Talk:Chain duality V (Ex)

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Latest revision as of 18:05, 29 August 2013

This is immediate from the definition that $<wikitex>; This is immediate from the definition that $T_{M,N}(\phi:TM\to N)=e_M\circ T(\phi)$. </wikitex>T_{M,N}(\phi:TM\to N)=e_M\circ T(\phi)$ .

Somewhat more elaborate, so that we really can see what's going on:

We turn the question around and show that $T_{M,TM}^{-1}(e_M)=\text{id}$ $T_{M,TM}^{-1}(e_M)=\text{id}$ . We will suppress all decorations $\Aa$ $\Aa$ occuring in tensor products and Hom-sets and look at a sequence of maps

$\displaystyle TM\otimes M = \text{Hom}(T^2M,M) \to T^2M\otimes TM = \text{Hom}(T^3M,TM)\to M\otimes TM = \text{Hom}(TM,TM),$ $\displaystyle TM\otimes M = \text{Hom}(T^2M,M) \to T^2M\otimes TM = \text{Hom}(T^3M,TM)\to M\otimes TM = \text{Hom}(TM,TM),$

given by

$\displaystyle \phi\mapsto T(\phi) \mapsto (e_{T(M)}\circ T)(\phi).$ $\displaystyle \phi\mapsto T(\phi) \mapsto (e_{T(M)}\circ T)(\phi).$

This sequence of maps is precisely the previously mentioned $T_{M,TM}^{-1}$ $T_{M,TM}^{-1}$ . (which has hopefully been treated in the solution to exercise 10 on L-groups) Applying this map to $e_M$ $e_M$ , we get $(e_{T(M)}\circ T)(e_M)$ $(e_{T(M)}\circ T)(e_M)$ , which is equal to the identity map, as part of its very definition.

Talk:Chain duality V (Ex)

Latest revision as of 18:05, 29 August 2013

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@@ Line 2: / Line 2: @@
 This is immediate from the definition that $T_{M,N}(\phi:TM\to N)=e_M\circ T(\phi)$.
 </wikitex>
+Somewhat more elaborate, so that we really can see what's going on:
+We turn the question around and show that <wikitex>$T_{M,TM}^{-1}(e_M)=\text{id}$.</wikitex> We will suppress all decorations <wikitex>$\Aa$</wikitex> occuring in tensor products and Hom-sets and look at a sequence of maps <wikitex>$$TM\otimes M = \text{Hom}(T^2M,M) \to T^2M\otimes TM = \text{Hom}(T^3M,TM)\to M\otimes TM = \text{Hom}(TM,TM),$$</wikitex> given by <wikitex>$$\phi\mapsto T(\phi) \mapsto (e_{T(M)}\circ T)(\phi).$$</wikitex> This sequence of maps is precisely the previously mentioned  <wikitex>$T_{M,TM}^{-1}$</wikitex>. (which has hopefully been treated in the solution to exercise 10 on L-groups) Applying this map to <wikitex>$e_M$</wikitex>, we get <wikitex>$(e_{T(M)}\circ T)(e_M)$</wikitex>, which is equal to the identity map, as part of its very definition.