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* 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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