## August 16, 2019

#### Posted by John Baez

I have a few questions about equalizers. I have my own reasons for wanting to know the answers, but I’ll admit right away that these questions are evil in the technical sense. So, investigating them requires a certain morbid curiosity… and have a feeling that some of you will be better at this than I am.

Here are the categories:

$Rex$ = [categories with finite colimits, functors preserving finite colimits]

$SMC$ = [symmetric monoidal categories, strong symmetric monoidal functors]

Both are brutally truncated stumps of very nice 2-categories!

My questions:

1) Does $Rex$ have equalizers?

2) Does $SMC$ have equalizers?

3) We can get a functor $F \colon Rex \to SMC$ by arbitrarily choosing an initial object for each category in $Rex$ to serve as the unit of the tensor product and arbitrarily choosing a binary coproduct for each pair of objects to serve as their tensor product. Do this. Does the resulting $F$ preserve equalizers?

You see, I have some equalizers that exist in $Rex$, and I know $F$ preserves them, but I know this by direct computation. If 1) were true I’d know these equalizers exist on general grounds, and if 3) were true I’d know they’re preserved on general grounds. This might help me avoid a bit of extra fiddliness as I walk on the dark side of mathematics and engage in some evil maneuvers… especially if there’s some reference I can just cite.

2) is just my morbid curiosity acting up. I’m pretty sure the category of symmetric monoidal categories and strict symmetric monoidal functors has equalizers, but strong?

Posted at August 16, 2019 7:31 AM UTC

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### Re: Evil Questions About Equalizers

Dear John,

(1) Consider the indiscrete category $A$ on two objects $0$ and $1$. Since it is equivalent to the terminal category it has finite colimits, and any functor from $A$ to $A$ preserves them. In particular, the functor $s:A \to A$ interchanging $0$ and $1$ preserves them. Any functor $F:X \to A$ equalising $s$ and the identity would have to factor through their equaliser in $Cat$, which is the empty category. But then $X$ would have to be empty itself. However no category with finite colimits can be empty – therefore the equaliser of $id$ and $s$ does not exist in $Rex$.

(2) The same example and argument, viewing the finitely cocomplete category as a symmetric monoidal one, shows that $SMC$ does not have equalisers.

(3) Well, your functor $F$ restricts to act on strict morphisms (if you view the objects of $Rex$ as coming equipped with choices of colimits). Restricted to strict morphisms, in the source and target categories, it does preserve equalisers, and is indeed induced by a map of $2$-monads. I haven’t thought about whether it preserves any equalisers of non-strict maps that happen to exist though – do you have some examples?

Note: if you view $Rex$ and $SMC$ as $2$-categories they do have many nice 2-dimensional limits (pie limits, flexible limits…) but not equalisers.

Best, John.

Posted by: John Bourke on August 16, 2019 3:57 PM | Permalink | Reply to this

### Re: Evil Questions About Equalizers

Thanks for deftly solving 1) and 2), John! I was trying a similar trick but it was more complicated and I got bogged down. You solved it with minimal fuss!

Can you tackle 3) with equal finesse? Here I’m looking to find an equalizer that exists in $Rex$ that is not preserved by $\Phi \colon Rex \to SMC$.

You see, I have a bunch of equalizers that are preserved, and I think I know how to show it, but it would look a bit silly to prove this if in fact all equalizers (that exist) in $Rex$ are preserved. Sort of like proving “the sum of two even perfect numbers is even”.

(Actually I should check again to make sure the equalizers I think are preserved, really are.)

Posted by: John Baez on August 19, 2019 8:31 AM | Permalink | Reply to this

### Re: Evil Questions About Equalizers

It seems to me that what can be proved is the following, which I’m guessing will cover your case.

Prop: Any equaliser in $Rex$ which is preserved by the forgetful functor to $Cat$ is also preserved by your functor $\Phi:Rex \to SMC$.

Since your $\Phi$ commutes with the forgetful functors to $\Cat$, to prove the above it suffices to show (*) that $U:SMC \to \Cat$ reflects equalisers: i.e., given a fork in $SMC$ sent by $U$ to an equaliser in $Cat$ then the original fork in $SMC$ is an equaliser.

Now (*) does appear to me to be true (I haven’t checked completely) if still slightly fiddly to prove carefully. I thought about more generally using 2-monads, their algebras and pseudomaps.

Hope that helps.

Posted by: John Bourke on August 20, 2019 10:35 AM | Permalink | Reply to this

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