* Products redone
@ 2000-05-17 20:09 Peter Freyd
0 siblings, 0 replies; only message in thread
From: Peter Freyd @ 2000-05-17 20:09 UTC (permalink / raw)
To: categories
Kai Bruennler asked:
Is there a binary product in the category of sets and functions
that is "strictly associative", i.e.
A x (B x C) = (A x B) x C and
the associativity isomorphisms are equal to the identity?
I gave a construction using the axiom of choice. Well, we can do
better. Fix on an ordered-pair construction, say Weiner's, and denote
its values as <x,y>. Let l be the left-coordinate function, r the
right-coordinate function.
Fix on a sequence of finite ordinals, say von Neumann's. Define,
inductively, an n-TUPLE as something of the form <x,y> where y is
a (n-1)-tuple if n>1 else any old set if n=1. (There are no
0-tuples.) Note that the same set can be an n-tuple for any number of
values for n. But if x is an n-tuple and we are given n then
each of it's n coordinates is well-defined:
x_1 = lx
x_2 = l(rx)
...
x_{n-1} = l(r(r(...r(rx)...)))
x_n = r(r(r(...r(rx)...)))
where there are i-1 applications of r used for x_i, one
application of l used for x_i if i<n and no application of l
used for x_n.
A trick of this construction of binary products is that we define
products twice. Given n sets a_1, a_2,...,a_n define their n-fold
PRE-PRODUCT as the set of pairs of the form <x,n> where x is an
n-tuple such that x_i \in a_i each relevant i.
If two sets x and y arise as pre-products, that is, if there are
positive ordinals m and n and sets a_1, a_2,..., a_m, b_1, b_2,
...,b_n such that x is the pre-product of the a's and y is the
product of the b's, then we define their PRODUCT as the (m+n)-fold
pre-product of a_1, a_2,..., a_m, b_1, b_2,...,b_n.
Define F(x) = x if x is a pre-product else { <y,1> | y \in x}.
Define the product of arbitrary x and y as the product of F(x)
and F(y).
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