--- a/blob1.tex Thu Jun 26 17:56:20 2008 +0000
+++ b/blob1.tex Fri Jun 27 04:24:25 2008 +0000
@@ -611,11 +611,11 @@
\nn{Should say something stronger about uniqueness.
Something like: there is
-a contractible subcomplex of the complex of chain maps
+a contractible subcomplex of the complex of chain maps
$CD_*(X) \otimes \bc_*(X) \to \bc_*(X)$ (0-cells are the maps, 1-cells are homotopies, etc.),
and all choices in the construction lie in the 0-cells of this
contractible subcomplex.
-Or maybe better to say any two choices are homotopic, and
+Or maybe better to say any two choices are homotopic, and
any two homotopies and second order homotopic, and so on.}
\nn{Also need to say something about associativity.
@@ -669,7 +669,7 @@
The strategy for the proof of Proposition \ref{CDprop} is as follows.
We will identify a subcomplex
\[
- G_* \sub CD_*(X) \otimes \bc_*(X)
+ G_* \sub CD_*(X) \otimes \bc_*(X)
\]
on which the evaluation map is uniquely determined (up to homotopy) by the conditions
in \ref{CDprop}.
@@ -678,12 +678,12 @@
\nn{need to be more specific here}
Let $p$ be a singular cell in $CD_*(X)$ and $b$ be a blob diagram in $\bc_*(X)$.
-Roughly speaking, $p\otimes b$ is in $G_*$ if each component $V$ of the support of $p$
+Roughly speaking, $p\otimes b$ is in $G_*$ if each component $V$ of the support of $p$
intersects at most one blob $B$ of $b$.
Since $V \cup B$ might not itself be a ball, we need a more careful and complicated definition.
Choose a metric for $X$.
-We define $p\otimes b$ to be in $G_*$ if there exist $\epsilon > 0$ such that
-$N_\epsilon(b) \cup \supp(p)$ is a union of balls, where $N_\epsilon(b)$ denotes the epsilon
+We define $p\otimes b$ to be in $G_*$ if there exist $\epsilon > 0$ such that
+$\supp(p) \cup N_\epsilon(b)$ is a union of balls, where $N_\epsilon(b)$ denotes the epsilon
neighborhood of the support of $b$.
\nn{maybe also require that $N_\delta(b)$ is a union of balls for all $\delta<\epsilon$.}
@@ -704,15 +704,16 @@
let $p\otimes b$ be a generator of $G_k$.
Let $C \sub X$ be a union of balls (as described above) containing $\supp(p)\cup\supp(b)$.
There is a factorization $p = p' \circ g$, where $g\in \Diff(X)$ and $p'$ is a family of diffeomorphisms which is the identity outside of $C$.
+\scott{Shouldn't this be $p = g\circ p'$?}
Let $b = b'\bullet b''$, where $b' \in \bc_*(C)$ and $b'' \in \bc_0(X\setmin C)$.
-We may assume inductively that $e_X(\bd(p\otimes b))$ has the form $x\bullet g(b'')$, where
+We may assume inductively \scott{why? I don't get this.} that $e_X(\bd(p\otimes b))$ has the form $x\bullet g(b'')$, where
$x \in \bc_*(g(C))$.
Since $\bc_*(g(C))$ is contractible, there exists $y \in \bc_*(g(C))$ such that $\bd y = x$.
\nn{need to say more if degree of $x$ is 0}
Define $e_X(p\otimes b) = y\bullet g(b'')$.
We now show that $e_X$ on $G_*$ is, up to homotopy, independent of the various choices made.
-If we make a different series of choice of the chain $y$ in the previous paragraph,
+If we make a different series of choice of the chain $y$ in the previous paragraph,
we can inductively construct a homotopy between the two sets of choices,
again relying on the contractibility of $\bc_*(g(G))$.
A similar argument shows that this homotopy is unique up to second order homotopy, and so on.
@@ -726,25 +727,25 @@
a homotopy (rel $\bd x$) to $x' \in G_*$, and further that $x'$ and
this homotopy are unique up to iterated homotopy.
