From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/5898 Path: news.gmane.org!not-for-mail From: Marco Grandis Newsgroups: gmane.science.mathematics.categories Subject: Re: Equality again Date: Tue, 1 Jun 2010 08:36:30 +0200 Message-ID: References: Reply-To: Marco Grandis NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 (Apple Message framework v752.2) Content-Type: text/plain; charset=US-ASCII; delsp=yes; format=flowed Content-Transfer-Encoding: 7bit X-Trace: dough.gmane.org 1275395829 25205 80.91.229.12 (1 Jun 2010 12:37:09 GMT) X-Complaints-To: usenet@dough.gmane.org NNTP-Posting-Date: Tue, 1 Jun 2010 12:37:09 +0000 (UTC) To: "Prof. Peter Johnstone" , categories@mta.ca Original-X-From: categories@mta.ca Tue Jun 01 14:37:08 2010 connect(): No such file or directory 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 4.69) (envelope-from ) id 1OJQiR-0008Qz-IM for gsmc-categories@m.gmane.org; Tue, 01 Jun 2010 14:37:07 +0200 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1OJQGh-0002Ks-Di for categories-list@mta.ca; Tue, 01 Jun 2010 09:08:27 -0300 Original-Sender: categories@mta.ca Precedence: bulk Xref: news.gmane.org gmane.science.mathematics.categories:5898 Archived-At: On 27 May 2010, at 13:30, Prof. Peter Johnstone wrote: > > TeX provides a command \doteq for an equality sign with a dot over it; > this is used in other areas of mathematics to mean "is approximately > equal to", but as far as I know it hasn't yet been used by category- > theorists. Perhaps we could use it to mean "is canonically > isomorphic to"? > > I'd also like to use it (or something like it) between pairs of > morphisms, meaning that (they are not equal but) they become equal > when composed with the appropriate canonical isomorphisms (to which > I can't be bothered to give names) in order to match up their domains > and codomains. (Of course, this is simply saying that they are > canonically isomorphic as objects of the functor category [2,C], > where C is the category in which they live.) > > Peter Johnstone Dear Peter, Isn't this very dangerous? 1. First, I think you are referring to some (specified) *coherent* (contractible) system of isomorphisms, otherwise you can easily prove that 1 = - 1 (see an example below). 2. Even in that case, we know that coherence can be a delicate thing. Let us take the cartesian product in Set (or the tensor product in a symmetric monoidal category). Would you write XxY =. YxX for the symmetry isomorphism s? Then by XxX =. XxX do you mean s or the identity? For XxXxX =. XxXxX we have six permutations of variables, generated by sxX and Xxs; and so on. I hope nobody will suggest some complicated trick to account for this; transpositions and permutations are already there, known to everybody; but we have to name them. 3. Coming back to point 1, "canonical" isomorphisms need not be coherent. There are a lot of such situations; I like to refer to the induced isomorphisms in homological algebra, because much of my early work was linked with that. A is an abelian group (or an object of an abelian category, or something more general that we do not need to consider here); X is a sublattice of the (modular) lattice of subobjects of A. We consider the subquotients H/K of A, where H and K belong to X. Then the canonical isomorphisms between these subquotients (induced by idA) are coherent if and only if X is distributive. (This is what I am calling now a "coherence theorem for homological algebra"; it applies to all the usual systems that produce spectral sequences, and is the reason "why" one cannot make errors when using canonical isomorphisms there.) An easy example of non-coherence can be built in the group A = ZxZ, taking for X the whole lattice of subgroups, obviously not distributive. Then Zx0 is canonically isomorphic to A/diagonal, and the latter is canonically isomorphic to 0xZ. Now, Zx0 and 0xZ are not canonically isomorphic, as already remarked in Mac Lane's "Homology". But notice that the composite of these isomorphisms is (x, 0) |--> (0, -x), while when you go through A/codiagonal, you get the opposite isomorphism, (x, 0) |--> (0, x). Best regards Marco Grandis [For admin and other information see: http://www.mta.ca/~cat-dist/ ]