Are (commutative) squares in some sense universal among edge-symmetric double categories?

Question

Definitions:
Given a category $\mathcal{C}$, and a double category $\mathbb{D}$, i.e. a category internal to $Cat$ with an “objects category” $\mathcal{D}_0$ and a “morphisms category” $\mathcal{D}_1$, call $\mathbb{D}$ a “double category over $\mathcal{C}$” if “$\mathcal{D}_0$ is $\mathcal{C}$”, i.e. if there are functors $\mathbf{F}: \mathcal{C} \to \mathcal{D}_0$, $\mathbf{F}^{-1}: \mathcal{D}_0 \to \mathcal{C}$, such that $\mathbf{F}^{-1} \circ \mathbf{F} = \mathbf{Id}_{\mathcal{C}}$ and $\mathbf{F} \circ \mathbf{F}^{-1} = \mathbf{Id}_{\mathcal{D}_0}$ (call this condition “strictly invertible”).

(The use of equality / "strict" invertibility is deliberate, a natural isomorphism is too weak here and incompatible with the motivation for the question.)

The morphisms of $\mathcal{D}_0$ are called “vertical morphisms”, the objects of $\mathcal{D}_1$ are called “horizontal morphisms”. We can swap the roles of these two “morphisms” and still get valid categories, $\tilde{\mathcal{D}}_0$ whose objects are the objects of $\mathcal{D}_0$ and whose morphisms are the objects of $\mathcal{D}_1$, and $\tilde{\mathcal{D}}_1$ whose objects are the morphisms of $\mathcal{D}_0$ and whose morphisms are the morphisms of $\mathcal{D}_1$. Moreover, the combination of $\tilde{\mathcal{D}}_0$ and $\tilde{\mathcal{D}}_1$ is again a double category, $\tilde{\mathbb{D}}$, called the “transpose” of $\mathbb{D}$.

A double category $\mathbb{D}$ is called “edge-symmetric” if, given its transpose $\tilde{\mathbb{D}}$, one has strictly invertible functors between the objects category $\mathcal{D}_0$ of $\mathbb{D}$ and the objects category $\tilde{\mathcal{D}}_0$ of the transpose $\tilde{\mathbb{D}}$. (Presumably this definition is equivalent to the condition that there are strictly invertible double functors between $\mathbb{D}$ and its transpose $\tilde{\mathbb{D}}$, although I haven’t checked.)

Given a category $\mathcal{C}$, and a double category $\mathbb{D}$, unsurprisingly say that $\mathbb{D}$ is “an edge-symmetric double category over $\mathcal{C}$” when $\mathbb{D}$ is both “edge-symmetric” and a “double category over $\mathcal{C}$”.

For an arbitrary category $\mathcal{C}$, the class $\operatorname{Sq}(\mathcal{C})$ of squares of morphisms in $\mathcal{C}$ form the 2-cells (morphisms of the morphism category) of an edge-symmetric double category over $\mathcal{C}$. Likewise, the class $\operatorname{ComSq}(\mathcal{C})$ of commutative squares of morphisms in $\mathcal{C}$ also induces an edge-symmetric double category over $\mathcal{C}$.

Question: For an arbitrary category $\mathcal{C}$, is there any sense in which either $\operatorname{ComSq}(\mathcal{C})$ or $\operatorname{Sq}(\mathcal{C})$ is universal among edge-symmetric double categories over $\mathcal{C}$?

What about among edge-symmetric double categories over $\mathcal{C}$ with additional extra properties, e.g. that are both left-connected and right-connected?

If I had to guess, it seems like any such universal property would be "right universal", i.e. make $\operatorname{Sq}(\mathcal{C})$ / $\operatorname{CommSq}(\mathcal{C})$ the terminal object in some construction such that given an arbitrary edge-symmetric double category $\mathbb{D}$ over $\mathcal{C}$, there should always be a unique double functor from $\mathbb{D}$ to $\operatorname{Sq}(\mathcal{C})$ / $\operatorname{CommSq}(\mathcal{C})$.

Following the description of "right universal" properties from section 3.8 of Bergman's Cook's Tour of Other Universal Constructions, the “minimal information guaranteed to be associated” to a 2-cell in a general double category is its source horizontal morphism and target horizontal morphism. But in an edge-symmetric double category, the minimal information guaranteed to be associated to a 2-cell is not only the source and target horizontal morphisms, but also the source and target vertical morphisms for the corresponding 2-cell in the transpose double category. This seems very similar to the minimal information guaranteed to be associated to a square of morphisms in $\mathcal{C}$, i.e. to a 2-cell in $\operatorname{Sq}(\mathcal{C})$ considered as a double category. Of course, commutative squares are better behaved than arbitrary squares, so it would be nice to restrict to $\operatorname{ComSq}(\mathcal{C})$, but I don’t see why that should always be possible.

