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1. • CommentRowNumber1.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 9th 2016
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   Author: Karol Szumiło
   Format: MarkdownItexLet $J$ be a category and $C, D \colon J \to \mathsf{Cat}$ two strict functors. Let $f \colon C \to D$ be a strictly natural transformation such that $f_j$ is an equivalence for every $j \in J$. Can we always pick a pseudo-natural $g \colon D \to C$ such that $g_j$ is an inverse of $f_j$ for every $j \in J$? I believe I can prove this when $J$ is inverse and perhaps the argument generalizes to Reedy categories too, but it is somehow more involved than I expected. Are there any references for results of this sort?

   Let $J$ be a category and $C, D \colon J \to \mathsf{Cat}$ two strict functors. Let $f \colon C \to D$ be a strictly natural transformation such that $f_j$ is an equivalence for every $j \in J$. Can we always pick a pseudo-natural $g \colon D \to C$ such that $g_j$ is an inverse of $f_j$ for every $j \in J$?

   I believe I can prove this when $J$ is inverse and perhaps the argument generalizes to Reedy categories too, but it is somehow more involved than I expected. Are there any references for results of this sort?

2. • CommentRowNumber2.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 9th 2016
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   Author: Karol Szumiło
   Format: MarkdownItexIt seems that the general statement follows from existence of a projectively cofibrant replacement $\hat D$ of $D$ such that strict transformations out of $\hat D$ correspond to pseudo-natural transformations out of $D$. Still, it would be nice to have a reference.

   It seems that the general statement follows from existence of a projectively cofibrant replacement $\hat D$ of $D$ such that strict transformations out of $\hat D$ correspond to pseudo-natural transformations out of $D$. Still, it would be nice to have a reference.

3. • CommentRowNumber3.
   • CommentAuthorRichard Williamson
   • CommentTimeNov 9th 2016
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   Author: Richard Williamson
   Format: MarkdownItexI don't know about a reference. But the argument seems very easy to me. Let $a : j_{0} \rightarrow j_{1}$ be an arrow of $J$. Let $f^{-1}(j_{0})$ be an inverse of $f(j_{0})$ (up to natural isomorphism, of course), and define $f^{-1}(j_{1})$ similarly. We then have: $f(j_{1}) \circ C(a) \circ f^{-1}(j_{0}) = D(a) \circ f(j_{0}) \circ f^{-1}(j_{0}) \simeq D(a)$, where $\simeq$ denotes natural isomorphism. In short: $f(j_{1}) \circ C(a) \circ f^{-1}(j_{0}) \simeq D(a)$. Hence $f^{-1}(j_{1}) \circ f(j_{1}) \circ C(a) \circ f^{-1}(j_{0}) \simeq f^{-1}(j_{1}) \circ D(a)$, and thus $C(a) \circ f^{-1}(j_{0}) \simeq f^{-1}(j_{1}) \circ D(a)$, as required. I haven't attempted to check the coherence conditions, but I would certainly expect them to hold.

   I don’t know about a reference. But the argument seems very easy to me. Let $a : j_{0} \rightarrow j_{1}$ be an arrow of $J$. Let $f^{-1}(j_{0})$ be an inverse of $f(j_{0})$ (up to natural isomorphism, of course), and define $f^{-1}(j_{1})$ similarly. We then have: $f(j_{1}) \circ C(a) \circ f^{-1}(j_{0}) = D(a) \circ f(j_{0}) \circ f^{-1}(j_{0}) \simeq D(a)$, where $\simeq$ denotes natural isomorphism. In short: $f(j_{1}) \circ C(a) \circ f^{-1}(j_{0}) \simeq D(a)$. Hence $f^{-1}(j_{1}) \circ f(j_{1}) \circ C(a) \circ f^{-1}(j_{0}) \simeq f^{-1}(j_{1}) \circ D(a)$, and thus $C(a) \circ f^{-1}(j_{0}) \simeq f^{-1}(j_{1}) \circ D(a)$, as required.

