x))) + c(r(\bd_t x)) \\
+	\bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) - \bd_t(e(x - r(x))) + \bd_t(c(r(x))) + 
+											e(\bd_t x - r(\bd_t x)) - c(r(\bd_t x)) \\
+			&= x - r(x) + \bd_t(c(r(x))) - c(r(\bd_t x)) \\
 			&= x - r(x) + r(x) \\
 			&= x.
 \end{align*}
 Here we have used the fact that $\bd_b(c(r(x))) = 0$ since $c(r(x))$ is a $0$-blob diagram, 
-as well as that $\bd_t(e(r(x))) = e(r(\bd_t x))$ 
-\nn{explain why this is true?} 
-and $c(r(\bd_t x)) - \bd_t(c(r(x))) = r(x)$ \nn{explain?}.
+as well as that $\bd_t(e(r(x))) = e(r(\bd_t x))$
+and $\bd_t(c(r(x))) - c(r(\bd_t x))  = r(x)$.
 
 For $x\in \btc_{00}$ we have
 \begin{align*}