[ 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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