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From: "Prof. Peter Johnstone" <P.T.Johnstone@dpmms.cam.ac.uk>
To: Categories list <categories@mta.ca>
Subject: Re: On defining *-autonomous categories
Date: Wed, 19 Dec 2007 12:04:56 +0000 (GMT)	[thread overview]
Message-ID: <E1J54IV-000315-LZ@mailserv.mta.ca> (raw)

This result is a special case of a general fact about adjunctions,
which appears (yes, really!) as Lemma A1.1.1 in the Elephant.
I don't claim any originality for it, but I did comment in the
text that it "seems not to be widely known".

Lemma: Let F: C --> D be a functor having a right adjoint G.
If there is any natural isomorphism between the composite FG and
the identity functor on D, then the counit of the adjunction is
an isomorphism.

Proof: One can transport the comonad structure on FG across the
isomorphism, to obtain a comonad structure on 1_D. But the monoid
of natural endomorphisms of the identity functor on any category
is commutative, so the counit and comultiplication of this comonad
must be inverse isomorphisms. Transporting back again, the counit
of (F -| G) is an isomorphism.

Peter Johnstone

On Tue, 18 Dec 2007, Robin Houston wrote:

> On Mon, Dec 17, 2007 at 02:29:17PM -0400, Peter Selinger wrote:
>> On the other hand, none of the above matters in some sense, due to a
>> theorem proved last year by Robin Houston (it was previously unknown
>> at least to me):
>>
>>  Theorem. Let C be a symmetric monoidal category, and let D be an object.
>>  If there *exists* a natural isomorphism f : A -> (A -o D) -o D,
>>  then the *canonical* natural transformation g : A -> (A -o D) -o D
>>  (coming from the symmetric monoidal structure) is an isomorphism
>>  (although it may in general be different from f).
>
> I suspect I may not have been the first to notice this, since it's
> fairly easy, though I'm not aware that it's ever been mentioned
> in print. (If anyone knows otherwise, I'd be interested to hear.)
>
>
> Claim: Let C be a symmetric monoidal closed category, and
> let D be an object of C. Then the following are equivalent:
>
> 1. There exists a natural isomorphism A = (A -o D) -o D,
> 2. The functor (- -o D) is full and faithful,
> 3. The canonical natural transformation A -> (A -o D) -o D is
>   invertible.
>
>
> Lemma 1. Let G: X -> Y and H: Y -> Z be functors. If HG is faithful
> then G is faithful; if HG is full and G is essentially surjective,
> then H is full.
>
> Proof: easy.
>
> Lemma 2. Let C be a category, and F: C -> C an endofunctor. If FF
> is naturally isomorphic to 1_C, then F is an equivalence.
>
> Proof: Certainly F is essentially surjective, since every object
> X in C is naturally isomorphic to FFX. It then follows by Lemma 1
> that F is full and faithful.
>
>
> For the implication 1 => 2, take F = (- -o D) in Lemma 2.
>
>
> For the implication 2 => 3, consider the following sequence of
> isomorphisms natural in A \in C:
>
>    C(A, A)
> -> C(A -o D,  A -o D)       ;apply the functor (- -o D)
> =  C((A -o D) @ A, D)       ;closure
> =  C(A @ (A -o D), D)       ;symmetry of tensor
> =  C(A,  (A -o D) -o D)     ;closure again
>
> By definition, this natural transformation corresponds under Yoneda
> to the canonical A -> (A -o D) -o D. Since (- -o D) is full and faithful,
> the first step is invertible; the others are necessarily so. Therefore,
> by the Yoneda correspondence, the canonical natural transformation with
> components A -> (A -o D) -o D is invertible.
>
> [ Robin Cockett pointed out to me that this can be decomposed into
>  two steps, one easy and one well-known:
>   a) The functor (- -o D) is self-adjoint on the right.
>   b) For any adjunction, the left adjoint is full and faithful
>      if and only if the unit of the adjunction is invertible. ]
>
>
> Finally, the implication 3 => 1 is immediate.
>
> Robin
>
>
>




             reply	other threads:[~2007-12-19 12:04 UTC|newest]

Thread overview: 5+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2007-12-19 12:04 Prof. Peter Johnstone [this message]
  -- strict thread matches above, loose matches on Subject: below --
2007-12-18 14:27 Robin Houston
2007-12-18 13:08 Robin Houston
2007-12-17 18:29 Peter Selinger
2007-12-17  2:45 Michael Barr

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