From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/4367 Path: news.gmane.org!not-for-mail From: Janus Newsgroups: gmane.science.mathematics.categories Subject: Re: Equational correspondence and equational embedding Date: Tue, 15 Apr 2008 12:53:55 -0300 Message-ID: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-Trace: ger.gmane.org 1241019900 12936 80.91.229.2 (29 Apr 2009 15:45:00 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:45:00 +0000 (UTC) Cc: categories@mta.ca Original-X-From: rrosebru@mta.ca Wed Apr 16 09:36:53 2008 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Wed, 16 Apr 2008 09:36:53 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1Jm6jo-0005F9-0c for categories-list@mta.ca; Wed, 16 Apr 2008 09:27:44 -0300 Content-Disposition: inline Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 16 Original-Lines: 29 Xref: news.gmane.org gmane.science.mathematics.categories:4367 Archived-At: Dear Prof. Waddler, I don't know if the following proof is correct, please, let me know if I made any mistake: By hypothesis there exists k:S->T and h:k(S)->S such that h(k(S)) =3D_S s for all s in S and s =3D_S s' implies k(s) =3D_T k(s') for all s, s= ' in S On the other hand, there exists f:T->S and g:f(T)->S such that g(f(t)) =3D_T t for all t in T and t=3D_T t' implies f(t) =3D_S f(t') for all t, t' in T Assume that there exists t in T\k(S) =3D> |T| > |S| because k is injective and S =3D dom(k). Hence, there exists t' not equal to t such that f(t) =3D_S f(t') which is absurdum. So, there not exists t in T\k(S), then T =3D k(S). So, k and k^-1 constitutes an equational correspondence. Yours, Alejandro --=20 Alejandro D=EDaz-Caro Homepage: http://www.fceia.unr.edu.ar/~diazcaro Weblog: http://computacioncuantica.exactas.org