From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/1016 Path: news.gmane.org!not-for-mail From: "Dr. P.T. Johnstone" Newsgroups: gmane.science.mathematics.categories Subject: Re: Pullback preserving functor Date: Fri, 29 Jan 1999 15:35:47 +0000 (GMT) Message-ID: References: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1241017478 28996 80.91.229.2 (29 Apr 2009 15:04:38 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:04:38 +0000 (UTC) To: categories@mta.ca Original-X-From: cat-dist Fri Jan 29 23:05:04 1999 Original-Received: (from Majordom@localhost) by mailserv.mta.ca (8.8.8/8.8.8) id WAA03223 for categories-list; Fri, 29 Jan 1999 22:22:07 -0400 (AST) X-Authentication-Warning: mailserv.mta.ca: Majordom set sender to cat-dist@mta.ca using -f In-Reply-To: from "Hongseok Yang" at Jan 28, 99 11:13:03 pm X-Mailer: ELM [version 2.4 PL25] Original-Sender: cat-dist@mta.ca Precedence: bulk Original-Lines: 36 Xref: news.gmane.org gmane.science.mathematics.categories:1016 Archived-At: > Would someone let me know the answer and the proof or counter example of > the following question? > > Suppose the category C has a pullback for every pair of morphism > (f : X -> Y, g : W -> Y). Let K be the full subcategory of the functor > category Func(C,Set) whose objects are pullback perserving functors. > Is K ccc? (If so, how I can show this?) The answer is no. First note that K is closed under products in the functor category. Also, it contains all the representable functors; so, if it were cartesian closed, the exponential G^F would have to be given by G^F(c) \cong nat((c,-),G^F) \cong nat((c,-)\times F,G) i.e. K would have to be closed under exponentials in [C,Set]. However, it isn't in general. For a simple counterexample, let C be the category with five objects a,b,c,d,e and six non-identity morphisms a --> b, a --> c, b --> d, c --> d, a --> d, a --> e ; note that C has just one nontrivial pullback square a -----> b | | | | v v c -----> d Let F be the functor given by F(a) = F(b) = F(c) = F(d) = \emptyset, F(e) = {*}, and let G be F + F. Then (taking the above definition of G^F) G^F(a) has two elements, but G^F(b), G^F(c) and G^F(d) are singletons. Peter Johnstone