217 \end{proof} |
217 \end{proof} |
218 |
218 |
219 \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
219 \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
220 We now prove that $K_*$ is an exact functor. |
220 We now prove that $K_*$ is an exact functor. |
221 |
221 |
222 %\todo{p. 1478 of scott's notes} |
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223 Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules |
222 Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules |
224 \begin{equation*} |
223 \begin{equation*} |
225 M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M) |
224 M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M) |
226 \end{equation*} |
225 \end{equation*} |
227 is exact. For completeness we'll explain this below. |
226 is exact. For completeness we'll explain this below. |
228 |
227 |
229 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
228 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
230 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
229 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor, so $\hat{f}(a \tensor k \tensor b) = a \tensor f(k) \tensor b$, and similarly for $\hat{g}$. |
231 Most of what we need to check is easy. |
230 Most of what we need to check is easy. |
232 \nn{don't we need to consider sums here, e.g. $\sum_i(a_i\ot k_i\ot b_i)$ ?} |
231 Suppose we have $\sum_i (a_i \tensor k_i \tensor b_i) \in \ker(C \tensor K \tensor C \to K)$, assuming without loss of generality that $\{a_i \tensor b_i\}_i$ is linearly independent in $C \tensor C$, and $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$. We must then have $f(k_i) = 0 \in E$ for each $i$, which implies $k_i=0$ itself. |
233 If $(a \tensor k \tensor b) \in \ker(C \tensor K \tensor C \to K)$, to have $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$, we must have $f(k) = 0 \in E$, which implies $k=0$ itself. If $(a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, clearly |
232 If $\sum_i (a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, again by assuming the set $\{a_i \tensor b_i\}_i$ is linearly independent we can deduce that each |
234 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
233 $e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$, and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$. |
235 If $\hat{g}(a \tensor e \tensor b) = 0$, then $g(e) = 0$, so $e = f(\widetilde{e})$ for some $\widetilde{e} \in K$, and $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
234 If $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$, then each $g(e_i) = 0$, so $e_i = f(\widetilde{e_i})$ for some $\widetilde{e_i} \in K$, and $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$. |
236 Finally, the interesting step is in checking that any $q = \sum_i a_i \tensor q_i \tensor b_i$ such that $\sum_i a_i q_i b_i = 0$ is in the image of $\ker(C \tensor E \tensor C \to C)$ under $\hat{g}$. |
235 Finally, the interesting step is in checking that any $q = \sum_i a_i \tensor q_i \tensor b_i$ such that $\sum_i a_i q_i b_i = 0$ is in the image of $\ker(C \tensor E \tensor C \to C)$ under $\hat{g}$. |
237 For each $i$, we can find $\widetilde{q_i}$ so $g(\widetilde{q_i}) = q_i$. However $\sum_i a_i \widetilde{q_i} b_i$ need not be zero. |
236 For each $i$, we can find $\widetilde{q_i}$ so $g(\widetilde{q_i}) = q_i$. However $\sum_i a_i \widetilde{q_i} b_i$ need not be zero. |
238 Consider then $$\widetilde{q} = \sum_i (a_i \tensor \widetilde{q_i} \tensor b_i) - 1 \tensor (\sum_i a_i \widetilde{q_i} b_i) \tensor 1.$$ Certainly |
237 Consider then $$\widetilde{q} = \sum_i (a_i \tensor \widetilde{q_i} \tensor b_i) - 1 \tensor (\sum_i a_i \widetilde{q_i} b_i) \tensor 1.$$ Certainly |
239 $\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further, |
238 $\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further, |
240 \begin{align*} |
239 \begin{align*} |
253 \tensor C \tensor \ker(C \tensor C \to M)$) are all still exact. |
252 \tensor C \tensor \ker(C \tensor C \to M)$) are all still exact. |
254 |
253 |
255 Finally, then we see that the functor $K_*$ is simply an (infinite) |
254 Finally, then we see that the functor $K_*$ is simply an (infinite) |
256 direct sum of copies of this sort of functor. The direct sum is indexed by |
255 direct sum of copies of this sort of functor. The direct sum is indexed by |
257 configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors, |
256 configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors, |
258 with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$, and all other labelled points corresponding |
257 with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$ (depending on whether they contain a marked point $p_i$), and all other labelled points corresponding |
259 to tensor factors of $C$. |
258 to tensor factors of $C$ and $M$. |
260 \end{proof} |
259 \end{proof} |
261 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
260 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
262 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
261 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
263 |
262 |
264 We define a map $\ev: K_0(M) \to M$. If $x \in K_0(M)$ has the label $m \in M$ at $*$, and labels $c_i \in C$ at the other labeled points of $S^1$, reading clockwise from $*$, |
263 We define a map $\ev: K_0(M) \to M$. If $x \in K_0(M)$ has the label $m \in M$ at $*$, and labels $c_i \in C$ at the other labeled points of $S^1$, reading clockwise from $*$, |