7
$\begingroup$

$\newcommand{\cat}{\mathsf} \newcommand{\fun}{\mathrm{Fun}} \newcommand{\calg}{\mathrm{CAlg}}$ Let $\cat C^\otimes,\cat D^\otimes, \cat E^\otimes$ be symmetric monoidal $\infty$-categories, and $p^\otimes: \cat D^\otimes \to \cat E^\otimes$ a map of $\infty$-operads (aka a lax symmetric monoidal functor).

Assume $p: \cat D\to \cat E$ is a cartesian fibration.

I'd like to know under what natural conditions $\fun^{lax}(\cat C^\otimes,\cat D^\otimes)\to \fun^{lax}(\cat C^\otimes,\cat E^\otimes)$ is still a cartesian fibration, and more specifically if there are references regarding this type of situation.

A few remarks:

  • In the cases I'm interested in, $p^\otimes$ is a map of symmetric monoidal $\infty$-categories, so a strict symmetric monoidal functor. If that's necessary to give an interesting statement, I'm willing to assume it.
  • Still in those cases, $\cat{D,E}$ are presentable with a compatible tensor product, and I can reduce to the case where $\cat C$ is small, so using Day-convolution, one can reduce to the question of whether $\calg(\cat D)\to \calg(\cat E)$ is still a cartesian fibration, i.e. to the case $\cat C = N(Fin_*)$.
  • Finally, I'm also in a situation where $p$ is a "monoidal fibration", by which I mean that if $x\to y$ is a $p$-cartesian edge, then so is $x\otimes z\to y\otimes z$; I think this can be relevant - and hopefully this, together with the first bullet point, should be enough.

Has something like this been written up anywhere ?

EDIT : I wrote down what I think is a complete proof, and indeed items 1 and 3 are the ones that make it work. But the proof is long for this sort of technical result and I'm still interested in references so as not to lengthen what I'm writing too much.

$\endgroup$
2
  • $\begingroup$ You might want to check out Nikolaus’ paper on stable oo-operads. The content around Corollary 3.7 might give you what you want, but I’m not sure off the top of my head. $\endgroup$ Commented Sep 15, 2020 at 18:16
  • $\begingroup$ @LiamKeenan : Hi Liam; thanks for the reference ! Unfortunately I just had a look at it and it seems like there isn't anything like what I'm after in that paper - Nikolaus looks at how the functorialities of $Fun(A,B)$ are compatible with the operadic structure, but not (as far as I can tell) with questions of (co)cartesian-ness $\endgroup$ Commented Sep 15, 2020 at 18:27

