Talk:Chain duality V (Ex)

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This is immediate from the definition that 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}. We will suppress all decorations \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),
given by
\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}. (which has hopefully been treated in the solution to exercise 10 on L-groups) Applying this map to e_M, we get (e_{T(M)}\circ T)(e_M), which is equal to the identity map, as part of its very definition.
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