Skip to main content
1 of 4
Bugs Bunny
  • 12.3k
  • 1
  • 30
  • 65

As a discerning voice, let me say that it is true.

Take abelian $A$. Let $B:=A$ with forgotten enrichment in abelian groups. Then the identity functor is equivalence, but $B$ is not abelian because it is not even additive.

It is all in your definition, doc!!

Bugs Bunny
  • 12.3k
  • 1
  • 30
  • 65