5
$\begingroup$

Question I. Is the following statement, inspired by this one, true?

Universal Property I. The symmetric monoidal structure on the category $\mathsf{Sets}_*$ of pointed sets and pointed maps between them is uniquely determined by the following requirements:

  1. Two-Sided Preservation of Colimits. The smash product functor $$\wedge\colon\mathsf{Sets}_*\times\mathsf{Sets}_*\to\mathsf{Sets}_*$$ preserves colimits separately in each variable.
  2. The Unit Object Is $S^0$. We have $\mathbb{1}_{\mathsf{Sets}_{*}}=S^0$.

If so, how can it be proved?

Question II. Is the following statement, inspired by Gepner–Groth–Nikolaus's Universality of multiplicative infinite loop space machines, true?

Universal Property II. The symmetric monoidal structure on the category $\mathsf{Sets}_*$ of pointed sets and pointed maps between them is the unique symmetric monoidal structure on $\mathsf{Sets}_*$ such that the free pointed set functor $$(-)^+\colon\mathsf{Sets}\to\mathsf{Sets}_*$$ admits a symmetric monoidal structure.

Again, if this statement is true, how can we prove it?

Note: Everything here is taken $1$-categorically―there is no $\infty$-categorical stuff in sight.

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ A note on asking two questions at once: I'm asking two questions at once here since I expect there might be an easy reference or proof for them, and if someone knows the answer to one of the questions, then there's a good chance that they'll also know the answer for the other one. If this proves to be false, I'll split the question in two. $\endgroup$
    – Emily
    Commented Mar 7 at 19:54
  • 1
    $\begingroup$ Since you mention explicitly that you don't want $\infty$-categories in the picture it might help to make explicit what "unique" means. The answer to your question as I understand it is yes, in a suitable $(2,1)$-categorical sense, but I just want to make sure that you allow this $\endgroup$
    – Maxime Ramzi
    Commented Mar 8 at 4:19
  • $\begingroup$ @MaximeRamzi Yep, a $(2,1)$-categorical (or $2$-categorical) universal property for those would be lovely! $\endgroup$
    – Emily
    Commented Mar 8 at 16:59
  • 1
    $\begingroup$ Emily: that would indeed determine the functor $\wedge$, but not necessarily the monoidal structure, which is more than just that functor (by the way, the reason I haven't written an answer to your question yet is that I'm thinking about II : I've been trying to figure out whether one needs to assume that the monoidal structure preserves colimits in each variable or whether it follows from $(-)_+$ being strong monoidal; it seems like it does but I haven't been able to prove it yet. For UP I, I could already write an answer following the lines I suggested earlier) $\endgroup$
    – Maxime Ramzi
    Commented Mar 10 at 18:31
  • 1
    $\begingroup$ Of course, it's my pleasure :) I wrote an answer with what I know so far, hopefully someone else can chime in and figure out what I don't know $\endgroup$
    – Maxime Ramzi
    Commented Mar 10 at 19:18

1 Answer 1

Reset to default
1
$\begingroup$

Since I can't quite figure it out as I mentioned in the comments, let me already answer what I can - the answer to Question I is yes, and I'm enclined that it is so also for Question II but I'm missing an argument.

The point is that in Question II, one should be able to deduce that the monoidal structure preserves colimits in each variable: fix $X\in Set_*$, then it is of the form $(X_0)_+$ for some $X_0$, and for all $Y\in Set$, $X\otimes Y_+ \cong (X_0\times Y)_+ \cong X\wedge (Y_+)$. Since every $Y\in Set_*$ is of the form $(Y_0)_+$, this tells you that at least pointwise, $\otimes$ is given by the smash product. The same elementary argument shows that $X\otimes -$ preserves coproducts, but it seems a bit more subtle to properly show that it preserves pushouts although it should be doable (and I think elementary, once you spot it...)

Thus Question II should follow from Question I, and let me now focus on the latter.

The point that is made in the $\infty$-categorical proofs is that you can reinterpret "symmetric monoidal structure with 1." as "commutative algebra structure in $Pr^L_1$", the $(2,1)$-category of presentable $1$-categories equipped with the Lurie/Deligne/Kelley tensor product. From that perspective, together with the general theory of idempotent algebras, the claim follows from :

Theorem : $(Set_*,S^0)$ is an idempotent algebra in $Pr^L_1$, that is, the colimit-preserving functor $Set\to Set_*$ that picks out $S^0$ becomes an equivalence upon tensoring with $Set_*$.

Proof: The proof of that is exactly the same as that of HA 4.8.2.11, replacing each instance of "spaces" with "sets" and "$\infty$-category" with $1$-category. I can reproduce it with the relevant changes if you want, but for now let me just say that.

Here is a question to which I do not know the answer off the top of my head, and which might help with Question II: is the same result true if we replace $Pr^L_1$ with the category of categories admitting coproducts and coproduct-preserving functors ? Therein, $Set$ still turns out to be free on a point (it is surprising that $Set$ is both a free cocompletion and a free coproduct-completion), so it wouldn't be the most surprising thing but I'm not sure.

If not, then some extra argument is needed for Question II. Note that adding colimit-preservation of course makes it correct because then it satisfies the hypotheses of Question I :)

(and also note, as a side point, that because everything is presentable, Two-sided colimit-preservation is equivalent to monoidal closedness - and without presentability, it is implied by closedness)

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you so much, Maxime, and sorry for taking a bit to reply! It's taking a little longer than I expected for me to digest and explore a couple points in your answer. Would it be okay to get back to you in a bit, possibly with a couple questions, as I go making my way through these proofs and your answer? $\endgroup$
    – Emily
    Commented Mar 14 at 11:34
  • 1
    $\begingroup$ Of course, no worries and no hurry :) $\endgroup$
    – Maxime Ramzi
    Commented Mar 14 at 12:16
  • $\begingroup$ @MaximeRanzi Thank you so much again, Maxime :) $\endgroup$
    – Emily
    Commented Mar 14 at 13:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .