
diff -r 9744833c9b90 -r 700ac2678d00 text/hochschild.tex
--- a/text/hochschild.tex	Tue Jul 29 22:37:25 2008 +0000
+++ b/text/hochschild.tex	Mon Aug 04 20:34:48 2008 +0000
@@ -1,444 +1,449 @@
-In this section we analyze the blob complex in dimension $n=1$
-and find that for $S^1$ the homology of the blob complex is the
-Hochschild homology of the category (algebroid) that we started with.
-\nn{or maybe say here that the complexes are quasi-isomorphic?  in general,
-should perhaps put more emphasis on the complexes and less on the homology.}
-
-Notation: $HB_i(X) = H_i(\bc_*(X))$.
-
-Let us first note that there is no loss of generality in assuming that our system of
-fields comes from a category.
-(Or maybe (???) there {\it is} a loss of generality.
-Given any system of fields, $A(I; a, b) = \cC(I; a, b)/U(I; a, b)$ can be
-thought of as the morphisms of a 1-category $C$.
-More specifically, the objects of $C$ are $\cC(pt)$, the morphisms from $a$ to $b$
-are $A(I; a, b)$, and composition is given by gluing.
-If we instead take our fields to be $C$-pictures, the $\cC(pt)$ does not change
-and neither does $A(I; a, b) = HB_0(I; a, b)$.
-But what about $HB_i(I; a, b)$ for $i > 0$?
-Might these higher blob homology groups be different?
-Seems unlikely, but I don't feel like trying to prove it at the moment.
-In any case, we'll concentrate on the case of fields based on 1-category
-pictures for the rest of this section.)
-
-(Another question: $\bc_*(I)$ is an $A_\infty$-category.
-How general of an $A_\infty$-category is it?
-Given an arbitrary $A_\infty$-category can one find fields and local relations so
-that $\bc_*(I)$ is in some sense equivalent to the original $A_\infty$-category?
-Probably not, unless we generalize to the case where $n$-morphisms are complexes.)
-
-Continuing...
-
-Let $C$ be a *-1-category.
-Then specializing the definitions from above to the case $n=1$ we have:
-\begin{itemize}
-\item $\cC(pt) = \ob(C)$ .
-\item Let $R$ be a 1-manifold and $c \in \cC(\bd R)$.
-Then an element of $\cC(R; c)$ is a collection of (transversely oriented)
-points in the interior
-of $R$, each labeled by a morphism of $C$.
-The intervals between the points are labeled by objects of $C$, consistent with
-the boundary condition $c$ and the domains and ranges of the point labels.
-\item There is an evaluation map $e: \cC(I; a, b) \to \mor(a, b)$ given by
-composing the morphism labels of the points.
-Note that we also need the * of *-1-category here in order to make all the morphisms point
-the same way.
-\item For $x \in \mor(a, b)$ let $\chi(x) \in \cC(I; a, b)$ be the field with a single
-point (at some standard location) labeled by $x$.
-Then the kernel of the evaluation map $U(I; a, b)$ is generated by things of the
-form $y - \chi(e(y))$.
-Thus we can, if we choose, restrict the blob twig labels to things of this form.
-\end{itemize}
-
-We want to show that $HB_*(S^1)$ is naturally isomorphic to the
-Hochschild homology of $C$.
-\nn{Or better that the complexes are homotopic
-or quasi-isomorphic.}
-In order to prove this we will need to extend the blob complex to allow points to also
-be labeled by elements of $C$-$C$-bimodules.
-%Given an interval (1-ball) so labeled, there is an evaluation map to some tensor product
-%(over $C$) of $C$-$C$-bimodules.
-%Define the local relations $U(I; a, b)$ to be the direct sum of the kernels of these maps.
-%Now we can define the blob complex for $S^1$.
-%This complex is the sum of complexes with a fixed cyclic tuple of bimodules present.
-%If $M$ is a $C$-$C$-bimodule, let $G_*(M)$ denote the summand of $\bc_*(S^1)$ corresponding
-%to the cyclic 1-tuple $(M)$.
-%In other words, $G_*(M)$ is a blob-like complex where exactly one point is labeled
-%by an element of $M$ and the remaining points are labeled by morphisms of $C$.
-%It's clear that $G_*(C)$ is isomorphic to the original bimodule-less
-%blob complex for $S^1$.
-%\nn{Is it really so clear?  Should say more.}
-
-%\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 $K_*(S^1, (p_i), (M_i))$.
-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 $K_*(M) = K_*(S^1, (*), (M))$, 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 $M \to K_*(M)$ 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?
-Do we need a map from hoch to blob?
-Does the above exactness and contractibility guarantee such a map without writing it
-down explicitly?
-Probably it's worth writing down an explicit map even if we don't need to.}
-
-
-We claim that
-\begin{thm}
-The blob complex $\bc_*(S^1; C)$ on the circle is quasi-isomorphic to the
-usual Hochschild complex for $C$.
-\end{thm}
-
-This follows from two results. First, we see that
-\begin{lem}
-\label{lem:module-blob}%
-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 $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual
-Hochschild complex of $M$.
-\end{prop}
-\begin{proof}
-Recall that the usual Hochschild complex of $M$ is uniquely determined,
-up to quasi-isomorphism, by the following properties:
-\begin{enumerate}
-\item \label{item:hochschild-additive}%
-$HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$.
-\item \label{item:hochschild-exact}%
-An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an
-exact sequence $0 \to HC_*(M_1) \into HC_*(M_2) \onto HC_*(M_3) \to 0$.
-\item \label{item:hochschild-coinvariants}%
-$HH_0(M)$ is isomorphic to the coinvariants of $M$, $\coinv(M) =
-M/\langle cm-mc \rangle$.
-\item \label{item:hochschild-free}%
-$HC_*(C\otimes C)$ (the free $C$-$C$-bimodule with one generator) is contractible; that is,
-quasi-isomorphic to its $0$-th homology (which in turn, by \ref{item:hochschild-coinvariants}, is just $C$) via the quotient map $HC_0 \onto HH_0$.
-\end{enumerate}
-(Together, these just say that Hochschild homology is `the derived functor of coinvariants'.)
-We'll first recall why these properties are characteristic.
-
-Take some $C$-$C$ bimodule $M$, and choose a free resolution
-\begin{equation*}
-\cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0.
-\end{equation*}
-We will show that for any functor $\cP$ satisfying properties
-\ref{item:hochschild-additive}, \ref{item:hochschild-exact},
-\ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there
-is a quasi-isomorphism
-$$\cP_*(M) \iso \coinv(F_*).$$
-%
-Observe that there's a quotient map $\pi: F_0 \onto M$, and by
-construction the cone of the chain map $\pi: F_* \to M$ is acyclic. Now
-construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by
-$i+j$. We have two chain maps
-\begin{align*}
-\cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\
-\intertext{and}
-\cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \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
-$$\cP_*(M) \quismto \coinv(F_*).$$
-
-%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 $K_*$ has precisely these required properties.
-\begin{lem}
-\label{lem:hochschild-additive}%
-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 $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$.
-\end{lem}
-\begin{lem}
-\label{lem:hochschild-coinvariants}%
-$H_0(K_*(M))$ is isomorphic to the coinvariants of $M$.
-\end{lem}
-\begin{lem}
-\label{lem:hochschild-free}%
-$K_*(C\otimes C)$ is quasi-isomorphic to $H_0(K_*(C \otimes C)) \iso C$.
-\end{lem}
-
-The remainder of this section is devoted to proving Lemmas
-\ref{lem:module-blob},
-\ref{lem:hochschild-exact}, \ref{lem:hochschild-coinvariants} and
-\ref{lem:hochschild-free}.
-\end{proof}
-
-\begin{proof}[Proof of Lemma \ref{lem:module-blob}]
-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 left inverse $s: \bc_*(S^1) \to K_*(C)$ 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 *.
-Let $x \in \bc_*(S^1)$.
-Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in
-$x$ with $s(y)$.
-It is easy to check that $s$ is a chain map and $s \circ i = \id$.
-
-Let $L_*^\ep \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points
-in a neighborhood $B_\ep$ of $*$, except perhaps $*$, and $B_\ep$ is either disjoint from or contained in every blob.
-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}
-
-We define a degree $1$ chain map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. Let $x \in L_*^\ep$ be a blob diagram.
-If $*$ is not contained in any twig blob, we define $j_\ep(x)$ by adding $B_\ep$ as a new twig blob, with label $y - s(y)$ where $y$ is the restriction
-of $x$ to $B_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$,
-write $y_i$ for the restriction of $z_i$ to $B_\ep$, and let
-$x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin B_\ep$,
-and have an additional blob $B_\ep$ with label $y_i - s(y_i)$.
-Define $j_\ep(x) = \sum x_i$.
-\todo{need to check signs coming from blob complex differential}
-\todo{finish this}
-\end{proof}
-\begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}]
-We now prove that $K_*$ is an exact functor.
-
-%\todo{p. 1478 of scott's notes}
-Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules
-\begin{equation*}
-M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M)
-\end{equation*}
-is exact. For completeness we'll explain this below.
-
-Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$
-We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor.
-Most of what we need to check is easy.
-If $(a \tensor k \tensor b) \in \ker(C \tensor K \tensor C \to K)$, to have $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$, we must have $f(k) = 0 \in E$, so
-be $k=0$ itself. If $(a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, clearly
-$e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$.
-If $\hat{g}(a \tensor e \tensor b) = 0$, then $g(e) = 0$, so $e = f(\widetilde{e})$ for some $\widetilde{e} \in K$, and $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$.
-Finally, the interesting step is in checking that any $q = \sum_i a_i \tensor q_i \tensor b_i$ such that $\sum_i a_i q_i b_i = 0$ is in the image of $\ker(C \tensor E \tensor C \to C)$ under $\hat{g}$.
-For each $i$, we can find $\widetilde{q_i}$ so $g(\widetilde{q_i}) = q_i$. However $\sum_i a_i \widetilde{q_i} b_i$ need not be zero.
-Consider then $$\widetilde{q} = \sum_i (a_i \tensor \widetilde{q_i} \tensor b_i) - 1 \tensor (\sum_i a_i \widetilde{q_i} b_i) \tensor 1.$$ Certainly
-$\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further,
-\begin{align*}
-\hat{g}(\widetilde{q}) & = \sum_i (a_i \tensor g(\widetilde{q_i}) \tensor b_i) - 1 \tensor (\sum_i a_i g(\widetilde{q_i}) b_i) \tensor 1 \\
-                       & = q - 0
-\end{align*}
-(here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$).
-
-Identical arguments show that the functors
-\begin{equation}
-\label{eq:ker-functor}%
-M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M)
-\end{equation}
-are all exact too. Moreover, tensor products of such functors with each
-other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M)
-\tensor C \tensor \ker(C \tensor C \to M)$) are all still exact.
-
-Finally, then we see that the functor $K_*$ is simply an (infinite)
-direct sum of copies of this sort of functor. The direct sum is indexed by
-configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors,
-with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$, and all other labelled points corresponding
-to tensor factors of $C$.
-\end{proof}
-\begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}]
-We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$.
-
-We define a map $\ev: K_0(M) \to M$. If $x \in K_0(M)$ has the label $m \in M$ at $*$, and labels $c_i \in C$ at the other labeled points of $S^1$, reading clockwise from $*$,
-we set $\ev(x) = m c_1 \cdots c_k$. We can think of this as $\ev : M \tensor C^{\tensor k} \to M$, for each direct summand of $K_0(M)$ indexed by a configuration of labeled points.
-There is a quotient map $\pi: M \to \coinv{M}$, and the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; if $y \in K_1(M)$, the blob in $y$ either contains $*$ or does not. If it doesn't, then
-suppose $y$ has label $m$ at $*$, labels $c_i$ at other labeled points outside the blob, and the field inside the blob is a sum, with the $j$-th term having
-labeled points $d_{j,i}$. Then $\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \in \ker(\DirectSum_k C^{\tensor k} \to C)$, and so
-$\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$
-Similarly, if $*$ is contained in the blob, then the blob label is a sum, with the $j$-th term have labelled points $d_{j,i}$ to the left of $*$, $m_j$ at $*$, and $d_{j,i}'$ to the right of $*$,
-and there are labels $c_i$ at the labeled points outside the blob. We know that
-$$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$
-and so
-\begin{align*}
-\ev(\bdy y) & = \sum_j m_j d_{j,1}' \cdots d_{j,k'_j}' c_1 \cdots c_k d_{j,1} \cdots d_{j,k_j} \\
-            & = \sum_j d_{j,1} \cdots d_{j,k_j} m_j d_{j,1}' \cdots d_{j,k'_j}' c_1 \cdots c_k \\
-            & = 0
-\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{}
-\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
-$$K_*(C \tensor C) \quismto K'_* \htpyto K''_* \quismto C.$$
-
-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 $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism.
-
-Fix a small $\ep > 0$.
-Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$.
-Let $K_*^\ep \sub K_*(C\otimes C)$ 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 only labeled point inside $B_\ep$ is $*$.
-%and the two boundary points of $B_\ep$ are not labeled points of $b$.
-For a field $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$
-labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$.
-(See Figure \ref{fig:sy}.) Note that $y - s_\ep(y) \in U(B_\ep)$. We can think of
-$\sigma_\ep$ as a chain map $K_*^\ep \to K_*^\ep$ given by replacing the restriction $y$ to $B_\ep$ 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{align*}
-y & = \mathfig{0.2}{hochschild/y} &
-s_\ep(y) & = \mathfig{0.2}{hochschild/sy}
-\end{align*}
-\caption{Defining $s_\ep$.}
-\label{fig:sy}
-\end{figure}
-
-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.
-Let $y_i$ be the restriction of $z_i$ to $B_\ep$.
-Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin B_\ep$,
-and have an additional blob $B_\ep$ with label $y_i - s_\ep(y_i)$.
-Define $j_\ep(x) = \sum x_i$.
-\nn{need to check signs coming from blob complex differential}
-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.
-}
-If $j_\ep$ were defined on all of $K_*(C\otimes C)$, this 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 $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$.
-Instead, we'll be less ambitious and just show that
-$K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$.
-
-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 $K'_*$, 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 $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$.
-
-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 $K''_*$ 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.
-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 $*$.
-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{disjunion}).
-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) $h: K'_* \to K'_*$ such that
-for all $x \in K'_*$ we have
-\eq{
-    x - \bd h(x) - h(\bd x) \in K''_* .
-}
-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 $x \in K'_k$.
-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 $B$.
-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 $\bd y = x' - x'' - h_{l-1}\bd x'$.
-Define $h_k(x) = y \bullet p$.
-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 $K''_*$.
-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}
-\end{proof}
-
-We can also describe explicitly a map from the standard Hochschild
-complex to the blob complex on the circle. \nn{What properties does this
-map have?}
-
-\begin{figure}%
-$$\mathfig{0.6}{barycentric/barycentric}$$
-\caption{The Hochschild chain $a \tensor b \tensor c$ is sent to
-the sum of six blob $2$-chains, corresponding to a barycentric subdivision of a $2$-simplex.}
-\label{fig:Hochschild-example}%
-\end{figure}
-
-As an example, Figure \ref{fig:Hochschild-example} shows the image of the Hochschild chain $a \tensor b \tensor c$. Only the $0$-cells are shown explicitly.
-The edges marked $x, y$ and $z$ carry the $1$-chains
-\begin{align*}
-x & = \mathfig{0.1}{barycentric/ux} & u_x = \mathfig{0.1}{barycentric/ux_ca} - \mathfig{0.1}{barycentric/ux_c-a} \\
-y & = \mathfig{0.1}{barycentric/uy} & u_y = \mathfig{0.1}{barycentric/uy_cab} - \mathfig{0.1}{barycentric/uy_ca-b} \\
-z & = \mathfig{0.1}{barycentric/uz} & u_z = \mathfig{0.1}{barycentric/uz_c-a-b} - \mathfig{0.1}{barycentric/uz_cab}
-\end{align*}
-and the $2$-chain labelled $A$ is
-\begin{equation*}
-A = \mathfig{0.1}{barycentric/Ax}+\mathfig{0.1}{barycentric/Ay}.
-\end{equation*}
-Note that we then have
-\begin{equation*}
-\bdy A = x+y+z.
-\end{equation*}
-
-In general, the Hochschild chain $\Tensor_{i=1}^n a_i$ is sent to the sum of $n!$ blob $(n-1)$-chains, indexed by permutations,
-$$\phi\left(\Tensor_{i=1}^n a_i\right) = \sum_{\pi} \phi^\pi(a_1, \ldots, a_n)$$
-with ... (hmmm, problems making this precise; you need to decide where to put the labels, but then it's hard to make an honest chain map!)