Note: I asked this question previously on Math.SE, but it went unanswered for two weeks. I remain unsure whether this question should be considered "research-level" and appropriate for MathOverflow. This question is essentially a follow-up to / clarification of a previous question that is fairly basic. If nothing else, it seems like double categories remain a developing area of research, and that the only major monograph about them was written by a leading active researcher in the field to collect the most important current results.

• Marco Grandis, Higher dimensional categories: from double to multiple categories, World Scientific, 2019, doi:10.1142/11406

Related question: Is it possible to approach higher categories from the point of view of the arrow functor?

Answer 1

The answer to this question for (not necessarily commutative squares) may be given implicitly by the following remark in the last paragraph of section 2 of Ronald Brown and Ghafar H. Mosa, "Double categories, 2-categories, thin structures and connections":

In fact $\operatorname{Sq} : \mathbf{Cat} \to \mathbf{ESDblCat}$ is right adjoint to the forgetful functor $\operatorname{edge} : \mathbf{ESDblCat} \to \mathbf{Cat}$ assigning to every (small) edge symmetric double category $\mathbb{D}$ its edge category $\mathcal{D}_0$. The unit of this adjunction gives [the double functor] $\mathbb{D} \to \operatorname{Sq}(\mathcal{D}_0)$ which assigns to each element of $\operatorname{Mor}(\mathcal{D}_1)$ its shell, namely its square of boundary edges.

I have changed the notation from the original paper to match this question; any errors are mine alone. Note also that this paragraph is just stated as fact and provided without proof. I would like to figure out the details of the proof some day (assuming it is actually true).

Finally, note that the definition of "edge-symmetric double category" given in that paper is weaker than the definition I gave above, because it just requires that $\operatorname{Mor}(\mathcal{D}_0) = \operatorname{Ob}(\mathcal{D}_1)$ (and consequently also $\mathcal{D}_0 = \tilde{\mathcal{D}}_0$ and $\mathcal{D}_1 = \tilde{\mathcal{D}}_1$). The definition I gave in the question I gave above may be overly strong (or at least stronger than needed), and the two versions I gave above may not be equivalent.

Let $\mathbf{ESDblCat}(\mathcal{C})$ denote the category ("of double categories over $\mathcal{C}$")

• whose objects are edge-symmetric (in the sense of Brown, Mosa) double categories $\mathbb{D}$ with objects category equal to $\mathcal{C}$,
  • (The extra generality where the objects category is isomorphic to $\mathcal{C}$ is easy to get just by composing the functors defining the double category structure of $\mathbb{D}$ with the suitable invertible functors.)
• whose morphisms are double functors whose "objects category parts" / "edge functors" are always equal to the identity functor $\operatorname{Id}_{\mathcal{C}}$.

Then I claim that the following is a corollary of the adjunction claimed (without proof) by Brown and Mosa:

The double category $\operatorname{Sq}(\mathcal{C})$ is terminal in $\mathbf{ESDblCat}(\mathcal{C})$.

I will try to type up later more of my notes that convinced me this is likely true without being a full proof. Some initial observations though:

• Seemingly $\operatorname{edge} \circ \operatorname{Sq} = \operatorname{Id}_{\mathbf{Cat}}$.
• Based on the above, the counit of the adjunction would be a natural transformation $\operatorname{Id}_{\mathbf{Cat}} \Rightarrow \operatorname{Id}_{\mathbf{Cat}}$. I am assuming that this natural transformation is the identity natural transformation, but I haven't proved that it is. (Mostly because I can't think of another natural transformation $\operatorname{Id}_{\mathbf{Cat}} \Rightarrow \operatorname{Id}_{\mathbf{Cat}}$.)
• It seems to be true that the unique functor $\mathcal{C} \to \mathcal{X}$ that the adjunction's "right universal" property guarantees is $\operatorname{edge}(G)$ where $G: \mathbb{D} \to \operatorname{Sq}(X)$ is a double functor.
• So seemingly $\operatorname{edge}(G)$ is unique with property that $\operatorname{Sq}(\operatorname{edge}(G)) \circ \eta_{\mathbb{D}} = G$, where the double functor $\eta_{\mathbb{D}}: \mathbb{D} \to \operatorname{Sq}(\operatorname{edge}(\mathbb{D}))$ is one of the components of the unit natural transformation.
• If we restrict to double functors where $\operatorname{edge}(G) = \operatorname{Id}_{\mathcal{C}}$ (which also requires restricting to categories $\mathcal{X} = \mathcal{C}$), then always $\operatorname{Sq}(\operatorname{edge}(G)) = \operatorname{Sq}(\operatorname{Id}_{\mathcal{C}}) = \operatorname{Id}_{\operatorname{Sq}(\mathcal{C})}$, hence we get that all such double functors $G: \mathbb{D} \to \operatorname{Sq}(\mathcal{C})$ are equal to $\eta_{\mathbb{D}}$, i.e. $\eta_{\mathbb{D}}$ is the unique such double functor and thus a terminal morphism in $\mathbf{ESDblCat}(\mathcal{C})$.