[Note from moderator: resent with a subject line] Dear categorists, this is a remark about the very first proposition in the "elephant", followed by some questions. Lemma 1.1.1 says: If F is left adjoint to G : D --> C and there is a natural isomorphism FG --> 1_D (not necessarily the counit), then G is full and faithful. The proof sketched there uses the transfer of the comonad structure via the given iso. Here is an alternative and more explicit proof: 1) By the hypothesis, there are natural isomorphisms D(-,-) = D(FG-,-) = C(G-,G-) so that the bimodule C(G-,G-) : D -|-> D has a biuniversal element u(x) : Gx --> Gx, for any x in D. 2) By the right and left universality, u(x) is right and left invertible, so that it is an iso. 3) (Right) universality of u(x) gives bijections D(x,y) --> C(Gx,Gy) ; f |--> Gf.u(x) which factor through the arrow map of G and the composition by u(x); since the latter is a bijection by 2), the same holds for the former. Questions: 1) Is this proof correct? Is it "essentially the same" of the one in the book? 2) Anyway, the result states that in this case to say that "there is a natural isomorphism" is equivalent to say that "the canonical natural transformation (the counit) is an iso". Since many important categorical "exactness" conditions are expressed by requiring that some canonical transformations are iso (e.g. distributivity, Frobenius reciprocity and so on) one may wonder if also in these cases it is enough to require the existence of a natural isomorphism. I suppose that the answer is negative, but are there simple counter-examples? Best regards, Claudio [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Claudio, Regarding
2) Anyway, the result states that in this case to say that "there is a natural isomorphism" is equivalent to say that "the canonical natural transformation (the counit) is an iso". Since many important categorical "exactness" conditions are expressed by requiring that some canonical transformations are iso (e.g. distributivity, Frobenius reciprocity and so on) one may wonder if also in these cases it is enough to require the existence of a natural isomorphism. I suppose that the answer is negative, but are there simple counter-examples?
my immediate response, like yours, was that this couldn't be true. But in the case of distributivity, rather to my surprise, it turns out to be true: any category with finite products and coproducts and a natural family of isomorphisms XY+XZ~X(Y+Z) is in fact distributive. The proof, unfortunately, will not fit in this margin, but I have typed some notes which I can send you if you are interested. Regards, Steve Lack.
Best regards,
Claudio
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
On 20/11/2009, at 10:25 AM, claudio pisani wrote:
this is a remark about the very first proposition in the "elephant", followed by some questions. Lemma 1.1.1 says:
If F is left adjoint to G : D --> C and there is a natural isomorphism FG --> 1_D (not necessarily the counit), then G is full and faithful.
For anyone interested, I have put a string proof of this at <http://www.maths.mq.edu.au/~street/folkstring.pdf>. ==Ross [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Claudio,
If F is left adjoint to G : D --> C and there is a natural isomorphism FG --> 1_D (not necessarily the counit), then G is full and faithful.
A proof of the (dual) claim you find as the proof of Lemma 8.1 of my course on basic category theory www.mathematik.tu-darmstadt.de/~streicher/CTCL.pdf.gz on p.45. Thomas [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
participants (4)
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claudio pisani -
Ross Street -
Steve Lack -
Thomas Streicher