+In this section we analyze the blob complex in dimension $n=1$
+and find that for $S^1$ the homology of the blob complex is the
+Hochschild homology of the category (algebroid) that we started with.
+\nn{or maybe say here that the complexes are quasi-isomorphic?  in general,
+should perhaps put more emphasis on the complexes and less on the homology.}
+
+Notation: $HB_i(X) = H_i(\bc_*(X))$.
+
+Let us first note that there is no loss of generality in assuming that our system of
+fields comes from a category.
+(Or maybe (???) there {\it is} a loss of generality.
+Given any system of fields, $A(I; a, b) = \cC(I; a, b)/U(I; a, b)$ can be
+thought of as the morphisms of a 1-category $C$.
+More specifically, the objects of $C$ are $\cC(pt)$, the morphisms from $a$ to $b$
+are $A(I; a, b)$, and composition is given by gluing.
+If we instead take our fields to be $C$-pictures, the $\cC(pt)$ does not change
+and neither does $A(I; a, b) = HB_0(I; a, b)$.
+But what about $HB_i(I; a, b)$ for $i > 0$?
+Might these higher blob homology groups be different?
+Seems unlikely, but I don't feel like trying to prove it at the moment.
+In any case, we'll concentrate on the case of fields based on 1-category
+pictures for the rest of this section.)
+
+(Another question: $\bc_*(I)$ is an $A_\infty$-category.
+How general of an $A_\infty$-category is it?
+Given an arbitrary $A_\infty$-category can one find fields and local relations so
+that $\bc_*(I)$ is in some sense equivalent to the original $A_\infty$-category?
+Probably not, unless we generalize to the case where $n$-morphisms are complexes.)
+
+Continuing...
+
+Let $C$ be a *-1-category.
+Then specializing the definitions from above to the case $n=1$ we have:
+\begin{itemize}
+\item $\cC(pt) = \ob(C)$ .
+\item Let $R$ be a 1-manifold and $c \in \cC(\bd R)$.
+Then an element of $\cC(R; c)$ is a collection of (transversely oriented)
+points in the interior
+of $R$, each labeled by a morphism of $C$.
+The intervals between the points are labeled by objects of $C$, consistent with
+the boundary condition $c$ and the domains and ranges of the point labels.
+\item There is an evaluation map $e: \cC(I; a, b) \to \mor(a, b)$ given by
+composing the morphism labels of the points.
+Note that we also need the * of *-1-category here in order to make all the morphisms point
+the same way.
+\item For $x \in \mor(a, b)$ let $\chi(x) \in \cC(I; a, b)$ be the field with a single
+point (at some standard location) labeled by $x$.
+Then the kernel of the evaluation map $U(I; a, b)$ is generated by things of the
+form $y - \chi(e(y))$.
+Thus we can, if we choose, restrict the blob twig labels to things of this form.
+\end{itemize}
+
+We want to show that $HB_*(S^1)$ is naturally isomorphic to the
+Hochschild homology of $C$.
+\nn{Or better that the complexes are homotopic
+or quasi-isomorphic.}
+In order to prove this we will need to extend the blob complex to allow points to also
+be labeled by elements of $C$-$C$-bimodules.
+%Given an interval (1-ball) so labeled, there is an evaluation map to some tensor product
+%(over $C$) of $C$-$C$-bimodules.
+%Define the local relations $U(I; a, b)$ to be the direct sum of the kernels of these maps.
+%Now we can define the blob complex for $S^1$.
+%This complex is the sum of complexes with a fixed cyclic tuple of bimodules present.
+%If $M$ is a $C$-$C$-bimodule, let $G_*(M)$ denote the summand of $\bc_*(S^1)$ corresponding
+%to the cyclic 1-tuple $(M)$.
+%In other words, $G_*(M)$ is a blob-like complex where exactly one point is labeled
+%by an element of $M$ and the remaining points are labeled by morphisms of $C$.
+%It's clear that $G_*(C)$ is isomorphic to the original bimodule-less
+%blob complex for $S^1$.
+%\nn{Is it really so clear?  Should say more.}
+
+%\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 $K_*(S^1, (p_i), (M_i))$.
+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 $K_*(M) = K_*(S^1, (*), (M))$, 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 $M \to K_*(M)$ 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?
