Thursday, November 14, 2002

Pullback refresher! A pullback over (the diagram) A -f-> C <-g- B, is A <-x- P -y-> B such that f(x(p:P)) = g(y(p:P)) - the diagram commutes, e.g. in Set, P only contains elements that "wind up" at the same element of C <x;P;y> is universal among all such <x';P';y'>, i.e. any other such object and pair of arrows "factors" through <x;P;y> via a unique arrow P'-h->P. For example: again in Set, take P to be the subset of AxB for which x(a:A) = y(b:B), and take x and y to be the projections x: AxB -> A and y: AxB -> B. This satisfies 1 by definition. Before (or instead of) proving that it also satisfies 2, and that it's the only way to satisfy 2 (making it "the" pullback of this diagram), think about some other ways of satisfying 1. How about if P is an even smaller subset of AxB? That still satisfies 1, since everything in it still comes out equal "at the other end," in C. But this breaks 2. For example: our original restricted subset above is now a P' that has to factor uniquely through the new P. Now, since there are some elements in P' that "don't fit" in P, a function P'-h->P will have to map those elements to something else in P (i.e., h will no longer be 1-1). Which other elements to use? Well, there are a lot of choices - exactly what "unique" in 2 means there can't be. Okay, how about a "bigger" subset of AxB, but with x and y being "real" functions instead of projections, which happen to always map things in the new AxB onto members of A and B that "come out equal" (i.e., x and y create equivalence classes in the "bigger" AxB that map onto the "right" AxB)? This has the same problem as the preceding example: there are now many ways to map a given <x';P';y'> "through" our candidate P - just pick anything in each equivalence class and you're ok. Okay. How about a P that isn't a subset of AxB at all, just an arbitrary set P with one function P-x->A, and another function P-y->B. What characteristics would this set need to have, to satisfy the conditions? To satisfy 1, it couldn't have any elements p:P where f(x(p)) <> g(y(p)). And to satisfy 2, elements in P would have to "correspond to" pairs (a:A;b:B) which "come out equal" in C, in only one way. Basically, what a proof shows is that any set P that satisfies the conditions is isomorphic to the restricted subset of AxB we started with - meaning it doesn't have to "actually be" that set (whatever that means), but it has to be in a 1-1 correspondence with it. Next steps:     +:Cn->C --- d:C->Cn --- x:Cn->C        |                       |     colimit ---------------- limit        |                       |     pushout --------------- pullback        |                       |   coequalizer ------------ equalizer 12:52:01 PM

Could I possibly rely more on public broadcasting? No. This guy is doing some great experiments in the econ + psych spot. In particular demonstrating how a random quantity can act to scale an entire coherent network of relative values. A flurry of stuff comes to mind: a theoretical framework for all kinds of sales tactics (e.g. manipulating people's sense of value by "throwing anchors" at them in a deliberate way); the real difference between one person who has "money sense" and another who doesn't (e.g. being able to "self-anchor" in the presence of an impending value judgement - something I find myself doing as I get older, especially as I look back at periods of real profligacy and notice how my drifting sense of absolute monetary value played a part)... ...and beyond anything economic, it's a very nice example of some fundamentals of cognition, which lead you all the way out to, like, dynamic systems. Great stuff. (I'm not even high!) This paper is a nice summary - others may be even better; I haven't looked at everything. 11:58:41 AM