206 We must show that $\phi$ and $\phi'$ agree, up to homotopy, |
206 We must show that $\phi$ and $\phi'$ agree, up to homotopy, |
207 on the intersection of the subcomplexes on which they are defined. |
207 on the intersection of the subcomplexes on which they are defined. |
208 This is clear, since the acyclic subcomplexes $D(a)$ above used in the definition of |
208 This is clear, since the acyclic subcomplexes $D(a)$ above used in the definition of |
209 $\phi$ and $\phi'$ do not depend on the choice of cover. |
209 $\phi$ and $\phi'$ do not depend on the choice of cover. |
210 |
210 |
211 \nn{need to say (and justify) that we now have a map $\phi$ indep of choice of cover} |
211 %\nn{need to say (and justify) that we now have a map $\phi$ indep of choice of cover} |
212 |
212 |
213 We now show that $\phi\circ\psi$ and $\psi\circ\phi$ are homotopic to the identity. |
213 We now show that $\phi\circ\psi$ and $\psi\circ\phi$ are homotopic to the identity. |
214 |
214 |
215 $\psi\circ\phi$ is the identity on the nose. |
215 $\psi\circ\phi$ is the identity on the nose. |
216 $\phi$ takes a blob diagram $a$ and chops it into pieces |
216 $\phi$ takes a blob diagram $a$ and chops it into pieces |
217 according to some decomposition $K$ of $Y$. |
217 according to some decomposition $K$ of $Y$. |
218 $\psi$ glues those pieces back together, yielding the same $a$ we started with. |
218 $\psi$ glues those pieces back together, yielding the same $a$ we started with. |
219 |
219 |
220 $\phi\circ\psi$ is the identity up to homotopy by another MoAM argument... |
220 $\phi\circ\psi$ is the identity up to homotopy by another MoAM argument.... |
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221 |
221 |
222 |
222 This concludes the proof of Theorem \ref{product_thm}. |
223 This concludes the proof of Theorem \ref{product_thm}. |
223 \nn{at least I think it does; it's pretty rough at this point.} |
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224 \end{proof} |
224 \end{proof} |
225 |
225 |
226 \nn{need to say something about dim $< n$ above} |
226 \nn{need to say something about dim $< n$ above} |
227 |
227 |
228 \medskip |
228 \medskip |
235 Apply Theorem \ref{product_thm} with the fiber $F$ equal to a point. |
235 Apply Theorem \ref{product_thm} with the fiber $F$ equal to a point. |
236 \end{proof} |
236 \end{proof} |
237 |
237 |
238 \medskip |
238 \medskip |
239 |
239 |
240 \nn{To do: remark on the case of a nontrivial fiber bundle. |
240 Theorem \ref{product_thm} extends to the case of general fiber bundles |
241 I can think of two approaches. |
241 \[ |
242 In the first (slick but maybe a little too tautological), we generalize the |
242 F \to E \to Y . |
243 notion of an $n$-category to an $n$-category {\it over a space $B$}. |
243 \] |
244 (Should be able to find precedent for this in a paper of PT. |
244 We outline two approaches. |
245 This idea came up in a conversation with him, so maybe should site him.) |
245 |
246 In this generalization, we replace the categories of balls with the categories |
246 We can generalize the definition of a $k$-category by replacing the categories |
247 of balls equipped with maps to $B$. |
247 of $j$-balls ($j\le k$) with categories of $j$-balls $D$ equipped with a map $p:D\to Y$. |
248 A fiber bundle $F\to E\to B$ gives an example of such an $n$-category: |
248 \nn{need citation to other work that does this; Stolz and Teichner?} |
249 assign to $p:D\to B$ the blob complex $\bc_*(p^*(E))$. |
249 Call this a $k$-category over $Y$. |
250 We can do the colimit thing over $B$ with coefficients in a n-cat-over-B. |
250 A fiber bundle $F\to E\to Y$ gives an example of a $k$-category over $Y$: |
251 The proof below works essentially unchanged in this case to show that the colimit is the blob complex of the total space $E$. |
251 assign to $p:D\to Y$ the blob complex $\bc_*(p^*(E))$. |
252 } |
252 Let $\cF_E$ denote this $k$-category over $Y$. |
253 |
253 We can adapt the homotopy colimit construction (based decompositions of $Y$ into balls) to |
254 \nn{The second approach: Choose a decomposition $B = \cup X_i$ |
254 get a chain complex $\cF_E(Y)$. |
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255 The proof of Theorem \ref{product_thm} goes through essentially unchanged |
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256 to show that |
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257 \[ |
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258 \bc_*(E) \simeq \cF_E(Y) . |
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259 \] |
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260 |
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261 |
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262 |
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263 |
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264 \nn{The second approach: Choose a decomposition $Y = \cup X_i$ |
255 such that the restriction of $E$ to $X_i$ is a product $F\times X_i$. |
265 such that the restriction of $E$ to $X_i$ is a product $F\times X_i$. |
256 Choose the product structure as well. |
266 Choose the product structure as well. |
257 To each codim-1 face $D_i\cap D_j$ we have a bimodule ($S^0$-module). |
267 To each codim-1 face $D_i\cap D_j$ we have a bimodule ($S^0$-module). |
258 And more generally to each codim-$j$ face we have an $S^{j-1}$-module. |
268 And more generally to each codim-$j$ face we have an $S^{j-1}$-module. |
259 Decorate the decomposition with these modules and do the colimit. |
269 Decorate the decomposition with these modules and do the colimit. |
260 } |
270 } |
261 |
271 |
262 \nn{There is a version of this last construction for arbitrary maps $E \to B$ |
272 \nn{There is a version of this last construction for arbitrary maps $E \to Y$ |
263 (not necessarily a fibration).} |
273 (not necessarily a fibration).} |
264 |
274 |
265 |
275 |
266 |
276 |
267 \subsection{A gluing theorem} |
277 \subsection{A gluing theorem} |