+Do we need a map from hoch to blob?
+Does the above exactness and contractibility guarantee such a map without writing it
+down explicitly?
+Probably it's worth writing down an explicit map even if we don't need to.}
+
+
+We claim that
+\begin{thm}
+The blob complex $\bc_*(S^1; C)$ on the circle is quasi-isomorphic to the
+usual Hochschild complex for $C$.
+\end{thm}
+
+\nn{Note that since both complexes are free (in particular, projective),
+quasi-isomorphic implies homotopy equivalent.  
+This applies to the two claims below also.
+Thanks to Peter Teichner for pointing this out to me.}
+
+This follows from two results. First, we see that
+\begin{lem}
+\label{lem:module-blob}%
+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 $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual
+Hochschild complex of $M$.
+\end{prop}
+\begin{proof}
+Recall that the usual Hochschild complex of $M$ is uniquely determined,
+up to quasi-isomorphism, by the following properties:
+\begin{enumerate}
+\item \label{item:hochschild-additive}%
+$HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$.
+\item \label{item:hochschild-exact}%
+An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an
+exact sequence $0 \to HC_*(M_1) \into HC_*(M_2) \onto HC_*(M_3) \to 0$.
+\item \label{item:hochschild-coinvariants}%
+$HH_0(M)$ is isomorphic to the coinvariants of $M$, $\coinv(M) =
+M/\langle cm-mc \rangle$.
+\item \label{item:hochschild-free}%
+$HC_*(C\otimes C)$ (the free $C$-$C$-bimodule with one generator) is contractible; that is,
+quasi-isomorphic to its $0$-th homology (which in turn, by \ref{item:hochschild-coinvariants}, is just $C$) via the quotient map $HC_0 \onto HH_0$.
+\end{enumerate}
+(Together, these just say that Hochschild homology is `the derived functor of coinvariants'.)
+We'll first recall why these properties are characteristic.
+
+Take some $C$-$C$ bimodule $M$, and choose a free resolution
+\begin{equation*}
+\cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0.
+\end{equation*}
+We will show that for any functor $\cP$ satisfying properties
+\ref{item:hochschild-additive}, \ref{item:hochschild-exact},
+\ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there
+is a quasi-isomorphism
+$$\cP_*(M) \iso \coinv(F_*).$$
+%
+Observe that there's a quotient map $\pi: F_0 \onto M$, and by
+construction the cone of the chain map $\pi: F_* \to M$ is acyclic. Now
+construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by
+$i+j$. We have two chain maps
+\begin{align*}
+\cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\
+\intertext{and}
+\cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \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
+$$\cP_*(M) \quismto \coinv(F_*).$$
+
+%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 $K_*$ has precisely these required properties.
+\begin{lem}
+\label{lem:hochschild-additive}%
+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 $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$.
+\end{lem}
+\begin{lem}
+\label{lem:hochschild-coinvariants}%
+$H_0(K_*(M))$ is isomorphic to the coinvariants of $M$.
+\end{lem}
+\begin{lem}
+\label{lem:hochschild-free}%
+$K_*(C\otimes C)$ is quasi-isomorphic to $H_0(K_*(C \otimes C)) \iso C$.
+\end{lem}
+
+The remainder of this section is devoted to proving Lemmas
+\ref{lem:module-blob},
+\ref{lem:hochschild-exact}, \ref{lem:hochschild-coinvariants} and
+\ref{lem:hochschild-free}.
+\end{proof}
+
+\begin{proof}[Proof of Lemma \ref{lem:module-blob}]
+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 left inverse $s: \bc_*(S^1) \to K_*(C)$ 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 *.
+Let $x \in \bc_*(S^1)$.
+Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in
+$x$ with $s(y)$.
+It is easy to check that $s$ is a chain map and $s \circ i = \id$.
+
+Let $L_*^\ep \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points
+in a neighborhood $B_\ep$ of $*$, except perhaps $*$, and $B_\ep$ is either disjoint from or contained in every blob.
+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}
+
+We define a degree $1$ chain map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. Let $x \in L_*^\ep$ be a blob diagram.