-Given $k>0$ and a blob diagram $b$, we say that a cover of $X$ $\cU$ is $k$-compatible with
-$b$ if, for any $\{U_1, \ldots, U_k\} \sub \cU$, the union
+Given $k>0$ and a blob diagram $b$, we say that a cover $\cU$ of $X$ is $k$-compatible with
+$b$ if, for any $\{U_1, \ldots, U_k\} \sub \cU$, the union
$U_1\cup\cdots\cup U_k$ is a union of balls which satisfies the condition used to define $G_*$ above.
-Note that if a family of diffeomorphisms $p$ is adapted to
-$\cU$ and $b$ is a blob diagram occurring in $x$, then $p\otimes b \in G_*$.
+Note that if a family of diffeomorphisms $p$ is adapted to
+$\cU$ and $b$ is a blob diagram occurring in $x$ \scott{huh, what's $x$ here?}, then $p\otimes b \in G_*$.
\nn{maybe emphasize this more; it's one of the main ideas in the proof}
Let $k$ be the degree of $x$ and choose a cover $\cU$ of $X$ such that $\cU$ is
$k$-compatible with each of the (finitely many) blob diagrams occurring in $x$.
-We will use Lemma \ref{extension_lemma} with respect to the cover $\cU$ to
+We will use Lemma \ref{extension_lemma} with respect to the cover $\cU$ to
construct the homotopy to $G_*$.
First we will construct a homotopy $h \in G_*$ from $\bd x$ to a cycle $z$ such that
each family of diffeomorphisms $p$ occurring in $z$ is adapted to $\cU$.
Then we will construct a homotopy (rel boundary) $r$ from $x + h$ to $y$ such that
each family of diffeomorphisms $p$ occurring in $y$ is adapted to $\cU$.
This implies that $y \in G_*$.
-$r$ can also be thought of as a homotopy from $x$ to $y-h \in G_*$, and this is the homotopy we seek.
+The homotopy $r$ can also be thought of as a homotopy from $x$ to $y-h \in G_*$, and this is the homotopy we seek.
-We will define $h$ inductive on bidegrees $(0, k-1), (1, k-2), \ldots, (k-1, 0)$.
+We will define $h$ inductively on bidegrees $(0, k-1), (1, k-2), \ldots, (k-1, 0)$.
Define $h$ to be zero on bidegree $(0, k-1)$.
Let $p\otimes b$ be a generator occurring in $\bd x$ with bidegree $(1, k-2)$.
Using Lemma \ref{extension_lemma}, construct a homotopy $q$ from $p$ to $p'$ which is adapted to $\cU$.
@@ -761,7 +762,7 @@
The homotopy $r$ is constructed similarly.
-\nn{need to say something about uniqueness of $r$, $h$ etc.
+\nn{need to say something about uniqueness of $r$, $h$ etc.
postpone this until second draft.}
At this point, we have finished defining the evaluation map.
--- a/text/hochschild.tex Thu Jun 26 17:56:20 2008 +0000
+++ b/text/hochschild.tex Fri Jun 27 04:24:25 2008 +0000
@@ -71,18 +71,18 @@
%\nn{alternative to the above paragraph:}
Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$.
-We define a blob-like complex $F_*(S^1, (p_i), (M_i))$.
+We define a blob-like complex $K_*(S^1, (p_i), (M_i))$.
The fields have elements of $M_i$ labeling $p_i$ and elements of $C$ labeling
other points.
The blob twig labels lie in kernels of evaluation maps.
(The range of these evaluation maps is a tensor product (over $C$) of $M_i$'s.)
-Let $F_*(M) = F_*(S^1, (*), (M))$, where $* \in S^1$ is some standard base point.
-In other words, fields for $F_*(M)$ have an element of $M$ at the fixed point $*$
+Let $K_*(M) = K_*(S^1, (*), (M))$, where $* \in S^1$ is some standard base point.
+In other words, fields for $K_*(M)$ have an element of $M$ at the fixed point $*$
and elements of $C$ at variable other points.
