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Feb 25, 2010 at 17:58 vote accept Doctor Gibarian
Feb 25, 2010 at 17:57 vote accept Doctor Gibarian
Feb 25, 2010 at 17:57
Feb 25, 2010 at 17:15 answer added Reid Barton timeline score: 3
Feb 25, 2010 at 17:11 answer added Joel David Hamkins timeline score: 2
Feb 25, 2010 at 16:34 comment added Doctor Gibarian Mh. In that case the square giving raise to the pullback woludn't be commutative and therefore the pullback set would be empty. Isn't it? My question is: could we make that for prim. rec. functions (where all of them are total and we've not the empty set as a possible domain)? Have they pullbacks? F.G.Dorais: I wish I could find that info in the web you send, but I don't see it there. Thank you anyway.
Feb 25, 2010 at 16:22 comment added Tom Leinster Ximo, it looks to me as if your first paragraph is asking the following question. (If not, please clarify.) Let X, Y and Z be sets, and let f: X --> Z and g: Y --> Z be functions. Is it possible that the pullback of f with g is empty? Kevin answered "no, e.g. take X to be the one-element set". More specifically, you could take X and Y both to be one-element sets, Z to be a two-element set {z_1, z_2}, and f and g to be the functions with respective images z_1 and z_2.
Feb 25, 2010 at 15:40 comment added François G. Dorais Are you asking about the existence of pullbacks in the Lawvere Theory of primitive recursive functions? ncatlab.org/nlab/show/Lawvere+theory
Feb 25, 2010 at 15:37 comment added Doctor Gibarian Pullback to a point? You mean a degenerate case of a pullback? Makes it any sense? And..what if we haven't the empty set available?
Feb 25, 2010 at 15:24 comment added Kevin Buzzard "could this pullback set be the empty set in some cases?" Yes! Pullback to a point is just "fibre of the map" so sure a fibre can be empty.
Feb 25, 2010 at 15:15 history asked Doctor Gibarian CC BY-SA 2.5