Dear Neil, Steve, Steve: what you say about fibrations and opfibrations is completely correct. So if E ---> D is a bifibration, and D ---> C is a bifibration, then the composite E ---> C is a bifibration. It is moreover easy to check that if the individual parts of this satisfy Beck-Chevalley, then so too will the composite. Thus the composite of fibrations with sums is again a fibration with sums. For products, though, I think the situation is a bit more complex. I could not see how to derive the result by a simple duality. From scribbling on the back of an envelope, it does appear to be true that: - if E -p--> D and D ---q---> C are fibrations with products, then so too is E ---qp---> C The argument I have is basically the following. Given e in E with pe = d and qd = c, and some f: c ---> c' in C, we wish to describe the Pi of e along f; we'll write this as f_*(e). Well first we form f_*(d) in D, its pullback g : f^*f_*(d) ----> f_*(d) along f, and the counit k: f^*f_*(d) ---> d in the fibre over c. Using these data in D, we now form in E the pullback k^*(e) of e along k and then the Pi along g, so yielding e' = g_*k^*(e). Now pe' = f_*(d) and q(f_*(d)) = c' and it's now not so hard to check that e' is in fact the Pi of e along f, as desired. For the checking in the last step, I seemed to need a few times the Beck-Chevalley condition for products. So I doubt one could get away without that. I have not tried to verify whether the Pi's of the composite so defined themselves satisfy the Beck-Chevalley condition, but I would be amazed if this were not the case. Finally, one might ask the following question. Suppose that E ----> D and D ----> C are fibrations with products and sums, and that in each case the sums and products satisfy the distributivity axiom ("type-theoretic axiom of choice"). Does the composite fibration, which by the above again has sums and products, also satisfy the distributivity axiom? I do not know the answer to this, but I rather suspect so. It would be a largeish diagram chase. It would be nice to know if there was a more abstract reason why this is true. Richard On Sun, Jun 8, 2014, at 11:58 PM, Steve Vickers wrote:
Some thoughts:
* The result about composition of fibrations holds in any 2-category with comma objects and 2-pullbacks, not just Cat. (Think of the Chevalley criterion for fibrations.)
* By duality on 2-cells it thus also applies to opfibrations, and hence to bifibrations.
* It is bifibration structure that gives you the left adjoints you ask for.
* For the right adjoints, look at the dual 2-category, where your fibrations become bifibrations.
Hence it seems to me that your conjectures are all true, and even generalize widely.
Steve.
On 6 Jun 2014, at 10:47, Neil Ghani <neil.ghani@strath.ac.uk> wrote:
Dear All
We know that if p and q are fibrations, then their composition p.q is a fibration.
But what about quantification … that is if reindexing along every morphism has a right/left adjoint in p and q, then does reindexing along every morphism in p.q have a right/left adjoint? Under some circumstances?
Thanks for any thoughts Neil
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