   I haven’t attempted to check the coherence conditions, but I would certainly expect them to hold.

4. • CommentRowNumber4.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 10th 2016
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   Author: Karol Szumiło
   Format: MarkdownItexI doubt that this is coherent, but maybe it has a chance if we pick inverses to be adjoint inverses.

   I doubt that this is coherent, but maybe it has a chance if we pick inverses to be adjoint inverses.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeNov 10th 2016
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   Author: Mike Shulman
   Format: MarkdownItexThere is a more general fact: if $T$ is any 2-monad on a 2-category $K$, and $f:A\to B$ is a pseudo $T$-morphism whose underlying morphism in $K$ is an equivalence, then $f$ is an equivalence in $T$-$Alg_ps$. Choosing an adjoint pseudo-inverse $g$ to $f$, [[doctrinal adjunction]] tells us that since $f$ is a pseudo morphism, $g$ is a lax morphism and the adjunction lives in $T$-$Alg_lax$. Inspecting the construction of the lax structure on $g$, it's built out of the pseudo structure of $f$ and the unit and counit of the adjunction, so if the latter two are also invertible, $g$ is also pseudo. The inclusion $T$-$Alg_ps \to T$-$Alg_lax$ is locally fully faithful, so the adjunction also lives in $T$-$Alg_ps$.

   There is a more general fact: if $T$ is any 2-monad on a 2-category $K$, and $f:A\to B$ is a pseudo $T$-morphism whose underlying morphism in $K$ is an equivalence, then $f$ is an equivalence in $T$-$Alg_ps$. Choosing an adjoint pseudo-inverse $g$ to $f$, doctrinal adjunction tells us that since $f$ is a pseudo morphism, $g$ is a lax morphism and the adjunction lives in $T$-$Alg_lax$. Inspecting the construction of the lax structure on $g$, it’s built out of the pseudo structure of $f$ and the unit and counit of the adjunction, so if the latter two are also invertible, $g$ is also pseudo. The inclusion $T$-$Alg_ps \to T$-$Alg_lax$ is locally fully faithful, so the adjunction also lives in $T$-$Alg_ps$.

6. • CommentRowNumber6.
   • CommentAuthorRichard Williamson
   • CommentTimeNov 10th 2016
   • (edited Nov 10th 2016)
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   Author: Richard Williamson
   Format: MarkdownItexKarol: I've now checked, and the construction I gave is definitely coherent. One makes use of the naturality of the isomorphism between $f \circ f^{-1}$ (and the other way around) and the identity functor. I don't know why you felt the coherence to be doubtful? To my intuition, it is 'obviously' coherent. One is making using only of given data of a categorical nature, which will always respect composition, etc, in the expected ways. Or, from a different point of view: it clearly works in the case that $f \circ f^{-1}$ (and the same the other way around) is _equal_ to the identity functor; one is only weakening up to a _natural_ isomorphism; and naturality should always allow one to obtain the coherence one needs (otherwise the coherence condition would be incorrect). I assume that the construction I gave is well-known. I must have re-constructed it to myself on three or four occasions over the years.

   Karol: I’ve now checked, and the construction I gave is definitely coherent. One makes use of the naturality of the isomorphism between $f \circ f^{-1}$ (and the other way around) and the identity functor.

   I don’t know why you felt the coherence to be doubtful? To my intuition, it is ’obviously’ coherent. One is making using only of given data of a categorical nature, which will always respect composition, etc, in the expected ways. Or, from a different point of view: it clearly works in the case that $f \circ f^{-1}$ (and the same the other way around) is equal to the identity functor; one is only weakening up to a natural isomorphism; and naturality should always allow one to obtain the coherence one needs (otherwise the coherence condition would be incorrect).

   I assume that the construction I gave is well-known. I must have re-constructed it to myself on three or four occasions over the years.