\todo{Some orphaned questions:}
-\nn{Or maybe we should claim that $M \to F_*(M)$ is the/a derived coend.
-Or maybe that $F_*(M)$ is quasi-isomorphic (or perhaps homotopic) to the Hochschild
+\nn{Or maybe we should claim that $M \to K_*(M)$ is the/a derived coend.
+Or maybe that $K_*(M)$ is quasi-isomorphic (or perhaps homotopic) to the Hochschild
complex of $M$.}
\nn{What else needs to be said to establish quasi-isomorphism to Hochschild complex?
@@ -101,19 +101,19 @@
This follows from two results. First, we see that
\begin{lem}
\label{lem:module-blob}%
-The complex $F_*(C)$ (here $C$ is being thought of as a
+The complex $K_*(C)$ (here $C$ is being thought of as a
$C$-$C$-bimodule, not a category) is quasi-isomorphic to the blob complex
$\bc_*(S^1; C)$. (Proof later.)
\end{lem}
Next, we show that for any $C$-$C$-bimodule $M$,
\begin{prop}
-The complex $F_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual
+The complex $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual
Hochschild complex of $M$.
\end{prop}
\begin{proof}
-First, since we're working over $\Complex$, note that saying two complexes are quasi-isomorphic simply means they have isomorphic homologies.
-\todo{We really don't want to work over $\Complex$, though; it would be nice to talk about torsion!}
+%First, since we're working over $\Complex$, note that saying two complexes are quasi-isomorphic simply means they have isomorphic homologies.
+%\todo{We really don't want to work over $\Complex$, though; it would be nice to talk about torsion!}
Recall that the usual Hochschild complex of $M$ is uniquely determined, up to quasi-isomorphism, by the following properties:
\begin{enumerate}
@@ -129,47 +129,67 @@
$HH_0(M)$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$.
\end{enumerate}
(Together, these just say that Hochschild homology is `the derived functor of coinvariants'.)
-We'll first explain why these properties are characteristic. Take some
-$C$-$C$ bimodule $M$. If $M$ is free, that is, a direct sum of copies of
-$C \tensor C$, then properties \ref{item:hochschild-additive} and
-\ref{item:hochschild-free} determine $HC_*(M)$. Otherwise, choose some
-free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we
-have a short exact sequence $0 \to K \into F \onto M \to 0$, and hence a
-short exact sequence of complexes $0 \to HC_*(K) \into HC_*(F) \onto HC_*(M)
-\to 0$. Such a sequence gives a long exact sequence on homology
+We'll first recall why these properties are characteristic.
+
+Take some $C$-$C$ bimodule $M$, and choose a free resolution
\begin{equation*}
-%\begin{split}
-\cdots \to HH_{i+1}(F) \to HH_{i+1}(M) \to HH_i(K) \to HH_i(F) \to \cdots % \\
-%\cdots \to HH_1(F) \to HH_1(M) \to HH_0(K) \to HH_0(F) \to HH_0(M).
-%\end{split}
+\cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0.
\end{equation*}
-For any $i \geq 1$, $HH_{i+1}(F) = HH_i(F) = 0$, by properties
-\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so
-$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}.
+There's a quotient map $\pi: F_0 \onto M$, and by construction the cone of the chain map $\pi: F_j \to M$ is acyclic. Now construct the total complex
+$HC_i(F_j)$, with $i,j \geq 0$, graded by $i+j$.
+
+Observe that we have two chain maps
+\begin{align*}
+HC_i(F_j) & \xrightarrow{HC_i(\pi)} HC_i(M) \\
+\intertext{and}
+HC_i(F_j) & \xrightarrow{HC_0(F_j) \onto HH_0(F_j)} \operatorname{coinv}(F_j).
+\end{align*}
+The cone of each chain map is acyclic. In the first case, this is because the `rows' indexed by $i$ are acyclic since $HC_i$ is exact.
+In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free.
