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

Finally, then we see that the functor $K_*$ is simply an (infinite)

to tensor factors of $C$.

\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}]

\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!)