+If $*$ is not contained in any twig blob, we define $j_\ep(x)$ by adding $B_\ep$ as a new twig blob, with label $y - s(y)$ where $y$ is the restriction
+of $x$ to $B_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$,
+write $y_i$ for the restriction of $z_i$ to $B_\ep$, and let
+$x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin B_\ep$,
+and have an additional blob $B_\ep$ with label $y_i - s(y_i)$.
+Define $j_\ep(x) = \sum x_i$.
+\todo{need to check signs coming from blob complex differential}
+\todo{finish this}
+\end{proof}
+\begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}]
+We now prove that $K_*$ is an exact functor.
+
+%\todo{p. 1478 of scott's notes}
+Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules
+\begin{equation*}
+M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M)
+\end{equation*}
+is exact. For completeness we'll explain this below.
+
+Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$
+We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor.
+Most of what we need to check is easy.
+If $(a \tensor k \tensor b) \in \ker(C \tensor K \tensor C \to K)$, to have $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$, we must have $f(k) = 0 \in E$, so
+be $k=0$ itself. If $(a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, clearly
+$e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$.
+If $\hat{g}(a \tensor e \tensor b) = 0$, then $g(e) = 0$, so $e = f(\widetilde{e})$ for some $\widetilde{e} \in K$, and $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$.
+Finally, the interesting step is in checking that any $q = \sum_i a_i \tensor q_i \tensor b_i$ such that $\sum_i a_i q_i b_i = 0$ is in the image of $\ker(C \tensor E \tensor C \to C)$ under $\hat{g}$.
+For each $i$, we can find $\widetilde{q_i}$ so $g(\widetilde{q_i}) = q_i$. However $\sum_i a_i \widetilde{q_i} b_i$ need not be zero.
+Consider then $$\widetilde{q} = \sum_i (a_i \tensor \widetilde{q_i} \tensor b_i) - 1 \tensor (\sum_i a_i \widetilde{q_i} b_i) \tensor 1.$$ Certainly
+$\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further,
+\begin{align*}
+\hat{g}(\widetilde{q}) & = \sum_i (a_i \tensor g(\widetilde{q_i}) \tensor b_i) - 1 \tensor (\sum_i a_i g(\widetilde{q_i}) b_i) \tensor 1 \\
+                       & = q - 0
+\end{align*}
+(here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$).
+
+Identical arguments show that the functors
+\begin{equation}
+\label{eq:ker-functor}%
+M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M)
+\end{equation}
+are all exact too. Moreover, tensor products of such functors with each
+other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M)
+\tensor C \tensor \ker(C \tensor C \to M)$) are all still exact.
+
+Finally, then we see that the functor $K_*$ is simply an (infinite)
+direct sum of copies of this sort of functor. The direct sum is indexed by
+configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors,
+with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$, and all other labelled points corresponding
+to tensor factors of $C$.
+\end{proof}
+\begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}]
+We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$.
+
+We define a map $\ev: K_0(M) \to M$. If $x \in K_0(M)$ has the label $m \in M$ at $*$, and labels $c_i \in C$ at the other labeled points of $S^1$, reading clockwise from $*$,
+we set $\ev(x) = m c_1 \cdots c_k$. We can think of this as $\ev : M \tensor C^{\tensor k} \to M$, for each direct summand of $K_0(M)$ indexed by a configuration of labeled points.
+There is a quotient map $\pi: M \to \coinv{M}$, and the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; if $y \in K_1(M)$, the blob in $y$ either contains $*$ or does not. If it doesn't, then
+suppose $y$ has label $m$ at $*$, labels $c_i$ at other labeled points outside the blob, and the field inside the blob is a sum, with the $j$-th term having
+labeled points $d_{j,i}$. Then $\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \in \ker(\DirectSum_k C^{\tensor k} \to C)$, and so
+$\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$
+Similarly, if $*$ is contained in the blob, then the blob label is a sum, with the $j$-th term have labelled points $d_{j,i}$ to the left of $*$, $m_j$ at $*$, and $d_{j,i}'$ to the right of $*$,
+and there are labels $c_i$ at the labeled points outside the blob. We know that
+$$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$
+and so
+\begin{align*}
+\ev(\bdy y) & = \sum_j m_j d_{j,1}' \cdots d_{j,k'_j}' c_1 \cdots c_k d_{j,1} \cdots d_{j,k_j} \\
+            & = \sum_j d_{j,1} \cdots d_{j,k_j} m_j d_{j,1}' \cdots d_{j,k'_j}' c_1 \cdots c_k \\
+            & = 0
+\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{}
+\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
+$$K_*(C \tensor C) \quismto K'_* \htpyto K''_* \quismto C.$$
+
+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 $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism.
