text/evmap.tex
changeset 453 e88e44347b36
parent 447 ba4f86b15ff0
child 492 833bd74143a4
equal deleted inserted replaced
452:43fa3a30d89c 453:e88e44347b36
    34             \bc_*(Y)\ar[d]^{\gl} \\
    34             \bc_*(Y)\ar[d]^{\gl} \\
    35      CH_*(X\sgl, Y\sgl) \otimes \bc_*(X\sgl) \ar[r]_(.6){e_{X\sgl Y\sgl}}   & 	\bc_*(Y\sgl)  
    35      CH_*(X\sgl, Y\sgl) \otimes \bc_*(X\sgl) \ar[r]_(.6){e_{X\sgl Y\sgl}}   & 	\bc_*(Y\sgl)  
    36 }
    36 }
    37 \end{equation*}
    37 \end{equation*}
    38 \end{enumerate}
    38 \end{enumerate}
    39 Up to (iterated) homotopy, there is a unique family $\{e_{XY}\}$ of chain maps
    39 Moreover, for any $m \geq 0$, we can find a family of chain maps $\{e_{XY}\}$ 
    40 satisfying the above two conditions.
    40 satisfying the above two conditions which is $m$-connected. In particular, this means that the choice of chain map above is unique up to homotopy.
    41 \end{thm}
    41 \end{thm}
       
    42 \begin{rem}
       
    43 Note that the statement doesn't quite give uniqueness up to iterated homotopy. We fully expect that this should actually be the case, but haven't been able to prove this.
       
    44 \end{rem}
       
    45 
    42 
    46 
    43 Before giving the proof, we state the essential technical tool of Lemma \ref{extension_lemma}, 
    47 Before giving the proof, we state the essential technical tool of Lemma \ref{extension_lemma}, 
    44 and then give an outline of the method of proof.
    48 and then give an outline of the method of proof.
    45 
    49 
    46 Without loss of generality, we will assume $X = Y$.
    50 Without loss of generality, we will assume $X = Y$.
   343 for all $l\ge m$ and all $q\ot a$ which appear in the boundary of $p\ot b$.
   347 for all $l\ge m$ and all $q\ot a$ which appear in the boundary of $p\ot b$.
   344 
   348 
   345 
   349 
   346 \begin{proof}
   350 \begin{proof}
   347 
   351 
   348 There exists $\lambda > 0$ such that for every  subset $c$ of the blobs of $b$ $\Nbd_u(c)$ is homeomorphic to $|c|$ for all $u < \lambda$ .
   352 There exists $\lambda > 0$ such that for every  subset $c$ of the blobs of $b$ the set $\Nbd_u(c)$ is homeomorphic to $|c|$ for all $u < \lambda$ .
   349 (Here we are using the fact that the blobs are 
   353 (Here we are using the fact that the blobs are 
   350 piecewise smooth or piecewise-linear and that $\bd c$ is collared.)
   354 piecewise smooth or piecewise-linear and that $\bd c$ is collared.)
   351 We need to consider all such $c$ because all generators appearing in
   355 We need to consider all such $c$ because all generators appearing in
   352 iterated boundaries of $p\ot b$ must be in $G_*^{i,m}$.)
   356 iterated boundaries of $p\ot b$ must be in $G_*^{i,m}$.)
   353 
   357 
   580 On $G^{m+1}_* \sub G^m_*$ we have defined two maps, $e_m$ and $e_{m+1}$.
   584 On $G^{m+1}_* \sub G^m_*$ we have defined two maps, $e_m$ and $e_{m+1}$.
   581 One can similarly (to the proof of Lemma \ref{m_order_hty}) show that 
   585 One can similarly (to the proof of Lemma \ref{m_order_hty}) show that 
   582 these two maps agree up to $m$-th order homotopy.
   586 these two maps agree up to $m$-th order homotopy.
   583 More precisely, one can show that the subcomplex of maps containing the various
   587 More precisely, one can show that the subcomplex of maps containing the various
   584 $e_{m+1}$ candidates is contained in the corresponding subcomplex for $e_m$.
   588 $e_{m+1}$ candidates is contained in the corresponding subcomplex for $e_m$.
   585 \nn{now should remark that we have not, in fact, produced a contractible set of maps,
       
   586 but we have come very close}
       
   587 \nn{better: change statement of thm}
       
   588 
   589 
   589 \medskip
   590 \medskip
   590 
   591 
   591 Next we show that the action maps are compatible with gluing.
   592 Next we show that the action maps are compatible with gluing.
   592 Let $G^m_*$ and $\ol{G}^m_*$ be the complexes, as above, used for defining
   593 Let $G^m_*$ and $\ol{G}^m_*$ be the complexes, as above, used for defining