What happens if we drop the globularity of coherences in the definition of (pseudo) double categories?

Question

Double categories are categories internal to categories. A pseudo double category is a double category where composition in one direction is associative and unital only upto a coherence isomorphism. I have seen that the associator and the two unitors are defined to be globular. Further, pseudo functors of double categories are defined to have globular coherences. (See page 2 of Shulman's paper)

In most real life examples of double categories, the associators and unitors are indeed globular. This makes the definition practical. Since it is an additional condition to check, why do we need that assumption?

Is there a theoretical/nice/categorical reason why we should assume globularity for coherences?

Answer 1

There are a lot of possible answers one could give to this. One class of answers is that pseudo double categories, as usually defined, are the "weakening" of strict double categories in some standard 2-categorical sense. For instance, there is a 2-monad on the 2-category $\rm Cat^{\rightrightarrows}$ whose strict algebras are strict double categories and whose pseudoalgebras are (unbiased) pseudo double categories. Similarly, just as a category is equivalent to a simplicial object in Set satisfying the Segal conditions, a pseudo double category is equivalent to a simplicial object in Cat satisfying the Segal conditions (up to equivalence).

Another class of answers would point out practical ways in which non-globular constraints just wouldn't work. For instance, a non-globular isomorphism can't be whiskered: given horizontal arrows $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} D$, a non-globular 2-cell associativity isomorphism $f(gh) \cong (fg)h$ would have at the right a nonidentity vertical isomorphism $\delta:D\cong D$, and if we wanted to whisker this with $D \xrightarrow{k} E$ to get an isomorphism $(f(gh))k \cong ((fg)h)k$ we'd need a 2-cell isomorphism $k\cong k$ that is $\delta$ on the left, and in general there's no reason there should be such a thing. In particular, this means we wouldn't be able to state the pentagon axiom for the associator.

There are probably other classes of answers as well.