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Based on the discussion here I feel like there should be a bijection between pseudonatural transformations of pseudofunctors $J\to\mathcal{C}$ and pseudofunctors $J\times [1]\to\mathcal{C}$, at least morally ($[1]$ denotes the poset 0<1).

The map from 'homotopies' $J\times [1]\to\mathcal{C}$ to pseudonatural transformations works out nicely, but the other direction seems problematic. In particular, given $\alpha:F\Rightarrow G$, if we try to form a pseudofunctor $\tilde{\alpha}:J\times [1]\to\mathcal{C}$ it's clear that we should set $$ \tilde{\alpha}(\text{id}_j,0\to 1) = \alpha (j),\; \text{and}\quad \tilde{\alpha}(j'\to j,\text{id}_0)=F(j'\to j),\tilde{\alpha}(j'\to j,\text{id}_1) = G(g) $$ but what about $\tilde{\alpha}(j'\to j,0\to 1)$? The problem is pseudonatural transformations only tell how to fill squares, not the triangles individually (i.p. they don't give any diagonal 1-morphism in the middle). One could just choose either $\alpha(j)\circ F(g)$ or $G(g) \circ \alpha(j)$ but this won't result in a bijection.

Is the best one can hope for an equivalence modulo modification?

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  • $\begingroup$ Perhaps instead of cartesian product $J\times [1]$ you could look at the Gray tensor $J\otimes [1]$, ncatlab.org/nlab/show/Gray+tensor+product :) $\endgroup$
    – David Roberts ♦
    Commented Jun 6, 2011 at 5:22
  • $\begingroup$ @David: The Gray tensor product has more data than the cartesian product, which makes the problem worse, i.e. there are "more" p.functors $J\otimes [1]\to\mathcal{C}$ than p.functors $J\times [1]\to\mathcal{C}$, because $J\times [1]$ embeds in $J\otimes [1]$. $\endgroup$
    – Alan Wilder
    Commented Jun 6, 2011 at 19:57

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Choose in any case $\alpha(j) \circ F(g)$, then you get an equivalence of categories between the category of pseudofunctors $J \times [1] \to C$ and the category of pseudonatural transformations of pseudofunctors $J \to C$. Note that under this equivalence of categories, the modifcations of pseudonatural transformations just correspond to pseudonatural transformations of the corresponding homotopies, which could also serve as their motivation and definition.

I don't think that there is an isomorphism of categories here. You have also indicated a reason, namely otherwise we would lose the data of an isomorphism $\alpha(j) \circ F(g) \cong G(g) \circ \alpha(j)$.

You could also choose $G(g) \circ \alpha(j)$ in any case and get another equivalence.

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    $\begingroup$ The problem you identify is one of direction of the 2-arrow. I used to work with homotopy coherence and in its simplicial manifestation that gets around the problem by asking for both directions! There are still problems with the directions but they are resolved by the initial structures of the homotopy coherent nerve construction. (I can go into more detail if you want, but this is a comment not an answer.) $\endgroup$
    – Tim Porter
    Commented Jun 5, 2011 at 16:26
  • $\begingroup$ Thanks, Tim, can you go into more detail please? $\endgroup$
    – Alan Wilder
    Commented Jun 5, 2011 at 18:11
  • $\begingroup$ @Martin: I'm not sure what you mean by the middle paragraph, but assuming you meant a(j)o F(g)=>G(g)o a(j), you actually do have this data on both sides. Clearly it is part of the natural transformation data, and for a p.functor $h:J\times [1]\to C$, this isomorphism is produced by applying the compositors of $h$ to the equal factorizations [ (j'\to j,\text{id}_1)\circ (\text{id}_j, 0\to 1) = (\text{id}_{j'},0\to 1)\circ (j'\to j, \text{id}_0) ] in the middle we have $h(j'\to j,0\to 1)$, which is the datum that has no counterpart on the p.natural transformation side $\endgroup$
    – Alan Wilder
    Commented Jun 5, 2011 at 19:49
  • $\begingroup$ @Alan A homotopy coherent (hc) version of these ideas is obtained using the idea that the 'diagrams' used have to be hc, but a hc square does have the diagonal in it. (See page 366 of the version of the crossed menagerie (available from the n-Lab page of that name).) This use of the diagonal allows both conventions, but still hits a problem as it has to make a choice of `diagonal towards the sides' as against 'sides towards diagonal'. There are papers back in the 1980s(?) by Cordier and myself that take this apart. (It is also discussed in Kamps and Porter's book. (blatant advertisement. :-)) $\endgroup$
    – Tim Porter
    Commented Jun 6, 2011 at 7:00

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