+
+Because the cones are acyclic, the chain maps are quasi-isomorphisms. Composing one with the inverse of the other, we obtain the desired quasi-isomorphism
+$$HC_*(M) \iso \operatorname{coinv}(F_*).$$
-This tells us how to
-compute every homology group of $HC_*(M)$; we already know $HH_0(M)$
-(it's just coinvariants, by property \ref{item:hochschild-coinvariants}),
-and higher homology groups are determined by lower ones in $HC_*(K)$, and
-hence recursively as coinvariants of some other bimodule.
+%If $M$ is free, that is, a direct sum of copies of
+%$C \tensor C$, then properties \ref{item:hochschild-additive} and
+%\ref{item:hochschild-free} determine $HC_*(M)$. Otherwise, choose some
+%free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we
+%have a short exact sequence $0 \to K \into F \onto M \to 0$, and hence a
+%short exact sequence of complexes $0 \to HC_*(K) \into HC_*(F) \onto HC_*(M)
+%\to 0$. Such a sequence gives a long exact sequence on homology
+%\begin{equation*}
+%%\begin{split}
+%\cdots \to HH_{i+1}(F) \to HH_{i+1}(M) \to HH_i(K) \to HH_i(F) \to \cdots % \\
+%%\cdots \to HH_1(F) \to HH_1(M) \to HH_0(K) \to HH_0(F) \to HH_0(M).
+%%\end{split}
+%\end{equation*}
+%For any $i \geq 1$, $HH_{i+1}(F) = HH_i(F) = 0$, by properties
+%\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so
+%$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}.
+%
+%This tells us how to
+%compute every homology group of $HC_*(M)$; we already know $HH_0(M)$
+%(it's just coinvariants, by property \ref{item:hochschild-coinvariants}),
+%and higher homology groups are determined by lower ones in $HC_*(K)$, and
+%hence recursively as coinvariants of some other bimodule.
-The proposition then follows from the following lemmas, establishing that $F_*$ has precisely these required properties.
+The proposition then follows from the following lemmas, establishing that $K_*$ has precisely these required properties.
\begin{lem}
\label{lem:hochschild-additive}%
-Directly from the definition, $F_*(M_1 \oplus M_2) \cong F_*(M_1) \oplus F_*(M_2)$.
+Directly from the definition, $K_*(M_1 \oplus M_2) \cong K_*(M_1) \oplus K_*(M_2)$.
\end{lem}
\begin{lem}
\label{lem:hochschild-exact}%
An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an
-exact sequence $0 \to F_*(M_1) \into F_*(M_2) \onto F_*(M_3) \to 0$.
+exact sequence $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$.
\end{lem}
\begin{lem}
\label{lem:hochschild-free}%
-$F_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$.
+$K_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$.
\end{lem}
\begin{lem}
\label{lem:hochschild-coinvariants}%
-$H_0(F_*(M))$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$.
+$H_0(K_*(M))$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$.
\end{lem}
The remainder of this section is devoted to proving Lemmas
@@ -179,12 +199,12 @@
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:module-blob}]
-We show that $F_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$.
-$F_*(C)$ differs from $\bc_*(S^1)$ only in that the base point *
-is always a labeled point in $F_*(C)$, while in $\bc_*(S^1)$ it may or may not be.
-In other words, there is an inclusion map $i: F_*(C) \to \bc_*(S^1)$.
+We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$.
+$K_*(C)$ differs from $\bc_*(S^1)$ only in that the base point *
+is always a labeled point in $K_*(C)$, while in $\bc_*(S^1)$ it may or may not be.
+In other words, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$.
-We define a quasi-inverse \nn{right term?} $s: \bc_*(S^1) \to F_*(C)$ to the inclusion as follows.
+We define a quasi-inverse \nn{right term?} $s: \bc_*(S^1) \to K_*(C)$ to the inclusion as follows.
If $y$ is a field defined on a neighborhood of *, define $s(y) = y$ if
* is a labeled point in $y$.
Otherwise, define $s(y)$ to be the result of adding a label 1 (identity morphism) at *.
@@ -193,25 +213,25 @@
$x$ with $y$.
It is easy to check that $s$ is a chain map and $s \circ i = \id$.
