Question about Ext I heard that $Ext(M,N)$ is naturally isomorphic to $Ext(M^*\otimes N,1)$ where 1 is the trivial representation and $M,N$ some representations of a group $G$.
Can anyone explain why?
Is there an explicit construction of a map from one to the other or does it just follow from some general considerations about derived functors?
Thanks.
 A: This is probably typographic, but the original question writes

Ext(M,N)  is naturally isomorphic to Ext(M*⊗N,1)

when what is true (see other answers) is that Ext(M,N) is isom.
to Ext(M $\otimes$ N*,1).
By the way, it is often useful to view Ext(M,N)
as Ext(1,M* $\otimes$ N) instead, since this later ext-group
coincides with the group cohomology H(G,M* $\otimes$ N);
here H*(G,-) = derived functor of "G-fixed-points".
A: Perhaps the best way to think about this is as follows:  pick your favorite injective resolution for N and favorite projective resolution of M.  Then $\mathrm{Ext}(M,N)$ is given by taking Hom between these complexes (NOT chain maps, just all maps of representations between the underlying modules), and putting a differential on those in the usual way.
Now, use the usual identification of $\mathrm{Hom}(A,B)\cong \mathrm{Hom}(A\otimes B^*,1)$ on this complex.  So you see, it's the same as if we had tensored the projective resolution of $M$ with the dual of the injective resolution of $N$, which is a projective resolution of $N^*$, and then taken Hom to 1.  Of course, the tensor product of two projective resolutions is a projective resolution of the tensor product, so we see this complex also computes $\mathrm{Hom}(N^*\otimes M,1)$.`
It also follows by abstract nonsense in one line: isomorphic functors have isomorphic derived functors.
