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May 27, 2010 at 9:36 comment added Torsten Ekedahl The map $A^G_H \to A$ is a little bit tricky to define (and works only because $H$ ahs finite index in $G$). It depends on the fact that the coinduced and induced modules are isomorphic, i.e., $A^G_H$ is both right and left adjoint to the restriction functor. (Think of the case of the permutation module $k[G/H]$ which has a $G$-map $k \to k[G/H]$ mapping $1$ to the sum of the elements of $G/H$ and a $G$-map $k[G/H] \to k$ mapping all the elements to $1$.)
May 27, 2010 at 0:43 comment added user1832 Why $A^{G}_{H} \rightarrow A$ is a G-map? It seems just a H-map.
May 25, 2010 at 4:04 history answered Torsten Ekedahl CC BY-SA 2.5