text/a_inf_blob.tex
changeset 116 3f180943709f
parent 115 76f423a9c787
child 117 b62214646c4f
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    58 For $m$ sufficiently small there exist decompositions of $K$ of $Y$ into $k$-balls such that the
    58 For $m$ sufficiently small there exist decompositions of $K$ of $Y$ into $k$-balls such that the
    59 codimension 1 cells of $K\times F$ miss the blobs of $a$, and more generally such that $a$ is splittable along $K\times F$.
    59 codimension 1 cells of $K\times F$ miss the blobs of $a$, and more generally such that $a$ is splittable along $K\times F$.
    60 Let $D(a)$ denote the subcomplex of $\bc_*^\cF(Y)$ generated by all $(a, \bar{K})$
    60 Let $D(a)$ denote the subcomplex of $\bc_*^\cF(Y)$ generated by all $(a, \bar{K})$
    61 such that each $K_i$ has the aforementioned splittable property
    61 such that each $K_i$ has the aforementioned splittable property
    62 (see Subsection \ref{ss:ncat_fields}).
    62 (see Subsection \ref{ss:ncat_fields}).
    63 (By $(a, \bar{K})$ we really mean $(a', \bar{K})$, where $a^\sharp$ is 
    63 \nn{need to define $D(a)$ more clearly; also includes $(b_j, \bar{K})$ where
       
    64 $\bd(a) = \sum b_j$.}
       
    65 (By $(a, \bar{K})$ we really mean $(a^\sharp, \bar{K})$, where $a^\sharp$ is 
    64 $a$ split according to $K_0\times F$.
    66 $a$ split according to $K_0\times F$.
    65 To simplify notation we will just write plain $a$ instead of $a^\sharp$.)
    67 To simplify notation we will just write plain $a$ instead of $a^\sharp$.)
    66 Roughly speaking, $D(a)$ consists of filtration degree 0 stuff which glues up to give
    68 Roughly speaking, $D(a)$ consists of filtration degree 0 stuff which glues up to give
    67 $a$, filtration degree 1 stuff which makes all of the filtration degree 0 stuff homologous, 
    69 $a$, filtration degree 1 stuff which makes all of the filtration degree 0 stuff homologous, 
    68 filtration degree 2 stuff which kills the homology created by the 
    70 filtration degree 2 stuff which kills the homology created by the 
    74 \end{lemma}
    76 \end{lemma}
    75 
    77 
    76 \begin{proof}
    78 \begin{proof}
    77 We will prove acyclicity in the first couple of degrees, and \nn{in this draft, at least}
    79 We will prove acyclicity in the first couple of degrees, and \nn{in this draft, at least}
    78 leave the general case to the reader.
    80 leave the general case to the reader.
       
    81 
    79 Let $K$ and $K'$ be two decompositions of $Y$ compatible with $a$.
    82 Let $K$ and $K'$ be two decompositions of $Y$ compatible with $a$.
    80 We want to show that $(a, K)$ and $(a, K')$ are homologous
    83 We want to show that $(a, K)$ and $(a, K')$ are homologous via filtration degree 1 stuff.
    81 \nn{oops -- can't really ignore $\bd a$ like this}
    84 \nn{need to say this better; these two chains don't have the same boundary.}
       
    85 We might hope that $K$ and $K'$ have a common refinement, but this is not necessarily
       
    86 the case.
       
    87 (Consider the $x$-axis and the graph of $y = x^2\sin(1/x)$ in $\r^2$.)
       
    88 However, we {\it can} find another decomposition $L$ such that $L$ shares common
       
    89 refinements with both $K$ and $K'$.
       
    90 Let $KL$ and $K'L$ denote these two refinements.
       
    91 Then filtration degree 1 chains associated to the four anti-refinemnts
       
    92 $KL\to K$, $KL\to L$, $K'L\to L$ and $K'L\to K'$
       
    93 give the desired chain connecting $(a, K)$ and $(a, K')$
       
    94 (see Figure xxxx).
       
    95 
       
    96 Consider a different choice of decomposition $L'$ in place of $L$ above.
       
    97 This leads to a cycle consisting of filtration degree 1 stuff.
       
    98 We want to show that this cycle bounds a chain of filtration degree 2 stuff.
       
    99 Choose a decomposition $M$ which has common refinements with each of 
       
   100 $K$, $KL$, $L$, $K'L$, $K'$, $K'L'$, $L'$ and $KL'$.
       
   101 Then we have a filtration degree 2 chain, as shown in Figure yyyy, which does the trick.
       
   102 For example, ....
       
   103 
       
   104 
    82 \end{proof}
   105 \end{proof}
    83 
   106 
    84 
   107 
    85 \nn{....}
   108 \nn{....}
    86 \end{proof}
   109 \end{proof}