categories: Exact adjunctions

[grandis@dima.unige.it (Marco Grandis)]

There is a nice lemma on adjoint functors, purely 2-categorical, and
certainly known to various colleagues.

As I need it in a paper on homotopy, I would like to know if it is
*published with proof*, somewhere.
I also like to "advertise" it here, because I think it deserves to be known
more widely.

LEMMA.

If,in an adjunction, any one of the four natural transformations which
appear in the triangle identities is invertible, so are the other three.
[Proof below.]

A few years ago I was considering such adjunctions, which I was calling
"connections" because adjunctions between ordered sets ("covariant Galois
connections") are always of this type.
Renato Betti was also considering them, under the name of "exact
adjunctions" (which I now prefer, also because "connection" has already too
many meanings).
I learnt from him that "one condition is sufficient".

At my knowledge, the above result appears in two works. It is cited without
proof in

- R.Betti, Adjointness in descent theory,
JPAA 116 (1997), 41-47.

It also appears, with a proof and in a more general formulation (for
biadjoints), in a preprint:

- R.Betti, D. Schumacher and R. Street, Factorizations in bicategories,
Dip. Mat. Politecnico di Milano n. 22/R, 1999.

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Proof.

Write  F -| G  the adjunction;
u: 1 -> GF  the unit;  v: FG -> 1  the counit.

   The triangle identities say that

 (a)  vF.Fu = idF,    (b)  Gv.uG = idG.

   Assume that  Fu  is invertible, so that also  vF  is so and it is
sufficient to prove that:

(c) uG.Gv = id(GFG).

        This commposite occurs in the upper row of the following
commutative diagram
(vertical arrows down)

            Gv         uG
      GFG  ---->  G   ---->  GFG

  uGFG |          |uG         |uGFG

     GFGFG ----> GFG  ----> GFGFG
           GFGv       GFuG

   Now, the lower row is the identity, because  G(Fu)G  is invertible and
the "other composite", GF(Gv.uG)  is an identity, by (b). Since the
vertical arrow  uGFG  has a left inverse, by (b) again, "cancelling it from
the outer rectangle" we get (c).

[Perhaps it can be simplified; I did not spent much time for that.]

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