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%!TEX root = ../blob1.tex
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\section{The blob complex for $A_\infty$ $n$-categories}
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\label{sec:ainfblob}
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Given an $A_\infty$ $n$-category $\cC$ and an $n$-manifold $M$, we define the blob
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complex $\bc_*(M)$ to the be the colimit $\cC(M)$ of Section \ref{sec:ncats}.
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\nn{say something about this being anticlimatically tautological?}
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We will show below
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\nn{give ref}
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that this agrees (up to homotopy) with our original definition of the blob complex
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in the case of plain $n$-categories.
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When we need to distinguish between the new and old definitions, we will refer to the
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new-fangled and old-fashioned blob complex.
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\medskip
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Let $M^n = Y^k\times F^{n-k}$.
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Let $C$ be a plain $n$-category.
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Let $\cF$ be the $A_\infty$ $k$-category which assigns to a $k$-ball
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$X$ the old-fashioned blob complex $\bc_*(X\times F)$.
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\begin{thm}
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The old-fashioned blob complex $\bc_*^C(Y\times F)$ is homotopy equivalent to the
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new-fangled blob complex $\bc_*^\cF(Y)$.
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\end{thm}
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\begin{proof}
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We will use the concrete description of the colimit from Subsection \ref{ss:ncat_fields}.
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First we define a map from $\bc_*^\cF(Y)$ to $\bc_*^C(Y\times F)$.
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In filtration degree 0 we just glue together the various blob diagrams on $X\times F$
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(where $X$ is a component of a permissible decomposition of $Y$) to get a blob diagram on
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$Y\times F$.
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In filtration degrees 1 and higher we define the map to be zero.
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It is easy to check that this is a chain map.
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Next we define a map from $\phi: \bc_*^C(Y\times F) \to \bc_*^\cF(Y)$.
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Actually, we will define it on the homotopy equivalent subcomplex
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$\cS_* \sub \bc_*^C(Y\times F)$ generated by blob diagrams which are small with
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respect to some open cover
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of $Y\times F$.
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\nn{need reference to small blob lemma}
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We will have to show eventually that this is independent (up to homotopy) of the choice of cover.
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Also, for a fixed choice of cover we will only be able to define the map for blob degree less than
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some bound, but this bound goes to infinity as the cover become finer.
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Given a decomposition $K$ of $Y$ into $k$-balls $X_i$, let $K\times F$ denote the corresponding
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decomposition of $Y\times F$ into the pieces $X_i\times F$.
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%We will define $\phi$ inductively, starting at blob degree 0.
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%Given a 0-blob diagram $x$ on $Y\times F$, we can choose a decomposition $K$ of $Y$
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%such that $x$ is splittable with respect to $K\times F$.
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%This defines a filtration degree 0 element of $\bc_*^\cF(Y)$
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We will define $\phi$ using a variant of the method of acyclic models.
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Let $a\in S_m$ be a blob diagram on $Y\times F$.
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For $m$ sufficiently small there exist decompositions of $K$ of $Y$ into $k$-balls such that the
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codimension 1 cells of $K\times F$ miss the blobs of $a$, and more generally such that $a$ is splittable along $K\times F$.
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Let $D(a)$ denote the subcomplex of $\bc_*^\cF(Y)$ generated by all $(a, \bar{K})$
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such that each $K_i$ has the aforementioned splittable property
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(see Subsection \ref{ss:ncat_fields}).
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\nn{need to define $D(a)$ more clearly; also includes $(b_j, \bar{K})$ where
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$\bd(a) = \sum b_j$.}
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(By $(a, \bar{K})$ we really mean $(a^\sharp, \bar{K})$, where $a^\sharp$ is
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$a$ split according to $K_0\times F$.
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To simplify notation we will just write plain $a$ instead of $a^\sharp$.)
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Roughly speaking, $D(a)$ consists of filtration degree 0 stuff which glues up to give
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$a$, filtration degree 1 stuff which makes all of the filtration degree 0 stuff homologous,
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filtration degree 2 stuff which kills the homology created by the
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filtration degree 1 stuff, and so on.
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More formally,
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\begin{lemma}
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$D(a)$ is acyclic.
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\end{lemma}
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\begin{proof}
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We will prove acyclicity in the first couple of degrees, and \nn{in this draft, at least}
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leave the general case to the reader.
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Let $K$ and $K'$ be two decompositions of $Y$ compatible with $a$.
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We want to show that $(a, K)$ and $(a, K')$ are homologous via filtration degree 1 stuff.
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\nn{need to say this better; these two chains don't have the same boundary.}
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We might hope that $K$ and $K'$ have a common refinement, but this is not necessarily
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the case.
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(Consider the $x$-axis and the graph of $y = x^2\sin(1/x)$ in $\r^2$.)
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However, we {\it can} find another decomposition $L$ such that $L$ shares common
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refinements with both $K$ and $K'$.
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Let $KL$ and $K'L$ denote these two refinements.
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Then filtration degree 1 chains associated to the four anti-refinemnts
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$KL\to K$, $KL\to L$, $K'L\to L$ and $K'L\to K'$
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give the desired chain connecting $(a, K)$ and $(a, K')$
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(see Figure xxxx).
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Consider a different choice of decomposition $L'$ in place of $L$ above.
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This leads to a cycle consisting of filtration degree 1 stuff.
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We want to show that this cycle bounds a chain of filtration degree 2 stuff.
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Choose a decomposition $M$ which has common refinements with each of
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$K$, $KL$, $L$, $K'L$, $K'$, $K'L'$, $L'$ and $KL'$.
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Then we have a filtration degree 2 chain, as shown in Figure yyyy, which does the trick.
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For example, ....
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\end{proof}
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\nn{....}
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\end{proof}
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\nn{need to say something about dim $< n$ above}
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\medskip
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\hrule
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\medskip
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\nn{to be continued...}
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\medskip
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