* even more on inclusion maps
@ 2010-08-29 21:21 Peter Freyd
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From: Peter Freyd @ 2010-08-29 21:21 UTC (permalink / raw)
To: categories
The second construction I gave in my last note ("more on inclusion
maps") allowed a choice-free equivalence functor to one with an
inclusion-map structure. A simple argument shows that one could use
such not to find an inclusion structure on the original category
(which we know is not always possible) but at least to find a choice
nof monics, one in each set of names for a given subobject. And that of
course would imply the axiom of choice (see below).
Yikes.
In case one needs a proof, given a family of sets construct a category
as follows: for each S in the family let A_S B_S be names of
objects. The home-sets of the form (B_S, B_S) each have only one
element; S itself will be the hom-set (A_S, B_S), the complete group
of permutations of S will be the home-set (A_S, A_S), all other
hom-sets are empty. The composition of the endomorphism of A_S with
the maps to B_S is just what you would expect. All elements of
(A_S, B_S) name the same subject. Hence a choice of a monic from
each set-of-subobject-names yields a choice function for the
non-empty sets in the given family.
The second construction doesn't work. The much easier first
construction -- fortunately -- does work.
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2010-08-29 21:21 even more on inclusion maps Peter Freyd
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