Re: question about discrete op-fibrations

[Thorsten Palm <palm@iti.cs.tu-bs.de>]

[ To all, moderator in particular, and with apologies to David:
   Please ignore my previous message. It contains a seriously wrong
   piece of information. ]

Dear David,

This time the answer is "yes" for discrete C, "no" otherwise.

If C is discrete, the functor !_C : C-->1 is a discrete
op-fibration, so that the terminal objet (1,!_C) of Cat_{C/}
belongs to DOF(C).

Now let C contain a non-trivial morphism u : c_0-->c_1 and suppose
that (T,z) is terminal in DOF(C). Construct a category C_{c_1} by
freely adding to C an object d and a morphism v : d-->c_1. Then we
have an inclusion j : C-->C_{c_1}, which is a discrete op-fibration.
We obtain two morphisms (C_{c_1},j)-->(T,z) in DOF(C) by applying z on
the subcategory C and sending v to z(u) and the identity of z(c_1),
respectively. By terminality of (T,z) they have to be the same. But
since z is an op-fibration, z(u) cannot be an identity ---
contradiction.

(I hope Mark hasn't beaten me to it again.)

Thorsten


David Spivak hat am 01.02.12 geschrieben:

>
> I hope this isn't annoying, but what if I change the problem somewhat
> and take DOF(C) to be the full subcategory of Cat_{C/} spanned by the
> discrete opfibrations C-->D? Again I want to know whether DOF(C) has a
> terminal object. Under this definition, by setting C=empty-category we
> get DOF(C)=Cat, which does have a terminal object.
>







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