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(This question was asked in https://math.stackexchange.com/questions/741334/bicategorical-limits-with-parameters with no answer.)

Let $F(-,-)\colon \mathcal{A}\times \mathcal{B}\to \mathcal{C}$ be a pseudofunctor (other type of 2-functors may be considered as well) between bicategories. Suppose that $\mathcal{C}$ has all limits.

For fixed $A\in \mathcal{A}$, we compute the pseudolimit of $F(A,-)$ along $\mathcal{B}$. The limit is not unique but well-defined up to equivalences and the equivalences are unique up to a unique 2-cell.

The questions:

  1. Can we produce a pseudofunctor $\mathcal{A}\to \mathcal{C}$, uniquely defined up to natural transformations, and the natural transformations are unique up to a unique modification?

  2. For the induced pseudofunctor, we compute the limit again. Are the limits interchangeable?

These may be standard facts; for 1-categories, one can verify them by hand (see also Mac Lane).

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    $\begingroup$ A small note: The original functor you consider should be a bi-functor, i.e. a functor from the Gray tensor. Probably your construction will yield two limits which are isomorphic to the limit of the functor from the tensor. $\endgroup$
    – Adam Gal
    Commented Apr 10, 2014 at 19:49
  • $\begingroup$ @AdamGal Gray tensor product for bicategories? It seems that Gray tensor product is for 2-categories (I remember that the Cartesian product of bicategories is not closed but weak closed.) $\endgroup$
    – Ma Ming
    Commented Apr 12, 2014 at 9:41
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    $\begingroup$ There is a Gray tensor product for bicategories. However, at least if you're talking about the pseudo version of the Gray tensor product, it is equivalent to the cartesian product (in the appropriately weak sort of "equivalence" for bicategories, a.k.a. "biequivalence"). Thus, if you're talking about pseudofunctors (as you usually are when considering bicategories), there's no point to using the Gray tensor product instead of the cartesian product. $\endgroup$
    – Mike Shulman
    Commented Apr 12, 2014 at 23:33
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    $\begingroup$ As for your original questions, I think the answers are "of course", but I don't know whether anyone has sat down and written out the details. (-: $\endgroup$
    – Mike Shulman
    Commented Apr 12, 2014 at 23:36
  • $\begingroup$ By the way, arxiv.org/abs/1301.3191 may be helpful (see in particular 10.7 and 10.8), although I don't think it exactly answers your questions as posed without a bit more work. $\endgroup$
    – Mike Shulman
    Commented Jun 30, 2017 at 16:44

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