7. • CommentRowNumber7.
   • CommentAuthorTim_Porter
   • CommentTimeNov 10th 2016
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   Author: Tim_Porter
   Format: MarkdownItexI had not been following this discussion in detail but on glancing at it again, it looks more or less the argument that Jean-Marc Cordier and I used (which was a version of one of Boardman and Vogt) in our proof of Vogt's theorem. I may have put this as an example in my book with Heiner Kamps, although as I imply above the case you want is almost certainly `well known' from earlier.

   I had not been following this discussion in detail but on glancing at it again, it looks more or less the argument that Jean-Marc Cordier and I used (which was a version of one of Boardman and Vogt) in our proof of Vogt’s theorem. I may have put this as an example in my book with Heiner Kamps, although as I imply above the case you want is almost certainly ‘well known’ from earlier.

8. • CommentRowNumber8.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 11th 2016
   • (edited Nov 11th 2016)
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   Author: Karol Szumiło
   Format: MarkdownItexThank you all for the answers. Mike, this is really interesting but I guess it is a little too abstract compared to the situation I have in mind. Richard, something strange is going on here. I have also tried checking this and I'm pretty sure that your construction is not in general coherent. If I look at two arrows $j_0 \to j_i \to j_2$ in $J$ and attempt to check coherence for this composite, then I find myself having to cancel out isomorphisms $f_{j_1} g_{j_1} \cong \mathrm{id}$ and $id \cong g_{j_1} f_{j_1}$ in a particular way. This appears to be possible only if they satisfy a triangular identity. Of course, there is no trouble in arranging that which does provide an answer to my question. But are you really claiming that this is coherent no matter how you choose the isomorphisms? Tim, that's a very useful answer. Indeed, Theorem 1.1 of your _Vogt's theorem on categories of homotopy coherent diagrams_ seems to be a good reference for exactly this question.

   Thank you all for the answers.

   Mike, this is really interesting but I guess it is a little too abstract compared to the situation I have in mind.

   Richard, something strange is going on here. I have also tried checking this and I’m pretty sure that your construction is not in general coherent. If I look at two arrows $j_0 \to j_i \to j_2$ in $J$ and attempt to check coherence for this composite, then I find myself having to cancel out isomorphisms $f_{j_1} g_{j_1} \cong \mathrm{id}$ and $id \cong g_{j_1} f_{j_1}$ in a particular way. This appears to be possible only if they satisfy a triangular identity. Of course, there is no trouble in arranging that which does provide an answer to my question. But are you really claiming that this is coherent no matter how you choose the isomorphisms?

   Tim, that’s a very useful answer. Indeed, Theorem 1.1 of your Vogt’s theorem on categories of homotopy coherent diagrams seems to be a good reference for exactly this question.

9. • CommentRowNumber9.
   • CommentAuthorRichard Williamson
   • CommentTimeNov 11th 2016
   • (edited Nov 11th 2016)
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   Author: Richard Williamson
   Format: MarkdownItexI was about to write down my argument for coherence, when I realised that one does not need to worry about coherence at all: I believe that one gets a _strict_ natural transformation $g : D \rightarrow C$. Indeed, the construction can be viewed as follows. $f^{-1}(j_{1}) \circ D(a) = f^{-1}(j_{1}) \circ D(a) \circ f(j_{0}) \circ f^{-1}(j_{0})$ $= f^{-1}(j_{1}) \circ f(j_{1}) \circ C(a) \circ f^{-1}(j_{0})$ $= C(a) f^{-1}(j_{0})$ The first equality arises as follows. The naturality of the isomorphism $f(j_{0}) f^{-1}(j_{0}) \rightarrow id$ says exactly, in the case of the arrow $a$, that $D(a) = D(a) \circ f(j_{0}) \circ f^{-1}(j_{0})$, where this is actual equality. Similarly, the naturality of the isomorphism $id \rightarrow f(j_{1}) f^{-1}(j_{1})$ says exactly, in the case of the arrow $a$, that $C(a) = f(j_{1}) \circ f^{-1}(j_{1}) \circ C(a)$, where this is actual equality.

