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+%!TEX root = ../blob1.tex
+
+\section{Basic properties of the blob complex}
+\label{sec:basic-properties}
+
+\begin{prop} \label{disjunion}
+There is a natural isomorphism $\bc_*(X \du Y) \cong \bc_*(X) \otimes \bc_*(Y)$.
+\end{prop}
+\begin{proof}
+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.
+The two maps are inverses of each other.
+\nn{should probably say something about sign conventions for the differential
+in a tensor product of chain complexes; ask Scott}
+\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 you 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}
+
+Note that if 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.
+
+\medskip
+
+\nn{Maybe there is no longer a need to repeat the next couple of props here, since we also state them in the introduction.
+But I think it's worth saying that the Diff actions will be enhanced later.
+Maybe put that in the intro too.}
+
+As we noted above,
+\begin{prop}
+There is a natural isomorphism $H_0(\bc_*(X)) \cong A(X)$.
+\qed
+\end{prop}
+
+
+\begin{prop}
+For fixed fields ($n$-cat), $\bc_*$ is a functor from the category
+of $n$-manifolds and diffeomorphisms to the category of chain complexes and
+(chain map) isomorphisms.
+\qed
+\end{prop}
+
+In particular,
+\begin{prop}  \label{diff0prop}
+There is an action of $\Diff(X)$ on $\bc_*(X)$.
+\qed
+\end{prop}
+
+The above will be greatly strengthened in Section \ref{sec:evaluation}.
+
+\medskip
+
+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 $X\sgl$:
+
+\begin{prop}
+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.
+\qed
+\end{prop}
+
+The above map is very far from being an isomorphism, even on homology.
+This will be fixed in Section \ref{sec:gluing} below.
+
+\nn{Next para not need, since we already use bullet = gluing notation above(?)}
+
+An instance of gluing we will encounter frequently below is where $X = X_1 \du X_2$
+and $X\sgl = X_1 \cup_Y X_2$.
+(Typically one of $X_1$ or $X_2$ is a disjoint union of balls.)
+For $x_i \in \bc_*(X_i)$, we introduce the notation
+\eq{
+    x_1 \bullet x_2 \deq \gl(x_1 \otimes x_2) .
+}
+Note that we have resumed our habit of omitting boundary labels from the notation.
+
+
+