# pseudofunctors and pseudonatural transformations

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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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 :) –  David Roberts Jun 6 '11 at 5:22
@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 $J\otimes [1]$. –  Alan Wilder Jun 6 '11 at 19:57

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.
@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 –  Alan Wilder Jun 5 '11 at 19:49