On Mon, 28 May 2012 10:04:33 -0400
Brent Yorgey <
byorgey@seas.upenn.edu> wrote:
> On Mon, May 28, 2012 at 02:08:14PM +0200, Manfred Lotz wrote:
> > On Mon, 28 May 2012 13:34:01 +0200
> > Alessandro Pezzoni <
alessandro_pezzoni@lavabit.com> wrote:
> >
> > > On Mon, May 28, 2012 at 11:54:11AM +0200, Manfred Lotz wrote:
> > > > In the definition of a (mathematical) category it is said (among
> > > > other things), that for any object A there exists an identity
> > > > morphism:
> > > >
> > > > idA: A -> A and if f: A -> B for two objects A, B then
> > > >
> > > > idB . f = f and f . idA = f
> > > >
> > > > must hold.
> > > >
> > > > My question: Because I cannot think of any counterexample for
> > > > the last statement I would like to know if I just could omit
> > > > this from the definition and formulate this as a small theorem.
> > > >
> > > > Or does there exist a counterexample where all conditions of a
> > > > category hold but there exist two objects A, and B where we have
> > > > idB . f <> f and/or f .idA <> f?
> > >
> > > When you ask that
> > > idB . f = f and f . idA = f
> > > you are basically defining a left and a right identity,
> > > respectively.
> > >
> > > If I get your question correctly, you are asking if you can drop
> > > the axiom (requirement) of existence of an identity morphism for
> > > every object and deduce it from the other axioms, i.e. that the
> > > composition of morphisms is always well defined and that it is
> > > associative.
> > >
> >
> > No, I do not want to drop the requirement of existence of an
> > identity morphism. I only want to drop the axion that idB .f = f
> > and f . idA = f do hold because IMHO this follows readily from the
> > definition of what an identity morphism is all about.
>
> "Follows readily from the definition of what an identity morphism is
> all about" -- and what exactly is that defintion?
>