From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/2908 Path: news.gmane.org!not-for-mail From: Peter Freyd Newsgroups: gmane.science.mathematics.categories Subject: *-Autonomous Functor Categories, revision Date: Sat, 26 Nov 2005 07:50:40 -0500 (EST) Message-ID: <200511261250.jAQCoetu005155@saul.cis.upenn.edu> NNTP-Posting-Host: main.gmane.org X-Trace: ger.gmane.org 1241018978 6325 80.91.229.2 (29 Apr 2009 15:29:38 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:29:38 +0000 (UTC) To: categories@mta.ca Original-X-From: rrosebru@mta.ca Sun Nov 27 21:02:15 2005 -0400 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Sun, 27 Nov 2005 21:02:15 -0400 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.52) id 1EgXIS-00004d-Vh for categories-list@mta.ca; Sun, 27 Nov 2005 20:55:09 -0400 Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 53 Original-Lines: 49 Xref: news.gmane.org gmane.science.mathematics.categories:2908 Archived-At: Mike Barr has pointed out that the proof in my last posting of LEMMA: The object I = H^R is injective in *F*. doesn't work. (It was actually the fourth proof I had come up with. I wondered why it was so much simpler). So here's one that does work (and is just about as simple). Let O | H^R | H^A --> H^B --> T --> O be exact (all vertical arrows point down). We seek a retraction for H^R --> T. Since H^R is projective (as is any representable) we may choose a map H^R --> H^B to yield a commutative triangle. The full subcategory of representables is closed under finite limits, so let H^C --> H^R | | H^A --> H^B be a pullback in *F* and let B --> A | | R --> C be the corresponding pushout in the category of f.p R-modules. The map from H^C to T is the zero map and we use the hypothesis that H^R --> T is monic to infer that H^C --> H^R, hence R --> C, are zero maps. Let O --> K --> B --> A be exact. It is an exercise in abelian categories that R --> C = 0 implies K --> B --> R is epi. Now (finally using the projectivity of R) choose a retraction R --> K. The map H^A --> H^B --> H^K --> H^R is of course, a zero map and we may factor H^B --> H^K --> H^R as H^B --> T --> H^R. The map T --> H^R is easily checked to be the retraction we seek.