# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorRichard Williamson
• CommentTimeJul 6th 2020
• (edited Jul 6th 2020)

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.

• CommentRowNumber2.
• CommentAuthorRichard Williamson
• CommentTimeJul 6th 2020
• (edited Jul 6th 2020)

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.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeJul 6th 2020

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

1. 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.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeJul 7th 2020

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