categories: RE: A curiosity

[glenn <glenn@cua.edu>]

>===== Original Message From Michael Barr <barr@barrs.org> =====
>You may be amused trying to prove it (it is not easy), but a category in
>which every morphism has a unique quasi-inverse (AfE!g fgf=f) is a
>groupoid.  It is actually a theorem about semigroups, but the proof works
>just as well for categories.  (You can add an identity without affecting
>the hypothesis.)
>
>Michael

Dear category list,

Mike's exercise: Suppose C is a category with the following
property. For all f there exists a unique g such that fgf=f.
Prove: C is a groupoid.

Here's a short argument: Note that fgf=f implies we have two
idempotents, namely gfgf=gf and fgfg=fg. So as a special
case, suppose ff = f. The uniqueness of g such that fgf=f
together with the fact that fff=ff=f and f1f=ff=f implies
that f=g=1. So the only idempotents are the identity maps.
Therefore, also, gf=1 and fg=1.

Paul Glenn
Dept of Mathematics
Catholic Univ. of America
Washington, DC 20064