Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: practice: communication

Topic: Proof: the associativity square is a pullback => monoid i...

Eric M Downes (Jun 29 2024 at 04:08):

@John Baez said:

Here's a theorem I learned yesterday from Tom Leinster. Take the definition of monoid written out using commutative diagrams, where the associative law becomes a commutative square. What's a monoid where this commutative square is a pullback? It's a group.

I ran into this comment last week and decided to prove the theorem John mentions above without looking at Paolo's paper or looking up any of the stuff @Todd Trimble said about frobenius algebras, etc. It was a lot of fun, and I am happy to say I have worked pretty diligently at correctness, unlike my usual mistake-filled posts. I'm including it here in case anyone feels like giving feedback on proofs, proof writing, etc. or even just style.

That is no small investment of time! So I won't be offended if this just sits here; this was its own reward.

Here we prove a theorem, the result of which was relayed by Tom Leinster. (Baez, 2023) Specifically that a monoid object internal to a category is a group object when its association square is a pullback. Our proof proceeds by exploiting a symmetry of the pullback diagram.

Recall for a monoidal category $\sf C$ with finite product $\times$ and zero object $0$ its identity, the Lawvere syntax for a semigroup object $M\in {\sf C}_0$ is (MacLane, pp. 4, 170)

$(1)~~~~~~\mu\circ(id_M\times\mu)=\mu\circ(\mu\times id_M):M^3\to M$

$M$ is further a monoid when there exists $\eta:0\to M$ obeying

$(2)~~~~~~\mu\circ(id_M\times\eta)=\mu\circ(\eta\times id_M)=id_M:M\to M$

Where recall $0\times M\cong M\cong M\times 0$.

If equation (1) associativity, is a pullback square (MacLane p. 71), recall that any $L\in{\sf C}_0$ with morphisms to $M^2$ gives a unique morphism $\omega$:

$(3)~~~~~\forall L\in{\sf C}_0,~\forall\,\alpha,\beta:L\to M^2$
$~~~~~\mu\circ\alpha=\mu\circ\beta\implies\exists! \,\omega:L\to M^3$
$~~~~~\alpha=(\mu\times id_M)\circ \omega~~~~~\beta=(id_M\times\mu)\circ \omega$

This pullback diagram has a symmetry. Making this explicit defines a dual morphism $\omega^*$:

Diagram 1. The universal property of the pullback (to Left) implies a duality (to Right).

Even if the monoid is commutative, the fibration is not, so we uniquely determine a dual $\omega^*$ by applying an automorphic functor $^*$. We swap the positions of $\alpha,\beta:L\to M^2$ with respect to the only other asymmetric components of (1L): $\mu\times id_M,~id_M\times\mu$. The result is that duals $\alpha^*:=\beta,~ \beta^*:=\alpha$, and $\omega^*$ are determined by universal property of the pullback, and only those morphisms and objects depending on $\alpha,\beta,\omega$ are swapped by $^*$. Swapping twice returns us to the original diagram, so

$\omega^{**}=\omega$

Observing of left/right projections, that
$\pi_L\circ\beta=\pi_L\circ\omega, ~\pi_R\circ\alpha=\pi_R\circ\omega$
and vice versa for $\omega^*$, we expect to distill the adjointness between $\alpha,\beta$ into their common factor, the middle-projection $\pi_C\circ\omega$. An involution of $M$ capturing this duality for $(\omega,\omega^*)$ must satisfy

$(4)~~~~~~\xi_\omega:M\leadsto M;~~~~\xi_\omega\circ\pi_C\circ\omega^*=\pi_C\circ\omega(\alpha,\beta)~~~~\xi_\omega^*\circ\xi_\omega=id_M$
if it exists.

Dia. 2. Factoring the pullback UP morphism $\omega$ of the constant $\eta$ yields the inverse $\xi$, proving the monoid $M$ a group.

$\omega$ of the constant $\eta$ yields the inverse $\xi$, proving the monoid $M$ a group.">

We will exhibit a global $\xi$ by considering
$L=M,~\alpha=id_M\times\eta,~\beta=\eta\times id_M$

By eq (2) the monoid identity, the final product at $M$ is just the image of $id_M$ and the pullback square applies, so the common factor $\pi_C\circ\omega(\eta,\eta)$ will be relevant to any generalized element of $M$. WLOG we consider the commuting triangle involving $\eta\times id_M$. By the universal property of the product, we can factor $id_M$ out of the fibration on the right.

