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Stream: learning: questions

Topic: Categories generalizing groupoids and preorders

John Baez (Jan 16 2026 at 00:48):

Has anyone written about categories with this property?

Property A. Given morphisms $f: x \to y$ and $g: y \to x$, $f$ must be the inverse of $g$.

Note that any preorder has this property, and so does any groupoid.

More generally, I believe we can get a lot of categories with property A as follows. Take any preorder $P$ and any pseudofunctor $F: P \to \mathbf{Gpd}$, and do the covariant Grothendieck construction $\int F$. Then I believe $\int F$ has property A. The category $\int F$ is like a "bundle of groupoids over a preorder".

But I believe it's not true that any category with property A is given by the above construction! A counterexample would a category with only two objects $x$ and $y$, two morphisms $f, g \colon x \to y$, and identity morphisms for $x$ and $y$.

This category is not a preorder, but it has a weaker property shared by all preorders:

Property B. If there is a morphism $f : x \to y$ that is not an isomorphism, there is no morphism $g: x \to y$.

Has anyone studied categories with property B?

Kevin Carlson (Jan 16 2026 at 01:02):

Hm? Groupoids don't have Property A, do they? Just deloop any nontrivial group, and if you want $x\ne y$, cross this delooping with the walking iso.

Rémy Tuyéras (Jan 16 2026 at 01:44):

So Pro A $\Leftrightarrow$ endomorphisms are identities ?

Rémy Tuyéras (Jan 16 2026 at 04:34):

And if there is typo in Pro B (where $g:y \to x$), then it looks like Pro B $\Leftrightarrow$ endomorphisms are isomorphisms (by contradiction)

Rémy Tuyéras (Jan 16 2026 at 04:51):

So looks like we have an inclusion $\mathbf{Pro}(A) \hookrightarrow \mathbf{Pro}(B)$ and we have a functor (is it?) $\mathbf{Pro}(B) \to \mathbf{Pro}(A)$ which maps a category $C$ to a category $C'$ where $C'$ is defined as:

• Objects: $\mathsf{Obj}(C') = \mathsf{Obj}(C)$
• Arrows: for every arrow $f$ in $C(x,y)$, we impose that $C'(x,y)$ contains:

$[f] = \{ g ~| ~h \circ g = f \circ h'\,, \exists\, h, h' \textrm{ automorphisms }\}$

Here $g \in [f]$ defines a relation $g \sim f$ where $\sim$ is an equivalence:

• symmetry: use the inverse
• reflexive: $\mathsf{id}$
• transitive: transfer the automorphisms

and we have:

$[f] \circ [f'] = [f \circ f']$

John Baez (Jan 16 2026 at 05:55):

Kevin Carlson said:

Hm? Groupoids don't have Property A, do they? Just deloop any nontrivial group, and if you want $x\ne y$, cross this delooping with the walking iso.

Ugh, yeah. I had an idea but I didn't formalize it correctly. I guess I meant

Property A'. Given morphisms $f: x \to y$ and $g: y \to x$, $f$ and $g$ must be isomorphisms.

I think groupoids and preorders have this property, and more generally any category formed by the Grothendieck construction from a pseudofunctor $F \colon P \to \mathbf{Gpd}$ where $P$ is a preorder.

Morgan Rogers (he/him) (Jan 16 2026 at 08:51):

Every category has a preorder reflection. Property A' says that the functor from our category to its preorder reflection is conservative (reflects isomorphisms)

Morgan Rogers (he/him) (Jan 16 2026 at 08:56):

Isn't property B the contrapositive of A'?

Noam Zeilberger (Jan 16 2026 at 12:20):

Is EI-category what you have in mind?

Rémy Tuyéras (Jan 16 2026 at 12:34):

Noam Zeilberger said:

Is EI-category what you have in mind?

Yeah, without checking too much the details, it looks like we have Pro B $\Leftrightarrow$ endomorphisms are isos $\Leftrightarrow$ Pro A'

Rémy Tuyéras (Jan 16 2026 at 12:59):

John Baez said:

Kevin Carlson said:

Hm? Groupoids don't have Property A, do they? Just deloop any nontrivial group, and if you want $x\ne y$, cross this delooping with the walking iso.

