
diff -r 195b767cafdb -r d6466180cd66 text/hochschild.tex
--- a/text/hochschild.tex	Mon Apr 26 10:43:42 2010 -0700
+++ b/text/hochschild.tex	Mon Apr 26 21:54:41 2010 -0700
@@ -310,8 +310,34 @@
 \end{align*}
 where this time we use the fact that we're mapping to $\coinv{M}$, not just $M$.
 
-The map $\pi \compose \ev: H_0(K_*(M)) \to \coinv{M}$ is clearly surjective ($\ev$ surjects onto $M$); we now show that it's injective. \todo{}
+The map $\pi \compose \ev: H_0(K_*(M)) \to \coinv{M}$ is clearly surjective ($\ev$ surjects onto $M$); we now show that it's injective.
+This is equivalent to showing that 
+\[
+	\ev\inv(\ker(\pi)) \sub \bd K_1(M) .
+\]
+The above inclusion follows from
+\[
+	\ker(\ev) \sub \bd K_1(M)
+\]
+and
+\[
+	\ker(\pi) \sub \ev(\bd K_1(M)) .
+\]
+Let $x = \sum x_i$ be in the kernel of $\ev$, where each $x_i$ is a configuration of 
+labeled points in $S^1$.
+Since the sum is finite, we can find an interval (blob) $B$ in $S^1$
+such that for each $i$ the $C$-labeled points of $x_i$ all lie to the right of the 
+base point *.
+Let $y_i$ be the restriction of $x_i$ to $B$ and $y = \sum y_i$.
+Let $r$ be the ``empty" field on $S^1 \setmin B$.
+It follows that $y \in U(B)$ and 
+\[
+	\bd(B, y, r) = x .
+\]
+$\ker(\pi)$ is generated by elements of the form $cm - mc$.
+As shown in Figure \ref{fig:hochschild-1-chains}, $cm - mc$ lies in $\ev(\bd K_1(M))$.
 \end{proof}
+
 \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}]
 We show that $K_*(C\otimes C)$ is
 quasi-isomorphic to the 0-step complex $C$. We'll do this in steps, establishing quasi-isomorphisms and homotopy equivalences
@@ -334,7 +360,7 @@
 given by replacing the restriction $y$ to $N_\ep$ of each field
 appearing in an element of  $K_*^\ep$ with $s_\ep(y)$.
 Note that $\sigma_\ep(x) \in K'_*$.
-\begin{figure}[!ht]
+\begin{figure}[t]
 \begin{align*}
 y & = \mathfig{0.2}{hochschild/y} &
 s_\ep(y) & = \mathfig{0.2}{hochschild/sy}
@@ -413,7 +439,8 @@
 Let $x \in K'_k$.
 If $*$ is not contained in any of the blobs of $x$, then define $h_k(x) = 0$.
 Otherwise, let $B$ be the outermost blob of $x$ containing $*$.
-By xxxx above, $x = x' \bullet p$, where $x'$ is supported on $B$ and $p$ is supported away from $B$.
+We can decompose $x = x' \bullet p$, 
+where $x'$ is supported on $B$ and $p$ is supported away from $B$.
 So $x' \in G'_l$ for some $l \le k$.
 Choose $x'' \in G''_l$ such that $\bd x'' = \bd (x' - h_{l-1}\bd x')$.
 Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$.
@@ -456,7 +483,7 @@
 In degree 1, we send $m\ot a$ to the sum of two 1-blob diagrams
 as shown in Figure \ref{fig:hochschild-1-chains}.
 
-\begin{figure}[!ht]
+\begin{figure}[t]
 \begin{equation*}
 \mathfig{0.4}{hochschild/1-chains}
 \end{equation*}
@@ -467,7 +494,7 @@
 \label{fig:hochschild-1-chains}
 \end{figure}
 
-\begin{figure}[!ht]
+\begin{figure}[t]
 \begin{equation*}
 \mathfig{0.6}{hochschild/2-chains-0}
 \end{equation*}
@@ -478,7 +505,7 @@
 \label{fig:hochschild-2-chains}
 \end{figure}
 
-\begin{figure}[!ht]
+\begin{figure}[t]
 \begin{equation*}
 A = \mathfig{0.1}{hochschild/v_1} + \mathfig{0.1}{hochschild/v_2} + \mathfig{0.1}{hochschild/v_3} + \mathfig{0.1}{hochschild/v_4}
 \end{equation*}