-Let $G^\ep_* \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points
+Let $L_*^\ep \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points
in a neighborhood $B_\ep$ of *, except perhaps *.
-Note that for any chain $x \in \bc_*(S^1)$, $x \in G^\ep_*$ for sufficiently small $\ep$.
+Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$.
\nn{rest of argument goes similarly to above}
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}]
\todo{}
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:hochschild-free}]
-We show that $F_*(C\otimes C)$ is
+We show that $K_*(C\otimes C)$ is
quasi-isomorphic to the 0-step complex $C$.
-Let $F'_* \sub F_*(C\otimes C)$ be the subcomplex where the label of
+Let $K'_* \sub K_*(C\otimes C)$ be the subcomplex where the label of
the point $*$ is $1 \otimes 1 \in C\otimes C$.
-We will show that the inclusion $i: F'_* \to F_*(C\otimes C)$ is a quasi-isomorphism.
+We will show that the inclusion $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism.
Fix a small $\ep > 0$.
Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$.
-Let $F^\ep_* \sub F_*(C\otimes C)$ be the subcomplex
+Let $K_*^\ep \sub K_*(C\otimes C)$ be the subcomplex
generated by blob diagrams $b$ such that $B_\ep$ is either disjoint from
or contained in each blob of $b$, and the two boundary points of $B_\ep$ are not labeled points of $b$.
For a field (picture) $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$
@@ -220,8 +240,8 @@
Note that $y - s_\ep(y) \in U(B_\ep)$.
\nn{maybe it's simpler to assume that there are no labeled points, other than $*$, in $B_\ep$.}
-Define a degree 1 chain map $j_\ep : F^\ep_* \to F^\ep_*$ as follows.
-Let $x \in F^\ep_*$ be a blob diagram.
+Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows.
+Let $x \in K_*^\ep$ be a blob diagram.
If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $B_\ep$ to
$x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $B_\ep$.
If $*$ is contained in a twig blob $B$ with label $u = \sum z_i$, $j_\ep(x)$ is obtained as follows.
@@ -231,41 +251,41 @@
Define $j_\ep(x) = \sum x_i$.
\nn{need to check signs coming from blob complex differential}
-Note that if $x \in F'_* \cap F^\ep_*$ then $j_\ep(x) \in F'_*$ also.
+Note that if $x \in K'_* \cap K_*^\ep$ then $j_\ep(x) \in K'_*$ also.
The key property of $j_\ep$ is
\eq{
\bd j_\ep + j_\ep \bd = \id - \sigma_\ep ,
}
-where $\sigma_\ep : F^\ep_* \to F^\ep_*$ is given by replacing the restriction $y$ of each field
-mentioned in $x \in F^\ep_*$ with $s_\ep(y)$.
-Note that $\sigma_\ep(x) \in F'_*$.
+where $\sigma_\ep : K_*^\ep \to K_*^\ep$ is given by replacing the restriction $y$ of each field
+mentioned in $x \in K_*^\ep$ with $s_\ep(y)$.
+Note that $\sigma_\ep(x) \in K'_*$.
-If $j_\ep$ were defined on all of $F_*(C\otimes C)$, it would show that $\sigma_\ep$
-is a homotopy inverse to the inclusion $F'_* \to F_*(C\otimes C)$.
+If $j_\ep$ were defined on all of $K_*(C\otimes C)$, it would show that $\sigma_\ep$
+is a homotopy inverse to the inclusion $K'_* \to K_*(C\otimes C)$.
One strategy would be to try to stitch together various $j_\ep$ for progressively smaller
-$\ep$ and show that $F'_*$ is homotopy equivalent to $F_*(C\otimes C)$.
+$\ep$ and show that $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$.
Instead, we'll be less ambitious and just show that
-$F'_*$ is quasi-isomorphic to $F_*(C\otimes C)$.
+$K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$.
-If $x$ is a cycle in $F_*(C\otimes C)$, then for sufficiently small $\ep$ we have
-$x \in F_*^\ep$.
