--- a/text/hochschild.tex	Wed May 05 22:58:45 2010 -0700
+++ b/text/hochschild.tex	Fri May 07 11:18:39 2010 -0700
@@ -416,8 +416,7 @@
 and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence.
 For $G'_*$ the proof is the same as in (\ref{bcontract}), except that the splitting
 $G'_0 \to H_0(G'_*)$ concentrates the point labels at two points to the right and left of $*$.
-For $G''_*$ we note that any cycle is supported \nn{need to establish terminology for this; maybe
-in ``basic properties" section above} away from $*$.
+For $G''_*$ we note that any cycle is supported away from $*$.
 Thus any cycle lies in the image of the normal blob complex of a disjoint union
 of two intervals, which is contractible by (\ref{bcontract}) and (\ref{disj-union-contract}).
 Finally, it is easy to see that the inclusion
@@ -448,13 +447,25 @@
 This completes the proof that $i: K''_* \to K'_*$ is a homotopy equivalence.
 \nn{need to say above more clearly and settle on notation/terminology}
 
-Finally, we show that $K''_*$ is contractible.
-\nn{need to also show that $H_0$ is the right thing; easy, but I won't do it now}
-Let $x$ be a cycle in $K''_*$.
+Finally, we show that $K''_*$ is contractible with $H_0\cong C$.
+This is similar to the proof of Proposition \ref{bcontract}, but a bit more
+complicated since there is no single blob which contains the support of all blob diagrams
+in $K''_*$.
+Let $x$ be a cycle of degree greater than zero in $K''_*$.
 The union of the supports of the diagrams in $x$ does not contain $*$, so there exists a
 ball $B \subset S^1$ containing the union of the supports and not containing $*$.
-Adding $B$ as a blob to $x$ gives a contraction.
-\nn{need to say something else in degree zero}
+Adding $B$ as an outermost blob to each summand of $x$ gives a chain $y$ with $\bd y = x$.
+Thus $H_i(K''_*) \cong 0$ for $i> 0$ and $K''_*$ is contractible.
+
+To see that $H_0(K''_*) \cong C$, consider the map $p: K''_0 \to C$ which sends a 0-blob
+diagram to the product of its labeled points.
+$p$ is clearly surjective.
+It's also easy to see that $p(\bd K''_1) = 0$.
+Finally, if $p(y) = 0$ then there exists a blob $B \sub S^1$ which contains
+all of the labeled points (other than *) of all of the summands of $y$.
+This allows us to construct $x\in K''_1$ such that $\bd x = y$.
+(The label of $B$ is the restriction of $y$ to $B$.)
+It follows that $H_0(K''_*) \cong C$.
 \end{proof}
 
 \medskip