215 \item \nn{something about blob labels and vector space structure} |
215 \item \nn{something about blob labels and vector space structure} |
216 \item \nn{maybe also something about gluing} |
216 \item \nn{maybe also something about gluing} |
217 \end{itemize} |
217 \end{itemize} |
218 |
218 |
219 Next we define $\btc_*(X)$ to be the total complex of the double complex (denoted $\btc_{**}$) |
219 Next we define $\btc_*(X)$ to be the total complex of the double complex (denoted $\btc_{**}$) |
220 whose $(i,j)$ entry is $C_i(\BD_j)$, the singular $i$-chains on the space of $j$-blob diagrams. |
220 whose $(i,j)$ entry is $C_j(\BD_i)$, the singular $j$-chains on the space of $i$-blob diagrams. |
221 The horizontal boundary of the double complex, |
221 The vertical boundary of the double complex, |
222 denoted $\bd_t$, is the singular boundary, and the vertical boundary, denoted $\bd_b$, is |
222 denoted $\bd_t$, is the singular boundary, and the horizontal boundary, denoted $\bd_b$, is |
223 the blob boundary. |
223 the blob boundary. |
224 |
224 |
225 We will regard $\bc_*(X)$ as the subcomplex $\btc_{0*}(X) \sub \btc_{**}(X)$. |
225 We will regard $\bc_*(X)$ as the subcomplex $\btc_{*0}(X) \sub \btc_{**}(X)$. |
226 The main result of this subsection is |
226 The main result of this subsection is |
227 |
227 |
228 \begin{lemma} \label{lem:bt-btc} |
228 \begin{lemma} \label{lem:bt-btc} |
229 The inclusion $\bc_*(X) \sub \btc_*(X)$ is a homotopy equivalence |
229 The inclusion $\bc_*(X) \sub \btc_*(X)$ is a homotopy equivalence |
230 \end{lemma} |
230 \end{lemma} |
231 |
231 |
232 Before giving the proof we need a few preliminary results. |
232 Before giving the proof we need a few preliminary results. |
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233 |
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234 \begin{lemma} \label{bt-contract} |
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235 $\btc_*(B^n)$ is contractible (acyclic in positive degrees). |
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236 \end{lemma} |
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237 \begin{proof} |
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238 We will construct a contracting homotopy $h: \btc_*(B^n)\to \btc_*(B^n)$. |
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239 |
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240 We will assume a splitting $s:H_0(\btc_*(B^n))\to \btc_0(B^n)$ |
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241 of the quotient map $q:\btc_0(B^n)\to H_0(\btc_*(B^n))$. |
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242 Let $r = s\circ q$. |
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243 |
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244 For $x\in \btc_{ij}$ with $i\ge 1$ define |
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245 \[ |
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246 h(x) = e(x) , |
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247 \] |
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248 where |
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249 \[ |
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250 e: \btc_{ij}\to\btc_{i+1,j} |
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251 \] |
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252 adds an outermost blob, equal to all of $B^n$, to the $j$-parameter family of blob diagrams. |
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253 |
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254 A generator $y\in \btc_{0j}$ is a map $y:P\to \BD_0$, where $P$ is some $j$-dimensional polyhedron. |
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255 We define $r(y)\in \btc_{0j}$ to be the constant function $r\circ y : P\to \BD_0$. |
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256 Let $c(r(y))\in \btc_{0,j+1}$ be the constant map from the cone of $P$ to $\BD_0$ taking |
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257 the same value (i.e.\ $r(y(p))$ for any $p\in P$). |
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258 Let $e(y - r(y)) \in \btc_{1j}$ denote the $j$-parameter family of 1-blob diagrams |
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259 whose value at $p\in P$ is the blob $B^n$ with label $y(p) - r(y(p))$. |
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260 Now define, for $y\in \btc_{0j}$, |
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261 \[ |
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262 h(y) = e(y - r(y)) + c(r(y)) . |
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263 \] |
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264 \nn{up to sign, at least} |
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265 |
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266 We must now verify that $h$ does the job it was intended to do. |
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267 For $x\in \btc_{ij}$ with $i\ge 2$ we have |
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268 \nn{ignoring signs} |
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269 \begin{align*} |
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270 \bd h(x) + h(\bd x) &= \bd(e(x)) + e(\bd x) \\ |
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271 &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x) + e(\bd_t x) \\ |
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272 &= \bd_b(e(x)) + e(\bd_b x) \quad\quad\text{(since $\bd_t(e(x)) = e(\bd_t x)$)} \\ |
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273 &= x . |
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274 \end{align*} |
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275 For $x\in \btc_{1j}$ we have |
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276 \nn{ignoring signs} |
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277 \begin{align*} |
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278 \bd h(x) + h(\bd x) &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x - r(\bd_b x)) + c(r(\bd_b x)) + e(\bd_t x) \\ |
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279 &= \bd_b(e(x)) + e(\bd_b x) \quad\quad\text{(since $r(\bd_b x) = 0$)} \\ |
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280 &= x . |
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281 \end{align*} |
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282 For $x\in \btc_{0j}$ with $j\ge 1$ we have |
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283 \nn{ignoring signs} |
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284 \begin{align*} |
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285 \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(e(x - r(x))) + \bd_t(c(r(x))) + |
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286 e(\bd_t x - r(\bd_t x)) + c(r(\bd_t x)) \\ |
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287 &= x - r(x) + \bd_t(c(r(x))) + c(r(\bd_t x)) \\ |
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288 &= x - r(x) + r(x) \\ |
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289 &= x. |
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290 \end{align*} |
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291 For $x\in \btc_{00}$ we have |
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292 \nn{ignoring signs} |
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293 \begin{align*} |
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294 \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(c(r(x))) \\ |
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295 &= x - r(x) + r(x) - r(x)\\ |
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296 &= x - r(x). |
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297 \end{align*} |
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298 \end{proof} |
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299 |
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234 |
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235 |
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236 |
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237 |
304 |