The answer of Ben Webster, can be made easier. Consider the functor F : (A-mod) -> (A-mod) which maps any A-module on (0). Then, F is self-adjoint --- but a left adjoint to F ; and so, is a also a right adjoint to F. This is clear because for all A-modules N, M, one has Hom_A(0,N)=Hom_A(M,0). But, F is not an equivalence.
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The answer of Ben Webster, can be made easier. Consider the functor F : (A-mod) -> (A-mod) which maps any A-module on (0). Then, F is self-adjoint --- but not an equivalence. |
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