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1. • CommentRowNumber1.
   • CommentAuthorRichard Williamson
   • CommentTimeJul 6th 2020
   • (edited Jul 6th 2020)
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexGiven an (adjoint) equivalence in a 2-category, does anyone know if it is possible to replace the objects and/or 1-arrows in some reasonable way (up to equivalence/isomorphism) so that either the unit or the co-unit becomes an identity, not just a natural isomorphism? I don't have time to think about it just now, and maybe someone knows something off the top of their head.

   Given an (adjoint) equivalence in a 2-category, does anyone know if it is possible to replace the objects and/or 1-arrows in some reasonable way (up to equivalence/isomorphism) so that either the unit or the co-unit becomes an identity, not just a natural isomorphism? I don’t have time to think about it just now, and maybe someone knows something off the top of their head.

2. • CommentRowNumber2.
   • CommentAuthorRichard Williamson
   • CommentTimeJul 6th 2020
   • (edited Jul 6th 2020)
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexI guess this is fairly obvious actually. I'll drop adjointness for simplicitly. Suppose that we have $F: A \rightarrow B$ and $G: B \rightarrow A$, a natural isomorphism $\phi: GF \rightarrow id$, and a natural isomorphism $\psi: FG \rightarrow id$. Let $a$ be an object of $A$. Since $F$ is an equivalence, there must be an isomorphism $f_{a}$ in $B$ such that $G(f_{a})$ is equal (on the nose) to the isomorphism $\phi(a): GF(a) \rightarrow a$. We can then replace $F$ by $F'$, where $F'(a)$ is the target of $f_{f(A)}$, and $F'(g)$ for $g: a \rightarrow a'$ is $f_{f(a')} \circ F(g) \circ f_{f(a)}^{-1}$. Keeping $G$ the same, we then have that $GF'$ is on the nose equal to $id(A)$, and $F'G$ is still naturally isomorphic to $id$. If people agree, I'll add this to [[equivalence]] or some such page. If I have not made a mistake in the above, it must be in the literature somewhere; does anybody know of an explicit reference? Or if there is some canonical/abstract formalism which recovers the above, that would be very good to add too.

   I guess this is fairly obvious actually. I’ll drop adjointness for simplicitly. Suppose that we have $F: A \rightarrow B$ and $G: B \rightarrow A$, a natural isomorphism $\phi: GF \rightarrow id$, and a natural isomorphism $\psi: FG \rightarrow id$. Let $a$ be an object of $A$. Since $F$ is an equivalence, there must be an isomorphism $f_{a}$ in $B$ such that $G(f_{a})$ is equal (on the nose) to the isomorphism $\phi(a): GF(a) \rightarrow a$. We can then replace $F$ by $F'$, where $F'(a)$ is the target of $f_{f(A)}$, and $F'(g)$ for $g: a \rightarrow a'$ is $f_{f(a')} \circ F(g) \circ f_{f(a)}^{-1}$. Keeping $G$ the same, we then have that $GF'$ is on the nose equal to $id(A)$, and $F'G$ is still naturally isomorphic to $id$.

   If people agree, I’ll add this to equivalence or some such page. If I have not made a mistake in the above, it must be in the literature somewhere; does anybody know of an explicit reference? Or if there is some canonical/abstract formalism which recovers the above, that would be very good to add too.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeJul 6th 2020
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThat would only be true if you already knew there was an *object* $b$ such that $G(b)=a$ on the nose.

   That would only be true if you already knew there was an object $b$ such that $G(b)=a$ on the nose.

4. • CommentRowNumber4.
   • CommentAuthorRichard Williamson
   • CommentTimeJul 7th 2020
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexYes, thanks, good point! The functor $G$ would have to be assumed to be surjective-on-objects, which is probably fine in practise in most cases. I'll add something about this a little later to the nLab.

   Yes, thanks, good point! The functor $G$ would have to be assumed to be surjective-on-objects, which is probably fine in practise in most cases. I’ll add something about this a little later to the nLab.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeJul 7th 2020
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexNote that surjective-on-objects equivalences are the acyclic fibrations in the [[canonical model structure on Cat]].

   Note that surjective-on-objects equivalences are the acyclic fibrations in the canonical model structure on Cat.