===== 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