From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/5293 Path: news.gmane.org!not-for-mail From: claudio pisani Newsgroups: gmane.science.mathematics.categories Subject: (unknown) Date: Thu, 19 Nov 2009 23:25:51 +0000 (GMT) Message-ID: Reply-To: claudio pisani NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable X-Trace: ger.gmane.org 1258692476 13218 80.91.229.12 (20 Nov 2009 04:47:56 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Fri, 20 Nov 2009 04:47:56 +0000 (UTC) To: categories@mta.ca Original-X-From: categories@mta.ca Fri Nov 20 05:47:48 2009 Return-path: Envelope-to: gsmc-categories@m.gmane.org Original-Received: from [138.73.1.1] (helo=mailserv.mta.ca) by lo.gmane.org with esmtp (Exim 4.50) id 1NBLPN-0005E9-E3 for gsmc-categories@m.gmane.org; Fri, 20 Nov 2009 05:47:45 +0100 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1NBKwT-0007KM-GI for categories-list@mta.ca; Fri, 20 Nov 2009 00:17:53 -0400 Original-Sender: categories@mta.ca Precedence: bulk Xref: news.gmane.org gmane.science.mathematics.categories:5293 Archived-At: Dear categorists, this is a remark about the very first proposition in the "elephant", follow= ed by some questions. Lemma 1.1.1 says: If F is left adjoint to G : D --> C and there is a natural isomorphism=20 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.=20 Here is an alternative and more explicit proof: 1) By the hypothesis, there are natural isomorphisms D(-,-) =3D D(FG-,-) =3D 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) ;=A0 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 bo= ok? 2) Anyway, the result states that in this case to say that "there is a natu= ral isomorphism" is equivalent to say that "the canonical natural transform= ation (the counit) is an iso". Since many important categorical "exactness" conditions are expressed by re= quiring that some canonical transformations are iso (e.g. distributivity, F= robenius 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 t= he answer is negative, but are there simple counter-examples? Best regards, Claudio =0A=0A=0A [For admin and other information see: http://www.mta.ca/~cat-dist/ ]