Ugh, yeah. I had an idea but I didn't formalize it correctly. I guess I meant

Property A'. Given morphisms $f: x \to y$ and $g: y \to x$, $f$ and $g$ must be isomorphisms.

I think groupoids and preorders have this property, and more generally any category formed by the Grothendieck construction from a pseudofunctor $F \colon P \to \mathbf{Gpd}$ where $P$ is a preorder.

Just some intuitions here, nothing that I checked properly:

I would probably look into functors like $\mathbf{Pro}(B) \to \mathbf{Prop}(A)$, because they might hide some equivalence/adjunction which would help express any category $C$ in $\mathbf{Pro}(B)$ in terms of its corresponding category $C'$ in $\mathbf{Pro}(A)$, which could be close to a preorder, if not one.

And given how the quotient $C'$ is computed, the relationship that we would get might say that $C$ is linked to a function $\mathsf{Obj}(C') \to \mathbf{Aut}(C)$, which could be formally extended to some pseudofunctor functor $C' \to \mathbf{Gpd}$.

This could give you a good measurement of how far the construction $P \to \mathbf{Gpd}$ is from recovering all categories satisfying $\mathsf{Pro}(A)$, etc.

John Baez (Jan 17 2026 at 04:56):

Morgan Rogers (he/him) said:

Isn't property B the contrapositive of A'?

Haha, yes it is! As you can see, I haven't put much thought into this yet. Property A is stricter than I want - what I want is property A', which is the same as property B.

It would be nice to have a construction of all categories with property A' as some sort of hybrid of a preorder and a groupoid.

John Baez (Jan 17 2026 at 08:10):

Property B: given morphisms $f : x \to y$ and $g: y \to x$, both $f$ and $g$ are isomorphisms.

EI-category: every endomorphism is an isomorphism.

@Rémy Tuyéras suggested that these are equivalent. I easily saw that every category with property B is an EI-category, but I didn't believe the reverse implication until I looked at the nLab, which offers a proof.

The proof uses the (claimed) fact that isomorphisms obey the two-out-of-six property.

I say "claimed" because I still haven't shown to myself that they do!

(I haven't tried yet. It seems like a surprising fact, but someday it should seem obvious.)

John Baez (Jan 17 2026 at 08:29):

:shower: Okay, I took a shower and now it's obvious. :shower:

I'll pose it as a little self-contained puzzle for anyone just learning category theory, like me apparently:

Puzzle. Suppose $f: x \to y$ and $g: y \to x$ are morphisms such that $f g$ and $g f$ are isomorphisms. Show that $f$ and $g$ are isomorphisms.

I guess if you feel like posting a solution you should use a "spoiler warning" like this:

Jack Jia (Jan 18 2026 at 01:29):

John Baez said:

:shower: Okay, I took a shower and now it's obvious. :shower:

Puzzle. Suppose $f: x \to y$ and $g: y \to x$ are morphisms such that $f g$ and $g f$ are

It should be something like this?

John Baez (Jan 18 2026 at 06:17):

Why does $fg$ having a right inverse imply that $f$ has a right inverse? (Simple enough, but this is the key step in my opinion. Symmetry will then handle the left inverses, etc.)

John Baez (Jan 18 2026 at 06:31):

Okay, so it's [[EI categories]] that I'm interested in. I suppose I should say why:

Proposition The category of finite-dimensional simple algebras over a field (with algebra homomorphisms as morphisms) is an EI-category.

Proof. Suppose $A$ is a finite-dimensional simple algebra and $f\colon A \to A$ is an endomorphism. The kernel of $f$ is an ideal of $A$, but since $A$ is simple, by definition its only ideals are $\{0\}$ and $A$. Suppose $A$ has $1 \ne 0$. Then the kernel of $f$ can't be all of $A$ since $f(1) = 1 \ne 0$, so the kernel is $\{0\}$. Thus $f$ is one-to-one, so by dimension-counting it must also be onto. On the other hand if $1 = 0$ in $A$ then $A = \{0\}$ and every endomorphism is automatically an isomorphism. ▮

I hadn't expected to fuss over the algebra $\{0\}$, but according to the definition I just gave, a standard definition, it's simple! One might argue it's [[too simple to be simple]].

