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I think the OP meant “equivalent”, not “dual”, so just noting this example works for both.
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Jeremy Rickard
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The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respectively). However, there is an injective with endomorphism ring $\mathbb{Q}$, but no such projective, so the categories of projectives and injectives can't be equivalent or dual.

The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respectively). However, there is an injective with endomorphism ring $\mathbb{Q}$, but no such projective, so the categories of projectives and injectives can't be dual.

The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respectively). However, there is an injective with endomorphism ring $\mathbb{Q}$, but no such projective, so the categories of projectives and injectives can't be equivalent or dual.

Source Link
Jeremy Rickard
  • 35.2k
  • 2
  • 110
  • 151

The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respectively). However, there is an injective with endomorphism ring $\mathbb{Q}$, but no such projective, so the categories of projectives and injectives can't be dual.