From: Chris Heunen <heunen@math.ru.nl>
To: categories@mta.ca
Subject: Re: pushouts in REL
Date: Sat, 22 Aug 2009 11:00:54 +0200 [thread overview]
Message-ID: <E1Meq25-0007HW-IY@mailserv.mta.ca> (raw)
Dear all,
> let A and B be sets and R,S \inc A x B. Let A_0 be the subset of A consisting
> of all a such that (a,b) \in R iff (a,b) \in S. Then it seems to me that
> the inclusion function of A_0 into A is the equalizer of R and S.
I already answered Sergei privately, but this should be corrected
publicly: Rel does not have equalisers. In fact, it has almost no
equalisers.
For a counterexample, consider the sets A = {0,1} and B = {0}, and the
parallel relations R = A x B and S = {(0,0)}. Their equaliser, if it
existed, must be contained in T = {(0,0)}. But T' = {0} x A also
satisfies RT'=ST', but does not factor through any subrelation of T.
Best wishes,
Chris Heunen
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
next reply other threads:[~2009-08-22 9:00 UTC|newest]
Thread overview: 8+ messages / expand[flat|nested] mbox.gz Atom feed top
2009-08-22 9:00 Chris Heunen [this message]
-- strict thread matches above, loose matches on Subject: below --
2009-08-22 19:08 Vaughan Pratt
2009-08-21 23:45 Dusko Pavlovic
2009-08-21 22:02 Dominic Hughes
2009-08-21 21:36 Prof. Peter Johnstone
2009-08-21 20:38 Jamie Vicary
2009-08-20 21:38 Michael Barr
2009-08-20 10:01 soloviev
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