-(This is true for any chain in $F_*(C\otimes C)$, since chains are sums of
+If $x$ is a cycle in $K_*(C\otimes C)$, then for sufficiently small $\ep$ we have
+$x \in K_*^\ep$.
+(This is true for any chain in $K_*(C\otimes C)$, since chains are sums of
finitely many blob diagrams.)
-Then $x$ is homologous to $s_\ep(x)$, which is in $F'_*$, so the inclusion map
-$F'_* \sub F_*(C\otimes C)$ is surjective on homology.
-If $y \in F_*(C\otimes C)$ and $\bd y = x \in F'_*$, then $y \in F^\ep_*$ for some $\ep$
+Then $x$ is homologous to $s_\ep(x)$, which is in $K'_*$, so the inclusion map
+$K'_* \sub K_*(C\otimes C)$ is surjective on homology.
+If $y \in K_*(C\otimes C)$ and $\bd y = x \in K'_*$, then $y \in K_*^\ep$ for some $\ep$
and
\eq{
\bd y = \bd (\sigma_\ep(y) + j_\ep(x)) .
}
Since $\sigma_\ep(y) + j_\ep(x) \in F'$, it follows that the inclusion map is injective on homology.
-This completes the proof that $F'_*$ is quasi-isomorphic to $F_*(C\otimes C)$.
+This completes the proof that $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$.
-Let $F''_* \sub F'_*$ be the subcomplex of $F'_*$ where $*$ is not contained in any blob.
-We will show that the inclusion $i: F''_* \to F'_*$ is a homotopy equivalence.
+Let $K''_* \sub K'_*$ be the subcomplex of $K'_*$ where $*$ is not contained in any blob.
+We will show that the inclusion $i: K''_* \to K'_*$ is a homotopy equivalence.
-First, a lemma: Let $G''_*$ and $G'_*$ be defined the same as $F''_*$ and $F'_*$, except with
+First, a lemma: Let $G''_*$ and $G'_*$ be defined the same as $K''_*$ and $K'_*$, except with
$S^1$ replaced some (any) neighborhood of $* \in S^1$.
Then $G''_*$ and $G'_*$ are both contractible
and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence.
@@ -278,20 +298,20 @@
Actually, we need the further (easy) result that the inclusion
$G''_* \to G'_*$ induces an isomorphism on $H_0$.
-Next we construct a degree 1 map (homotopy) $h: F'_* \to F'_*$ such that
-for all $x \in F'_*$ we have
+Next we construct a degree 1 map (homotopy) $h: K'_* \to K'_*$ such that
+for all $x \in K'_*$ we have
\eq{
- x - \bd h(x) - h(\bd x) \in F''_* .
+ x - \bd h(x) - h(\bd x) \in K''_* .
}
-Since $F'_0 = F''_0$, we can take $h_0 = 0$.
-Let $x \in F'_1$, with single blob $B \sub S^1$.
-If $* \notin B$, then $x \in F''_1$ and we define $h_1(x) = 0$.
+Since $K'_0 = K''_0$, we can take $h_0 = 0$.
+Let $x \in K'_1$, with single blob $B \sub S^1$.
+If $* \notin B$, then $x \in K''_1$ and we define $h_1(x) = 0$.
If $* \in B$, then we work in the image of $G'_*$ and $G''_*$ (with respect to $B$).
Choose $x'' \in G''_1$ such that $\bd x'' = \bd x$.
Since $G'_*$ is contractible, there exists $y \in G'_2$ such that $\bd y = x - x''$.
Define $h_1(x) = y$.
The general case is similar, except that we have to take lower order homotopies into account.
-Let $x \in F'_k$.
+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 $B$.
@@ -299,12 +319,12 @@
Choose $x'' \in G''_l$ such that $\bd x'' = \bd (x' - h_{l-1}\bd x')$.
Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$.
Define $h_k(x) = y \bullet p$.
-This completes the proof that $i: F''_* \to F'_*$ is a homotopy equivalence.
+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 $F''_*$ is contractible.
+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 $F''_*$.
+Let $x$ be a cycle 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.