216 is exact. For completeness we'll explain this below. |
216 is exact. For completeness we'll explain this below. |
217 |
217 |
218 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
218 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
219 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
219 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
220 Most of what we need to check is easy. |
220 Most of what we need to check is easy. |
221 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 $f(k) = 0 \in E$, so |
221 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 |
222 be $k=0$ 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 |
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223 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
222 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
224 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 $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
223 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 $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
225 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 $\hat{g}$. |
224 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 $\hat{g}$. |
226 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. |
225 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. |
227 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 |
226 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 |
230 \hat{g}(\widetilde{q}) & = \sum_i (a_i \tensor g(\widetilde{q_i}) \tensor b_i) - 1 \tensor (\sum_i a_i g(\widetilde{q_i}) b_i) \tensor 1 \\ |
229 \hat{g}(\widetilde{q}) & = \sum_i (a_i \tensor g(\widetilde{q_i}) \tensor b_i) - 1 \tensor (\sum_i a_i g(\widetilde{q_i}) b_i) \tensor 1 \\ |
231 & = q - 0 |
230 & = q - 0 |
232 \end{align*} |
231 \end{align*} |
233 (here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$). |
232 (here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$). |
234 |
233 |
235 Identical arguments show that the functors |
234 Similar arguments show that the functors |
236 \begin{equation} |
235 \begin{equation} |
237 \label{eq:ker-functor}% |
236 \label{eq:ker-functor}% |
238 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
237 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
239 \end{equation} |
238 \end{equation} |
240 are all exact too. Moreover, tensor products of such functors with each |
239 are all exact too. Moreover, tensor products of such functors with each |