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Stream: theory: category theory

Topic: softened preorder

John Baez (Jan 21 2026 at 22:09):

Thanks to all your help with some questions of mine, I may be ready to prove a theorem. I'd be happy for feedback.

First a definition:

Definition. A softened preorder is a category such that given $f, g \colon x \to y$, there exists an automorphism $\alpha \colon y \to y$ such that $g = \alpha f$.

The idea is that if we can always take $\alpha = 1_y$ then our category is a preorder, but in general we are weakening or 'softening' the condition that there's at most one morphism from any object $x$ to any object $y$. Note that the roles of $x$ and $y$ are really symmetrical here, since $f = \alpha^{-1} g$.

Exercise. The opposite of a softened preorder may not be a softened preorder, but it is always an opsoftened preorder: given $f, g \colon x \to y$, there exists an automorphism $\alpha \colon x \to x$ such that $g = f \alpha$.

I'm happy to hear suggestions for better names for these kind of categories - and I know category theorists like nothing more than arguing about terminology, so consider this opportunity to argue my big gift to all of you. But I want to move on here....

Theorem. Suppose $P$ is a preorder, $F: P \to \mathbf{Gpd}$ is a pseudofunctor, and $C = \int F$ is the category built via the covariant Grothendieck construction from $F$. Then $C$ is a softened preorder. Conversely if $C$ is a softened preorder then $C \simeq \int F$ for some preorder $P$ and pseudofunctor $F: P \to \mathbf{Gpd}$.

Let me prove the first part now and try to prove the converse later.

John Baez (Jan 21 2026 at 22:31):

To reduce the hassle I will assume $F$ is a functor, not a pseudofunctor; I believe the pseudofunctor case works similarly.

Proof of forwards direction when $F$ is a functor. Suppose $P$ is a preorder, $F: P \to \mathbf{Gpd}$ is a functor, and $C = \int F$ is the category built via the covariant Grothendieck construction from $F$.

An object $x \in \int F$ is a pair consisting of an object $x_0 \in C$ together with an object $x_1 \in F(x_0)$. Similarly $y \in \int F$ is a pair $(y_0, y_1)$ where $y_1 \in F(y_0)$.

A morphism $f: x \to y$ in $\int F$ is then a morphism $f_0: x_0 \to y_0$ in $P$ (of which there's at most one) together with a morphism $f_1 : F(f_0)(x_1) \to y_1$.

Suppose we have another morphism $g: x \to y$. This is a morphism $g_0: x_0 \to y_0$ together with a morphism $g_1 : F(g_0)(x_1) \to y_1$.

But because $P$ is a preorder, $f_0 = g_0$, so $g_1: F(f_0)(x_1) \to y_0$.

Note that $f_1 : F(f_0)(x_1) \to y_1$ and $g_1: F(f_0)(x_1) \to y_1$ are morphisms with the same source and same target in the groupoid $F(y_0)$. We thus have

$g_1 = (g_1 f_1^{-1}) f_1$

I now claim that

$g = \alpha f$

for some morphism $\alpha : y \to y$ in $\int F$. This morphism $\alpha$ should itself be a pair of a morphism $\alpha_0 \colon y_0 \to y_0$ and a morphism $\alpha_1 \colon F(\alpha_0)(y_1) \to y_1$. Let's take $\alpha_0 = 1$, so we need $\alpha_1 \colon y_1 \to y_1$. And let's take

$\alpha_1 = g_1 f_1^{-1} \colon y_1 \to y_1$

So, we're taking $\alpha = (1, g_1 f_1^{-1})$ and I claim

$g = \alpha f$

John Baez (Jan 21 2026 at 22:46):

Everything here is a pair, and we're saying

$(g_0, g_1) = (1, g_1 f_1^{-1}) \circ (f_0, f_1)$

Composition at right is taking place in the Grothendieck construction $\int F$. Thus, the first component of each pair composes in the usual way in $P$, and for this we're claiming

$g_0 = f_0$

which we've seen is true because $P$ is a preorder. The second component composes in a more fancy way, and for this we're claiming

$g_1 = (g_1 f_1^{-1}) \circ F(\alpha_1)(f_1)$

which is true because $\alpha_1 = 1$ and $F$ is a functor so $F(\alpha_1)(f_1) = f_1$. $\qquad$ ▮

David Wärn (Jan 22 2026 at 09:27):

I don't think every softened preorder arises in this way? For any pair of softened preorders $C,D$, the cograph of the terminal profunctor from $C$ to $D$ seems to be a softened preorder. But if $C,D$ are $B\mathbb Z$ (or some other connected non-trivial groupoid) then the terminal profunctor isn't representable so the cograph shouldn't come from the Grothendieck construction. Maybe if you assume the automorphism $\alpha$ is unique in the definition of softened preorder.

Nathanael Arkor (Jan 22 2026 at 10:06):

If the automorphism is assumed unique, then $x$ is called prequasi-initial in Tholen's MacNeille completion of concrete categories with local properties. If $x$ is also weakly initial, then it is called quasi-initial. There are probably other concepts relevant to what you are studying in this paper.

John Baez (Jan 23 2026 at 00:58):

David Wärn said:

I don't think every softened preorder arises in this way?

Indeed, I think you're right. Is the following a counterexample? The category with only two objects $x,y$, a single morphism $f: x \to y$, the identity morphism $1_x : x \to x$, and two automorphisms $1_y, \alpha: y \to y$. (This forces $\alpha f = f$ and $\alpha^2 = 1_y$.)

So, I no longer have hopes for the converse here:

Theorem. Suppose $P$ is a preorder, $F: P \to \mathbf{Gpd}$ is a pseudofunctor, and $C = \int F$ is the category built via the covariant Grothendieck construction from $F$. Then $C$ is a softened preorder. Conversely if $C$ is a softened preorder then $C \simeq \int F$ for some preorder $P$ and pseudofunctor $F: P \to \mathbf{Gpd}$.

John Baez (Jan 23 2026 at 01:15):

By the way, I was trying to prove that the category of central finite-dimensional simple algebras over a field $k$ is a softened preorder. (Here $A$ is central if the center of $A$ is $k$.) This example is the only reason why I care about this topic.

I discovered to my delight that this result is true, and it follows right away from the Skolem-Noether theorem.

(I hadn't expected it to be a new result, so I'm not disappointed: I'm just happy that it was discovered by two bigshots.)

Alex Kreitzberg (Jan 23 2026 at 07:52):

It was useful seeing your thought process and how you asked for help in connection with the grothendieck construction, thank you for threading me in. You ended up with a nice example of how a condition on the domain of the pseudofunctor translates to a condition on the opfibration.

John Baez (Jan 23 2026 at 22:02):

Thanks! The good thing about doing math in public is that people can help out, and also people can watch and maybe learn something or at least enjoy watching someone flounder around and make mistakes.

I'm not quite done yet because I haven't gotten a "structure theorem" that describes all softened preorders. From my counterexamples so far, I'm guessing every softened preorder is a "quotient" of one obtained by the Grothendieck construction. There may be an adjunction between the 2-category of softened preorders and the 2-category of preorders equipped with pseudofunctors to $\mathsf{Gpd}$.

However, instead of investigating this general idea further right now, I think I'll turn toward the examples I'm actually interested in, and come back to this if it turns out to help.