   I was about to write down my argument for coherence, when I realised that one does not need to worry about coherence at all: I believe that one gets a strict natural transformation $g : D \rightarrow C$. Indeed, the construction can be viewed as follows.

   $f^{-1}(j_{1}) \circ D(a) = f^{-1}(j_{1}) \circ D(a) \circ f(j_{0}) \circ f^{-1}(j_{0})$ $= f^{-1}(j_{1}) \circ f(j_{1}) \circ C(a) \circ f^{-1}(j_{0})$ $= C(a) f^{-1}(j_{0})$

   The first equality arises as follows. The naturality of the isomorphism $f(j_{0}) f^{-1}(j_{0}) \rightarrow id$ says exactly, in the case of the arrow $a$, that $D(a) = D(a) \circ f(j_{0}) \circ f^{-1}(j_{0})$, where this is actual equality. Similarly, the naturality of the isomorphism $id \rightarrow f(j_{1}) f^{-1}(j_{1})$ says exactly, in the case of the arrow $a$, that $C(a) = f(j_{1}) \circ f^{-1}(j_{1}) \circ C(a)$, where this is actual equality.

10. • CommentRowNumber10.
    • CommentAuthorKarol Szumiło
    • CommentTimeNov 11th 2016
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    Author: Karol Szumiło
    Format: MarkdownItexIf $J$ is a group of order two, $C$ is the walking isomorphism with $J$ acting freely, $D = [0]$ (with necessarily trivial action) and $f \colon C \to D$ the unique (necessarily equivariant) equivalence, then $f$ has two inverses neither of which is equivariant.

    If $J$ is a group of order two, $C$ is the walking isomorphism with $J$ acting freely, $D = [0]$ (with necessarily trivial action) and $f \colon C \to D$ the unique (necessarily equivariant) equivalence, then $f$ has two inverses neither of which is equivariant.

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeNov 11th 2016
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    Author: Mike Shulman
    Format: MarkdownItexI agree with Karol: in general you cannot get a strict inverse, and you need adjoint inverses to make the inverse coherently natural. I think 2-monads are a very natural way of thinking about it, and one that's worth getting used to, even if they are a little more abstract. If you're going to do 2-category theory, which is what you're doing when you talk about pseudonatural transformations, then you should expect to benefit from learning some 2-categorical machinery.

    I agree with Karol: in general you cannot get a strict inverse, and you need adjoint inverses to make the inverse coherently natural.

    I think 2-monads are a very natural way of thinking about it, and one that’s worth getting used to, even if they are a little more abstract. If you’re going to do 2-category theory, which is what you’re doing when you talk about pseudonatural transformations, then you should expect to benefit from learning some 2-categorical machinery.

12. • CommentRowNumber12.
    • CommentAuthorRichard Williamson
    • CommentTimeNov 12th 2016
    • (edited Nov 12th 2016)
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    Author: Richard Williamson
    Format: MarkdownItexThanks very much for the nice example, Karol! You are completely correct, I was a bit quick in #9 and my earlier calculations concerning coherence; in the latter, I was trying to do something similar to the idea in #9. I agree now that one needs to choose the equivalences which can occur in the middle of a composition to be adjoint. The complete situation is as follows. 1) If one has a strictly commutative triangle $D(a) = D(a) \circ f(j_{0}) \circ f^{-1}(j_{0})$ and a strictly commutative triangle $C(a) = f(j_{1}) \circ f^{-1}(j_{1}) \circ C(a)$, then the proof that I have in #9 goes through to give a strict natural transformation $D \rightarrow C$. So in the case of Karol's original question in which the equivalences are in fact isomorphisms, we can apply this proof to construct a strict natural transformation of the kind he asked for. 2) If one has only a 2-cell with the boundary of the first of these triangles, and the same for the second triangle, and if $((id \Rightarrow f^{-1}(j_{1})f(j_{1})) \cdot f(j_{1})) \circ (f(j_{1}) \cdot (f(j_{1})f^{-1}(j_{1}) \Rightarrow id)) = f$, then the construction of #3 goes through, and one can prove that the coherence axioms are satisfied, so that one obtains a lax natural transformation $D \rightarrow C$. So in a generalisation of Karol's original question in which we have only an adjunction (not necessarily an equivalence) for every $f_{j}$, we can construct a lax natural transformation of the required kind. 3) If the 2-cells of 2) are in fact 2-isomorphisms, and if the same condition that $((id \Rightarrow f^{-1}(j_{1})f(j_{1})) \cdot f(j_{1})) \circ (f(j_{1}) \cdot (f(j_{1})f^{-1}(j_{1}) \Rightarrow id)) = f$ holds, then the argument of #3 again goes through, and one can prove that the coherence axioms are satisfied, so that one obtains a pseudo-natural transformation $D \rightarrow C$. So, for Karol's original question, choosing (as we always can) the equivalences which can occur in the middle of a composition to be adjoint equivalences, we can construct a pseudo-natural transformation of the kind he asked for.