+
+Fix a small $\ep > 0$.
+Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$.
+Let $K_*^\ep \sub K_*(C\otimes C)$ 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 only labeled point inside $B_\ep$ is $*$.
+%and the two boundary points of $B_\ep$ are not labeled points of $b$.
+For a field $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$
+labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$.
+(See Figure \ref{fig:sy}.) Note that $y - s_\ep(y) \in U(B_\ep)$. We can think of
+$\sigma_\ep$ as a chain map $K_*^\ep \to K_*^\ep$ given by replacing the restriction $y$ to $B_\ep$ 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{align*}
+y & = \mathfig{0.2}{hochschild/y} &
+s_\ep(y) & = \mathfig{0.2}{hochschild/sy}
+\end{align*}
+\caption{Defining $s_\ep$.}
+\label{fig:sy}
+\end{figure}
+
+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.
+Let $y_i$ be the restriction of $z_i$ to $B_\ep$.
+Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin B_\ep$,
+and have an additional blob $B_\ep$ with label $y_i - s_\ep(y_i)$.
+Define $j_\ep(x) = \sum x_i$.
+\nn{need to check signs coming from blob complex differential}
+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.
+}
+If $j_\ep$ were defined on all of $K_*(C\otimes C)$, this 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 $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$.
+Instead, we'll be less ambitious and just show that
+$K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$.
+
+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 $K'_*$, 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 $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$.
+
+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 $K''_*$ 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.
+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 $*$.
+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{disjunion}).
+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) $h: K'_* \to K'_*$ such that
+for all $x \in K'_*$ we have
+\eq{
+    x - \bd h(x) - h(\bd x) \in K''_* .
+}
+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 $x \in K'_k$.
+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 $B$.
+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 $\bd y = x' - x'' - h_{l-1}\bd x'$.
+Define $h_k(x) = y \bullet p$.
+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 $K''_*$.
+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}
+\end{proof}
+
+We can also describe explicitly a map from the standard Hochschild
+complex to the blob complex on the circle. \nn{What properties does this
+map have?}
+
+\begin{figure}%
+$$\mathfig{0.6}{barycentric/barycentric}$$
+\caption{The Hochschild chain $a \tensor b \tensor c$ is sent to
+the sum of six blob $2$-chains, corresponding to a barycentric subdivision of a $2$-simplex.}
+\label{fig:Hochschild-example}%
+\end{figure}
+
+As an example, Figure \ref{fig:Hochschild-example} shows the image of the Hochschild chain $a \tensor b \tensor c$. Only the $0$-cells are shown explicitly.
+The edges marked $x, y$ and $z$ carry the $1$-chains
+\begin{align*}
+x & = \mathfig{0.1}{barycentric/ux} & u_x = \mathfig{0.1}{barycentric/ux_ca} - \mathfig{0.1}{barycentric/ux_c-a} \\
+y & = \mathfig{0.1}{barycentric/uy} & u_y = \mathfig{0.1}{barycentric/uy_cab} - \mathfig{0.1}{barycentric/uy_ca-b} \\
+z & = \mathfig{0.1}{barycentric/uz} & u_z = \mathfig{0.1}{barycentric/uz_c-a-b} - \mathfig{0.1}{barycentric/uz_cab}
+\end{align*}
+and the $2$-chain labelled $A$ is
+\begin{equation*}
+A = \mathfig{0.1}{barycentric/Ax}+\mathfig{0.1}{barycentric/Ay}.
+\end{equation*}
+Note that we then have
+\begin{equation*}
+\bdy A = x+y+z.
+\end{equation*}
+
+In general, the Hochschild chain $\Tensor_{i=1}^n a_i$ is sent to the sum of $n!$ blob $(n-1)$-chains, indexed by permutations,
+$$\phi\left(\Tensor_{i=1}^n a_i\right) = \sum_{\pi} \phi^\pi(a_1, \ldots, a_n)$$
+with ... (hmmm, problems making this precise; you need to decide where to put the labels, but then it's hard to make an honest chain map!)