John Baez (Jan 18 2026 at 06:38):

It would be nice to generalize this to simple algebras over a commutative ring. I guess finite-dimensionality condition should get replaced by something like being [[Artinian]]?

It would also be nice to generalize this even further to some class of "simple monoids" in a suitably nice symmetric monoidal category, but for now I will just work with algebras over a field to keep things simple.

John Baez (Jan 18 2026 at 06:57):

The main thing is that we get an EI-category. Let $C$ be any EI-category. @Morgan Rogers (he/him) mentioned the preorder reflection $\underline{C}$ of $C$, and this is exactly the road I want to go down. Recall that $\underline{C}$ is the preorder with objects of $C$ as elements and $c \le c'$ iff there exists a morphism $f: c \to c'$ in $C$. There's an obvious functor

$\pi \colon C \to \underline{C}$

and Morgan noted that if $\pi(f)$ is an isomorphism $f$ must be too... given that $C$ is an EI-category.

Note the fibers of $\pi$ are groupoids, because $C$ is an EI-category.

I would like $\pi$ to be an opfibration and for it to arise via the Grothendieck construction from some pseudofunctor

$F \colon \underline{C} \to \mathbf{Gpd}$

but I may be wrong here.

John Baez (Jan 18 2026 at 07:02):

Since I'm pestering @Alex Kreitzberg to pose precise questions, here is one:

Question. Given an [[EI-category]] $C$ is the obvious functor from $C$ to its preorder reflection an opfibration?

Okay, I think the answer is no because we can take $C$ to the category with two objects $x$ and $y$ and two morphisms $f,g$ from $x$ to $y$ (together with identity morphisms). Neither $f$ nor $g$ can be a cocartesian lift of the unique arrow from $x$ to $y$ in the preorder reflection of $C$.

John Baez (Jan 18 2026 at 07:08):

However, I feel this situation doesn't happen in the kind of example I actually care about, where $C$ is the category of finite-dimensional simple algebras over a field. So it may be that I want some strengthening of the notion of EI-category, where:

given two morphisms $f, g \colon x \to y$, there exist automorphisms $\alpha \colon x \to x$, $\beta \colon y \to y$ such that $g = \beta f \alpha$.

I think this forces the functor from $C$ to its preorder reflection to be an opfibration.

Jack Jia (Jan 19 2026 at 00:47):

John Baez said:

Why does $fg$ having a right inverse imply that $f$ has a right inverse? (Simple enough, but this is the key step in my opinion. Symmetry will then handle the left inverses, etc.)

I don't quite see the difficulty in this step.

Also, since you are proving the two-out-of-six property, should you be dealing with $hg$ and $gf$ instead? The proof remains exactly the same though.

John Baez (Jan 19 2026 at 05:53):

I didn't say there was anything difficult, but it was a puzzle for beginners so a good solution will be one where it's easy for beginners to follow the key step (which you have now added).

The puzzle was not to prove the two-out-of-six property.

Joe Moeller (Jan 19 2026 at 21:18):

I'll just plug my paper Extensions of representation stable categories where I give conditions under which the Grothendieck construction produces an EI-category.

John Baez (Jan 19 2026 at 23:37):

Cool, that should be helpful! One thing I'm curious about is a characterization of categories obtained by applying the covariant Grothendieck construction to a pseudofunctor

$F : C \to \mathbf{Gpd}$

where $C$ is a preorder. Such categories are EI-categories, so your result may be illuminating. But not every EI-category is of this form!

I'm thinking that for a category $X$ to be of the form $\int F$ where $C$ is a preorder and $F: C \to \mathbf{Gpd}$, $X$ must have this property:

Property C. Given two morphisms $f, g: x \to y$, there exists an automorphism $\alpha : y \to y$ such that $g = \alpha \circ f$.

That is, I'm thinking this property is necessary; I'm not claiming it's sufficient.

Any category with property C is an EI-category, but not every EI-category has property C.

Notice that any preorder has property C, and so does any groupoid!

Ryan Wisnesky (Jan 20 2026 at 06:56):

fwiw, this condition comes up in practice all the time when formalizing database schemas as categories , because more constructions on El-categories are finite than in general. About 90% of the time we see users create schemas that are not El-categories in CQL it's a user error / modeling problem that leads to computational problems later on