    Thanks very much for the nice example, Karol!

    You are completely correct, I was a bit quick in #9 and my earlier calculations concerning coherence; in the latter, I was trying to do something similar to the idea in #9. I agree now that one needs to choose the equivalences which can occur in the middle of a composition to be adjoint.

    The complete situation is as follows.

    1) If one has a strictly commutative triangle $D(a) = D(a) \circ f(j_{0}) \circ f^{-1}(j_{0})$ and a strictly commutative triangle $C(a) = f(j_{1}) \circ f^{-1}(j_{1}) \circ C(a)$, then the proof that I have in #9 goes through to give a strict natural transformation $D \rightarrow C$. So in the case of Karol’s original question in which the equivalences are in fact isomorphisms, we can apply this proof to construct a strict natural transformation of the kind he asked for.

    2) If one has only a 2-cell with the boundary of the first of these triangles, and the same for the second triangle, and if $((id \Rightarrow f^{-1}(j_{1})f(j_{1})) \cdot f(j_{1})) \circ (f(j_{1}) \cdot (f(j_{1})f^{-1}(j_{1}) \Rightarrow id)) = f$, then the construction of #3 goes through, and one can prove that the coherence axioms are satisfied, so that one obtains a lax natural transformation $D \rightarrow C$. So in a generalisation of Karol’s original question in which we have only an adjunction (not necessarily an equivalence) for every $f_{j}$, we can construct a lax natural transformation of the required kind.

    3) If the 2-cells of 2) are in fact 2-isomorphisms, and if the same condition that $((id \Rightarrow f^{-1}(j_{1})f(j_{1})) \cdot f(j_{1})) \circ (f(j_{1}) \cdot (f(j_{1})f^{-1}(j_{1}) \Rightarrow id)) = f$ holds, then the argument of #3 again goes through, and one can prove that the coherence axioms are satisfied, so that one obtains a pseudo-natural transformation $D \rightarrow C$. So, for Karol’s original question, choosing (as we always can) the equivalences which can occur in the middle of a composition to be adjoint equivalences, we can construct a pseudo-natural transformation of the kind he asked for.

13. • CommentRowNumber13.
    • CommentAuthorKarol Szumiło
    • CommentTimeNov 14th 2016
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    Author: Karol Szumiło
    Format: MarkdownItexMike, I certainly appreciate learning about abstract perspectives on things. However, it is not necessarily appropriate to use such abstract concepts as a small part of an argument that doesn't involve a lot of 2-category theory otherwise. In fact, it turns out that what I needed is even much simpler than what I asked for here. Richard, I believe you are right. I'm glad that we came to an agreement.

    Mike, I certainly appreciate learning about abstract perspectives on things. However, it is not necessarily appropriate to use such abstract concepts as a small part of an argument that doesn’t involve a lot of 2-category theory otherwise. In fact, it turns out that what I needed is even much simpler than what I asked for here.

    Richard, I believe you are right. I’m glad that we came to an agreement.