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I know that whether something is obvious or not is highly subjective and depends on the perspective. Nonetheless, some things are collectively regarded to be less obvious than others.

I also know that left adjoints are unique up to a unique isomorphism. An isomorphism may not be obvious, however. Some less immediate examples from basic algebra are listed in https://math.stackexchange.com/a/1268003/229174 and some examples of isomorphic categories are listed in https://mathoverflow.net/questions/13995/nontrivial-isomorphisms-of-categories.

When talking about natural isomorphisms in a functor category, however, I don't know of an example where the isomorphism would need a lot of thought. For example, a free abelian group (left adjoint to $U: \mathbf{Ab} \to \mathbf{Set}$) can be defined either directly as a set of finite multisets or as the abelianization (left adjoint to $U: \mathbf{Ab} \to \mathbf{Grp}$) of the free group (left adjoint to $U: \mathbf{Grp} \to \mathbf{Set}$). The proof of their equivalence is both simple and intuitive, however.

I'm looking for counterintuitive examples to natural isomorphisms, ideally for left adjoints.

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    What's the question? Are you looking for such an example? – blargoner Jan 09 '22 at 18:33
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    I added a sentence in the end of the question to clarify that I'm looking for counterintuitive examples. – Ianis Vasilev Jan 09 '22 at 18:41
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    The forgetful functor from groups to sets might be what you want. Its left adjoint is the free group functor, but that has several constructions whose equivalence may not be obvious. The most familiar construction of the free group on a set $S$ is the set of reduced words on the alphabet $S\cup S^{-1}$. Another construction, given (if I remember correctly) in Lang's "Algebra", essentially follows the proof of the adjoint functor theorem. If I had to prove that these constructions are equivalent, I'd probably just check that they're both adjoint to "underlying set". – Andreas Blass Jan 10 '22 at 04:17

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