From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/5296 Path: news.gmane.org!not-for-mail From: Steve Lack Newsgroups: gmane.science.mathematics.categories Subject: Re: Question on adjoints Date: Mon, 23 Nov 2009 18:04:26 +1100 Message-ID: Reply-To: Steve Lack NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1258977771 26776 80.91.229.12 (23 Nov 2009 12:02:51 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Mon, 23 Nov 2009 12:02:51 +0000 (UTC) To: claudio pisani , categories Original-X-From: categories@mta.ca Mon Nov 23 13:02:44 2009 Return-path: Envelope-to: gsmc-categories@m.gmane.org Original-Received: from mailserv.mta.ca ([138.73.1.1]) by lo.gmane.org with esmtp (Exim 4.50) id 1NCXcx-0004IK-8s for gsmc-categories@m.gmane.org; Mon, 23 Nov 2009 13:02:43 +0100 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1NCX3M-0004sS-OL for categories-list@mta.ca; Mon, 23 Nov 2009 07:25:56 -0400 Original-Sender: categories@mta.ca Precedence: bulk Xref: news.gmane.org gmane.science.mathematics.categories:5296 Archived-At: 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/ ]