277 whose value at $p\in P$ is the blob $B^n$ with label $y(p) - r(y(p))$. |
277 whose value at $p\in P$ is the blob $B^n$ with label $y(p) - r(y(p))$. |
278 Now define, for $y\in \btc_{0j}$, |
278 Now define, for $y\in \btc_{0j}$, |
279 \[ |
279 \[ |
280 h(y) = e(y - r(y)) + c(r(y)) . |
280 h(y) = e(y - r(y)) + c(r(y)) . |
281 \] |
281 \] |
282 \nn{up to sign, at least} |
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283 |
282 |
284 We must now verify that $h$ does the job it was intended to do. |
283 We must now verify that $h$ does the job it was intended to do. |
285 For $x\in \btc_{ij}$ with $i\ge 2$ we have |
284 For $x\in \btc_{ij}$ with $i\ge 2$ we have |
286 \nn{ignoring signs} |
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287 \begin{align*} |
285 \begin{align*} |
288 \bd h(x) + h(\bd x) &= \bd(e(x)) + e(\bd x) \\ |
286 \bd h(x) + h(\bd x) &= \bd(e(x)) + e(\bd x) && \\ |
289 &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x) + e(\bd_t x) \\ |
287 &= \bd_b(e(x)) + (-1)^{i+1} \bd_t(e(x)) + e(\bd_b x) + (-1)^i e(\bd_t x) && \\ |
290 &= \bd_b(e(x)) + e(\bd_b x) \quad\quad\text{(since $\bd_t(e(x)) = e(\bd_t x)$)} \\ |
288 &= \bd_b(e(x)) + e(\bd_b x) && \text{(since $\bd_t(e(x)) = e(\bd_t x)$)} \\ |
291 &= x . |
289 &= x . && |
292 \end{align*} |
290 \end{align*} |
293 For $x\in \btc_{1j}$ we have |
291 For $x\in \btc_{1j}$ we have |
294 \nn{ignoring signs} |
|
295 \begin{align*} |
292 \begin{align*} |
296 \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) \\ |
293 \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) && \\ |
297 &= \bd_b(e(x)) + e(\bd_b x) \quad\quad\text{(since $r(\bd_b x) = 0$)} \\ |
294 &= \bd_b(e(x)) + e(\bd_b x) && \text{(since $r(\bd_b x) = 0$)} \\ |
298 &= x . |
295 &= x . && |
299 \end{align*} |
296 \end{align*} |
300 For $x\in \btc_{0j}$ with $j\ge 1$ we have |
297 For $x\in \btc_{0j}$ with $j\ge 1$ we have |
301 \nn{ignoring signs} |
|
302 \begin{align*} |
298 \begin{align*} |
303 \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(e(x - r(x))) + \bd_t(c(r(x))) + |
299 \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) - \bd_t(e(x - r(x))) - \bd_t(c(r(x))) + |
304 e(\bd_t x - r(\bd_t x)) + c(r(\bd_t x)) \\ |
300 e(\bd_t x - r(\bd_t x)) + c(r(\bd_t x)) \\ |
305 &= x - r(x) + \bd_t(c(r(x))) + c(r(\bd_t x)) \\ |
301 &= x - r(x) - \bd_t(c(r(x))) + c(r(\bd_t x)) \\ |
306 &= x - r(x) + r(x) \\ |
302 &= x - r(x) + r(x) \\ |
307 &= x. |
303 &= x. |
308 \end{align*} |
304 \end{align*} |
309 Here we have used the fact that $\bd_b(c(r(x))) = 0$ since $c(r(x))$ is a $0$-blob diagram, as well as that $\bd_t(e(r(x))) = e(r(\bd_t x))$ \nn{explain why this is true?} and $c(r(\bd_t x)) - \bd_t(c(r(x))) = r(x)$ \nn{explain?}. |
305 Here we have used the fact that $\bd_b(c(r(x))) = 0$ since $c(r(x))$ is a $0$-blob diagram, as well as that $\bd_t(e(r(x))) = e(r(\bd_t x))$ \nn{explain why this is true?} and $c(r(\bd_t x)) - \bd_t(c(r(x))) = r(x)$ \nn{explain?}. |
310 |
306 |
311 For $x\in \btc_{00}$ we have |
307 For $x\in \btc_{00}$ we have |
312 \nn{ignoring signs} |
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313 \begin{align*} |
308 \begin{align*} |
314 \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(c(r(x))) \\ |
309 \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(c(r(x))) \\ |
315 &= x - r(x) + r(x) - r(x)\\ |
310 &= x - r(x) + r(x) - r(x)\\ |
316 &= x - r(x). \qedhere |
311 &= x - r(x). \qedhere |
317 \end{align*} |
312 \end{align*} |