$\eta\times id_M\mapsto \eta~~~~~\mu\times id_M\mapsto \mu~~~~~\omega\mapsto(\pi_L\times\pi_C)\circ\omega$

Recalling eq (4) the constraints for $\xi_\omega$, we define $\xi\in{\sf C}_1$

$\xi:=\pi_C\circ\omega(\eta,\eta):M\to M$

and observe $\omega(\eta,\eta)^*=\omega(\eta,\eta) \implies \xi=\xi^*$ We refactor, distributing $\circ$ over $\times$:

$(\pi_L\times\pi_C)\circ\omega(\eta,\eta)=(id_M\times\xi)\circ\Delta$.

We invoke the unique morphism $M\overset{!}{\to}0$, recovering the standard syntactic diagram for the inversion morphism of a group object, shown in diagram (4), lower right. An analogous argument using $id_M\times\eta$ proves the same for multiplication of inversion on the left. We remark with pleasure how the existence of an involution on $M$ is reflected by the Yoneda lemma, in the $C_2$-symmetry of this pullback diagram. $\square$

References

• Baez, John. Aug 29, 2023 comment in “CT Representation Online” Category Theory Zulip
• MacLane, Saunders. Categories for the Working Mathematician. Springer 1998

Thank you for reading!

Ralph Sarkis (Jun 29 2024 at 07:26):

I will try to read this later, but I see that you are using the universal property of the product while it seems from the original thread that this result holds in any monoidal category.

Also, did you first try to prove it in the category $\mathbf{Set}$ ? I think there is a very short proof (2-3 lines), but I don't know if it can be generalized. (Sometimes proving things using the intuition in $\mathbf{Set}$ is a good idea, sometimes it is not.)

Eric M Downes (Jun 29 2024 at 13:00):

Oh great point on monoidal product! Lawvere theories involve (finite) product preserving functors from a 'syntactic category', and I just used that. Except that I call $\sf C$ a monoidal category, which is sloppy. I'll review how MacLane handles this, but IIRC the monoid object construction works in any monoidal category, so then its just a question of "can you have this kind of pullback?". It seems this particular pullback square can't be established for monoid objects in $\sf Ab$ for instance because there aren't enough abelian group homomorphisms $\alpha,\beta$ to specify all the multiplicative inverses in a field... which is comforting because $\xi$ isn't a group homomorphism over addition! :)

The first thing I did is try to disprove it in ${\sf Set}$, actually. I used the cancellative monoid $(\mathbb{N},+,0)$ which is where the comment "even if the monoid is commutative, the fibration is not" comes from. It clarified how the pullback relates to inverses. If I were being more pedagogical I'd have included that, and maybe should have anyway. Essentially, component-wise
$\alpha\mapsto(x,y)\land\beta\mapsto(z,w)\implies\omega\mapsto(x,u,w)$
so $u$ acts as an adjoint between the two coordinates; $xu=z, y=uw$. The dual $\omega^*$ then swaps the coordinate position relative to the $id\times\mu$ and you have $x=zu^*, u^*y=w$ exactly s it should be for the inverse.

Eric M Downes (Jun 29 2024 at 13:46):

For now I'll just claim to have proved this for categories with finite universal products $\times$, and will come back to the monoidal category question once any ither objections have been resolved.

Ralph Sarkis (Jun 29 2024 at 15:06):

Ralph Sarkis said:

I will try to read this later

Sorry, but there are several mistakes from the start which make it hard for me to follow.

Ralph Sarkis (Jun 29 2024 at 15:06):

Eric M Downes said:

dual morphism $\omega^*$

Duality refers to reversing the direction of arrows (going from $\mathsf{C}$ to $\mathsf{C}^{\mathrm{op}}$), so I would not use this word here, nor this notation. Maybe writing $\omega = \omega_{\alpha,\beta}$ and $\omega^* = \omega_{\beta,\alpha}$ is enough. You might say this will be weird to define a functor, but see my third comment.

the fibration is not

What is the fibration?

automorphic functor ${}^*$

This is not a well-defined operation. First, you have only defined this functor on morphisms that arise as $\omega$ for some $\alpha,\beta$, what does it do on other morphisms ? Moreover, if $\omega$ also makes that pullback diagram commute for $\alpha'$ and $\beta'$, is it possible that the two "duals" you have defined are different ? Using my notation, maybe $\omega_{\alpha,\beta} = \omega_{\alpha',\beta'}$ but $\omega_{\beta,\alpha} \neq \omega_{\beta',\alpha'}$, so $\omega^*$ could be two different things (or even more).

Ralph Sarkis (Jun 29 2024 at 15:20):

Let me give you the solution for $\mathbf{Set}$ that you may want to adapt (I haven't thought about this generalization too much).

