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In his paper Constructing symmetric monoidal bicategories, Shulman shows that if a symmetric monoidal double category has all companions and conjoints, then its horizontal bicategory is also symmetric monoidal.
It seems clear to me that it suffices for the double category to only have all companions and conjoints for isomorphisms (that is, it suffices for it to be isofibrant, rather than fibrant). This is simply because we only ever use companions and conjoints of the structural isomorphisms of the symmetric monoidal structure in the proof.
Now, if a vertical isomorphism and its inverse have companions, then the companion to will be a conjoint for . Therefore, it seems that in the key Theorem 4.6 of ibid., it suffices to assume that all vertical isomorphisms have companions, or in other words that the double category is left isofibrant (it's source map is an isofibration).
Is this reasoning correct, or have I missed something?
It might be worth pinging @Mike Shulman.
The non-symmetric version also follows from the work of Garner and Gurski, who merely ask for to be an isofibration.
(And for symmetry one does not need stronger assumptions.)
The reasoning about companions of tight isomorphisms is essentially Theorem 1.5 of Grandis–Paré's Adjoint for double categories
(I don't see that asking for the source functor to be an isofibration is the same as asking for every tight isomorphism to have a companion, though.)
Yes, see Remark 3.22 in my paper.
Mike Shulman said:
Yes, see Remark 3.22 in my paper.
Ah great, sorry I missed that!
Thanks all!