%!TEX root = ../blob1.tex

\section{Basic properties of the blob complex}

\label{sec:basic-properties}

Recall Property \ref{property:disjoint-union}, that there is a natural isomorphism $\bc_*(X \du Y) \cong \bc_*(X) \otimes \bc_*(Y)$.

\begin{proof}[Proof of Property \ref{property:disjoint-union}]

Given blob diagrams $b_1$ on $X$ and $b_2$ on $Y$, we can combine them

(putting the $b_1$ blobs before the $b_2$ blobs in the ordering) to get a

blob diagram $(b_1, b_2)$ on $X \du Y$.

Because of the blob reordering relations, all blob diagrams on $X \du Y$ arise this way.

In the other direction, any blob diagram on $X\du Y$ is equal (up to sign)

to one that puts $X$ blobs before $Y$ blobs in the ordering, and so determines

a pair of blob diagrams on $X$ and $Y$.

These two maps are compatible with our sign conventions.  (We follow the usual convention for tensors products of complexes, as in e.g. \cite{MR1438306}: $d(a \tensor b) = da \tensor b + (-1)^{\deg(a)} a \tensor db$.)

The two maps are inverses of each other.

\end{proof}

For the next proposition we will temporarily restore $n$-manifold boundary

conditions to the notation.

Suppose that for all $c \in \cC(\bd B^n)$

we have a splitting $s: H_0(\bc_*(B^n, c)) \to \bc_0(B^n; c)$

of the quotient map

$p: \bc_0(B^n; c) \to H_0(\bc_*(B^n, c))$.

For example, this is always the case if the coefficient ring is a field.

Then

\begin{prop} \label{bcontract}

For all $c \in \cC(\bd B^n)$ the natural map $p: \bc_*(B^n, c) \to H_0(\bc_*(B^n, c))$

is a chain homotopy equivalence

with inverse $s: H_0(\bc_*(B^n, c)) \to \bc_*(B^n; c)$.

Here we think of $H_0(\bc_*(B^n, c))$ as a 1-step complex concentrated in degree 0.

\end{prop}

\begin{proof}

By assumption $p\circ s = \id$, so all that remains is to find a degree 1 map

$h : \bc_*(B^n; c) \to \bc_*(B^n; c)$ such that $\bd h + h\bd = \id - s \circ p$.

For $i \ge 1$, define $h_i : \bc_i(B^n; c) \to \bc_{i+1}(B^n; c)$ by adding

an $(i{+}1)$-st blob equal to all of $B^n$.

In other words, add a new outermost blob which encloses all of the others.

Define $h_0 : \bc_0(B^n; c) \to \bc_1(B^n; c)$ by setting $h_0(x)$ equal to

the 1-blob with blob $B^n$ and label $x - s(p(x)) \in U(B^n; c)$.

\end{proof}

This proves Property \ref{property:contractibility} (the second half of the statement of this Property was immediate from the definitions).

Note that even when there is no splitting $s$, we can let $h_0 = 0$ and get a homotopy

equivalence to the 2-step complex $U(B^n; c) \to \cC(B^n; c)$.

For fields based on $n$-categories, $H_0(\bc_*(B^n; c)) \cong \mor(c', c'')$,

where $(c', c'')$ is some (any) splitting of $c$ into domain and range.

\begin{cor} \label{disj-union-contract}

If $X$ is a disjoint union of $n$-balls, then $\bc_*(X; c)$ is contractible.

\end{cor}

\begin{proof}

This follows from Properties \ref{property:disjoint-union} and \ref{property:contractibility}.

\end{proof}

Define the {\it support} of a blob diagram to be the union of all the 

blobs of the diagram.

Define the support of a linear combination of blob diagrams to be the union of the 

supports of the constituent diagrams.

For future use we prove the following lemma.

\begin{lemma} \label{support-shrink}

Let $L_* \sub \bc_*(X)$ be a subcomplex generated by some

subset of the blob diagrams on $X$, and let $f: L_* \to L_*$

be a chain map which does not increase supports and which induces an isomorphism on

$H_0(L_*)$.

Then $f$ is homotopic (in $\bc_*(X)$) to the identity $L_*\to L_*$.

\end{lemma}

\begin{proof}

We will use the method of acyclic models.

Let $b$ be a blob diagram of $L_*$, let $S\sub X$ be the support of $b$, and let

$r$ be the restriction of $b$ to $X\setminus S$.

Note that $S$ is a disjoint union of balls.

Assign to $b$ the acyclic (in positive degrees) subcomplex $T(b) \deq r\bullet\bc_*(S)$.

note that if a diagram $b'$ is part of $\bd b$ then $T(B') \sub T(b)$.

Both $f$ and the identity are compatible with $T$ (in the sense of acyclic models), 

so $f$ and the identity map are homotopic.

\end{proof}

For the next proposition we will temporarily restore $n$-manifold boundary

conditions to the notation.

Let $X$ be an $n$-manifold, $\bd X = Y \cup (-Y) \cup Z$.

Gluing the two copies of $Y$ together yields an $n$-manifold $X\sgl$

with boundary $Z\sgl$.

Given compatible fields (boundary conditions) $a$, $b$ and $c$ on $Y$, $-Y$ and $Z$,

we have the blob complex $\bc_*(X; a, b, c)$.

If $b = -a$ (the orientation reversal of $a$), then we can glue up blob diagrams on

$X$ to get blob diagrams on $X\sgl$. This proves Property \ref{property:gluing-map}, which we restate here in more detail.

\textbf{Property \ref{property:gluing-map}.}\emph{

There is a natural chain map

\eq{

    \gl: \bigoplus_a \bc_*(X; a, -a, c) \to \bc_*(X\sgl; c\sgl).

}

The sum is over all fields $a$ on $Y$ compatible at their

($n{-}2$-dimensional) boundaries with $c$.

`Natural' means natural with respect to the actions of diffeomorphisms.

}

This map is very far from being an isomorphism, even on homology.

We fix this deficit in Section \ref{sec:gluing} below.