Solution in Set

For any element $g \in M$, there are two functions $\alpha_g$ and $\beta_g$ from the singleton $\{\ast\}$ to $M^2$ which send the point $\ast$ to $(1,g)$ and $(g,1)$ respectively. Since $\mu(g,1) = g = \mu(1,g)$, we can apply the universal property of the pullback to get $\omega$ like so:
image.png

Now, what is the result of applying $\omega$ to $\ast$. It is a triple $(x,y,z) \in M^3$. The fact that region (a) commutes tells us two things: 1) the third coordinate of $(x,y,z)$ and the second coordinate of $(1,g)$ coincide, ie. $z=g$, and 2) the product of the first two coordinates of $\omega(\ast)$ gives the first coordinate of $\alpha_g(\ast)$, ie. $xy = 1$. Similarly, by region (b) commuting, we have $x = g$ and $yz =1$. This tells us that $y$ is the inverse of $x=z=g$. Since $g$ was arbitrary at the start, $M$ is a group.

Eric M Downes (Jun 29 2024 at 17:45):

Thanks for engaging Ralph!

Ralph Sarkis said:

Duality refers to reversing the direction of arrows

Thanks! I didn't know that dual always meant "op" now! (FWIW [[dual]] was/is used widely in algebra, geometry, and physics etc. as something like "$x^*,x$ are dual" $\sim$ "$x^*$ is the image of $x$ under an involution" but I am happy to change this; my audience is CTists after all!)

Maybe writing $\omega = \omega_{\alpha,\beta}$ and $\omega^* = \omega_{\beta,\alpha}$ is enough.

Yep, I like that. Keep It Simple. :)

What is the fibration?

I was thinking of $L\to M$ as a fibration and so $\mu\circ(\mu\times id)$ has a different "pullback fibration" than $\mu\circ(id\times\mu)$.

What term would you use for "$\mu\times id_M$ composes after $\omega$ differently than $id_M\times\mu$ does"?

This is not a well-defined operation.

Totally fair! I had described it in words, but I can certainly define this precisely on morphisms with a \begin{cases}... and show it is a functor explicitly.

Thanks again for your comments. I'll just edit the source, above, as reposting would be more annoying.

Eric M Downes (Jun 29 2024 at 17:50):

(As far as your solution in $\sf Set$, thank you! I'll certainly look at it. But at least at this point I still maintain that my proof is correct; right now I am just seeing a failure to communicate on my part, which is important and I want to fix. So thank you again for engaging.)

Todd Trimble (Jun 29 2024 at 17:58):

Ralph Sarkis said:

I will try to read this later, but I see that you are using the universal property of the product while it seems from the original thread that this result holds in any monoidal category.

Also, did you first try to prove it in the category $\mathbf{Set}$ ? I think there is a very short proof (2-3 lines), but I don't know if it can be generalized. (Sometimes proving things using the intuition in $\mathbf{Set}$ is a good idea, sometimes it is not.)

There's a general principle that for any statement that can be expressed in the language of limits, which applies in this case, the statement is true iff it holds in the category of sets. The idea is that the Yoneda embedding preserves and reflects limits, i.e., for any category $C$ in which the statement is interpreted, the collection of representable functors

$C \overset{C(c, -)}{\longrightarrow} \mathsf{Set}$

jointly preserves and reflects limits, so verification in sets suffices. This can be seen as a basic "completeness theorem".

Todd Trimble (Jun 29 2024 at 18:00):

(Sorry, I was interpreting this in the cartesian monoidal case. But the principle seems worth pointing out anyway.)

Eric M Downes (Jun 29 2024 at 19:23):

Todd Trimble said:

There's a general principle that for any statement that can be expressed in the language of limits, which applies in this case, the statement is true iff it holds in the category of sets.

That is fascinating, thank you! This seems closely related to the correspondence between algebraic theories of $A$ and functors from the category of $A$'s to $\sf Set$, but more general. If there is a phrase / theorem for this, I will add it to my reading list.

Eric M Downes (Jun 29 2024 at 19:33):

It sounds like a [[concrete category]] (although I was recommended recently not to think as much in terms of concreteness, as its not as broadly useful a concept.)

Anyway, back to specifying my alleged-functor that is a $C_2$-action on the pullback diagram for Ralph. I think it may be more parsimonious just to swap $\mu\times id_M,id_M\times\mu$ so I need to specify all the diagrams that tell us "wether $\mathsf{C}_1\ni f\cong id_M\times\mu$ or not".

Eric M Downes (Jun 30 2024 at 06:26):

Okay, good news I was able to simplify everything; we don't need a functor or even a group-action by $C_2$. Womp-womp news is that the proof (updated above) reduces to something that really is obviously set theory in disguise, as Todd was implying.

Eric M Downes (Jun 30 2024 at 07:08):

And yeah, the difference with Ralph's proof is pretty minimal. I take my source object as $M$ itself instead of $\set*$ but the effect is the same, to get pairs $(g,1), ~(1,g)$.