Talk:Chain duality V (Ex)

This is immediate from the definition that $T_{M,N}(\phi:TM\to N)=e_M\circ T(\phi)$$<wikitex>; This is immediate from the definition that $T_{M,N}(\phi:TM\to N)=e_M\circ T(\phi)$. </wikitex> Somewhat more elaborate, so that we really can see what's going on: We turn the question around and show that <wikitex>$T_{M,TM}^{-1}(e_M)=\text{id}$.</wikitex> We will suppress all decorations <wikitex>$\Aa$</wikitex> occuring in tensor products and Hom-sets and look at a sequence of maps <wikitex>$$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.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.