414 ($G''_*$ and $G'_*$ depend on $N$, but that is not reflected in the notation.) |
414 ($G''_*$ and $G'_*$ depend on $N$, but that is not reflected in the notation.) |
415 Then $G''_*$ and $G'_*$ are both contractible |
415 Then $G''_*$ and $G'_*$ are both contractible |
416 and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence. |
416 and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence. |
417 For $G'_*$ the proof is the same as in (\ref{bcontract}), except that the splitting |
417 For $G'_*$ the proof is the same as in (\ref{bcontract}), except that the splitting |
418 $G'_0 \to H_0(G'_*)$ concentrates the point labels at two points to the right and left of $*$. |
418 $G'_0 \to H_0(G'_*)$ concentrates the point labels at two points to the right and left of $*$. |
419 For $G''_*$ we note that any cycle is supported \nn{need to establish terminology for this; maybe |
419 For $G''_*$ we note that any cycle is supported away from $*$. |
420 in ``basic properties" section above} away from $*$. |
|
421 Thus any cycle lies in the image of the normal blob complex of a disjoint union |
420 Thus any cycle lies in the image of the normal blob complex of a disjoint union |
422 of two intervals, which is contractible by (\ref{bcontract}) and (\ref{disj-union-contract}). |
421 of two intervals, which is contractible by (\ref{bcontract}) and (\ref{disj-union-contract}). |
423 Finally, it is easy to see that the inclusion |
422 Finally, it is easy to see that the inclusion |
424 $G''_* \to G'_*$ induces an isomorphism on $H_0$. |
423 $G''_* \to G'_*$ induces an isomorphism on $H_0$. |
425 |
424 |
446 Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$. |
445 Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$. |
447 Define $h_k(x) = y \bullet p$. |
446 Define $h_k(x) = y \bullet p$. |
448 This completes the proof that $i: K''_* \to K'_*$ is a homotopy equivalence. |
447 This completes the proof that $i: K''_* \to K'_*$ is a homotopy equivalence. |
449 \nn{need to say above more clearly and settle on notation/terminology} |
448 \nn{need to say above more clearly and settle on notation/terminology} |
450 |
449 |
451 Finally, we show that $K''_*$ is contractible. |
450 Finally, we show that $K''_*$ is contractible with $H_0\cong C$. |
452 \nn{need to also show that $H_0$ is the right thing; easy, but I won't do it now} |
451 This is similar to the proof of Proposition \ref{bcontract}, but a bit more |
453 Let $x$ be a cycle in $K''_*$. |
452 complicated since there is no single blob which contains the support of all blob diagrams |
|
453 in $K''_*$. |
|
454 Let $x$ be a cycle of degree greater than zero in $K''_*$. |
454 The union of the supports of the diagrams in $x$ does not contain $*$, so there exists a |
455 The union of the supports of the diagrams in $x$ does not contain $*$, so there exists a |
455 ball $B \subset S^1$ containing the union of the supports and not containing $*$. |
456 ball $B \subset S^1$ containing the union of the supports and not containing $*$. |
456 Adding $B$ as a blob to $x$ gives a contraction. |
457 Adding $B$ as an outermost blob to each summand of $x$ gives a chain $y$ with $\bd y = x$. |
457 \nn{need to say something else in degree zero} |
458 Thus $H_i(K''_*) \cong 0$ for $i> 0$ and $K''_*$ is contractible. |
|
459 |
|
460 To see that $H_0(K''_*) \cong C$, consider the map $p: K''_0 \to C$ which sends a 0-blob |
|
461 diagram to the product of its labeled points. |
|
462 $p$ is clearly surjective. |
|
463 It's also easy to see that $p(\bd K''_1) = 0$. |
|
464 Finally, if $p(y) = 0$ then there exists a blob $B \sub S^1$ which contains |
|
465 all of the labeled points (other than *) of all of the summands of $y$. |
|
466 This allows us to construct $x\in K''_1$ such that $\bd x = y$. |
|
467 (The label of $B$ is the restriction of $y$ to $B$.) |
|
468 It follows that $H_0(K''_*) \cong C$. |
458 \end{proof} |
469 \end{proof} |
459 |
470 |
460 \medskip |
471 \medskip |
461 |
472 |
462 For purposes of illustration, we describe an explicit chain map |
473 For purposes of illustration, we describe an explicit chain map |