2 Answers 2

3
$\begingroup$

I don't know a reference but here is a not-too-long proof. The condition that $\mathsf{D} \to \mathsf{E}$ is a cartesian fibration implies that for every $\langle n \rangle \in \mathrm{Fin}_*$ the map $\mathsf{D}^{\otimes}_{\langle n\rangle} \to \mathsf{E}^{\otimes}_{\langle n\rangle}$ is a cartesian fibration and that for every inert map $\alpha : \langle n\rangle \to \langle m\rangle$ the transition functor $\alpha_*:\mathsf{D}^{\otimes}_{\langle n\rangle} \to \mathsf{D}^{\otimes}_{\langle m\rangle}$ sends $p^{\otimes}_{\langle n\rangle}$-cartesian edges to $p^{\otimes}_{\langle m\rangle}$-cartesian edges. Now apply (the dual of) [HTT, Corollary 4.3.1.15] to deduce that every $p^{\otimes}_{\langle n\rangle}$-cartesian edge in the fiber $\mathsf{E}^{\otimes}_{\langle n\rangle}$ is also $p^{\otimes}$-cartesian as an edge in $\mathsf{E}$ (note that being a cartesian edge is a form of a relative limit). This means that the map $p^{\otimes}: \mathsf{D} \to \mathsf{E}$, though possibly not a cartesian fibration itself, still admits cartesian lifts for a certain collection of edges in $\mathsf{E}$: all the edges which are contained in a fiber $\mathsf{E}^{\otimes}_{\langle n\rangle}$ for some $\langle n\rangle$. Otherwise put: all the arrows which map to an equivalence in $\mathrm{Fin}_*$. It then follows that the functor $$ \mathrm{Fun}_{\mathrm{Fin}_*}(\mathsf{C}^{\otimes},\mathsf{D}^{\otimes}) \to \mathrm{Fun}_{\mathrm{Fin}_*}(\mathsf{C}^{\otimes},\mathsf{E}^{\otimes}) $$ is a cartesian fibration, where $\mathrm{Fun}_{\mathrm{Fin}_*}$ denotes functors preserving the projection to $\mathrm{Fin}_*$ (but not necessarily preserving inert edges). Indeed, any natural transformation of functors $\mathsf{C}^{\otimes} \to \mathsf{E}^{\otimes}$ whose projection to $\mathrm{Fin}_*$ is constant consists object-wise of arrows in $\mathsf{E}^{\otimes}$ which admit cartesian lifts in $\mathsf{D}^{\otimes}$ by the above, and hence itself admits cartesian lifts as a natural transformation. By base change we then conclude that the functor $$ \mathrm{Fun}^{\mathrm{lax}/\mathsf{E}}(\mathsf{C}^{\otimes},\mathsf{D}^{\otimes}) \to \mathrm{Fun}^{\mathrm{lax}}(\mathsf{C}^{\otimes},\mathsf{E}^{\otimes}) $$ is a cartesian fibration, where $\mathrm{Fun}^{\mathrm{lax}/\mathsf{E}}$ here stands for the those functors $\mathsf{C}^{\otimes} \to \mathsf{D}^{\otimes}$ over $\mathrm{Fin}_*$ whose projection to $\mathsf{E}^{\otimes}$ preserves inert edges. To finish the proof it will suffice to show that for cartesian edge in $\mathrm{Fun}^{\mathrm{lax}/\mathsf{E}}(\mathsf{C}^{\otimes},\mathsf{D}^{\otimes})$, if its target preserves inert edges then its domain preserves inert edges. Given that inert edges in $\mathsf{D}^{\otimes}$ are exactly the cocartesian lifts of the inert edges in $\mathrm{Fin}_*$, this follows from the fact that inert transition functors $\alpha_*:\mathsf{D}^{\otimes}_{\langle n\rangle} \to \mathsf{D}^{\otimes}_{\langle m\rangle}$ sends $p^{\otimes}_{\langle n\rangle}$-cartesian edges to $p^{\otimes}_{\langle m\rangle}$-cartesian edges.

$\endgroup$
5
  • $\begingroup$ It seems like you're claiming that condition 3 is not necessary, is that correct ? or am I missing the point in your argument where you use that ? (in the argument I had in mind, I used it to prove that $p^\otimes_{\langle n \rangle}$-cartesian edges were $p^\otimes$-cartesian; which it seems you're proving more generally) $\endgroup$ Commented Sep 28, 2020 at 10:07
  • 1
    $\begingroup$ So I used the HTT reference for this, and looking at it again I see that for this one needs condition 1, i.e., that $p^{\otimes}$ preserves cocartesian edges, but not condition 3. That HTT reference (and especially the following particular case of it) is actually quite useful in practice. Applying its dual version with $K=\mathcal{E}_0 = \Delta^0$ one obtains the statement that for a map of cocartesian fibrations (which preserves cocartesian edges), an edge contained in a fiber is cartesian with respect to the induced map on fibers if and only if it is cartesian on the total spaces. $\endgroup$ Commented Sep 29, 2020 at 11:41
  • $\begingroup$ That does sound very useful ! I might change the way I wrote things up then (I ended up just writing it because there were no references - but the way I wrote it, I use condition 3, and it's longer than what you wrote) $\endgroup$ Commented Sep 29, 2020 at 11:44
  • 1
    $\begingroup$ Condition 3 will be needed if you wanted to prove that the induced map on categories of strong symmetric monoidal functors is also a cartesian fibration, see the last part of the proof above. $\endgroup$ Commented Sep 29, 2020 at 11:48
  • $\begingroup$ Yes, that's actually what I used (using the enveloppe of an $\infty$-operad, I reduced to the case of strong symmetric monoidal functors). But I only need the statement about categories of lax symmetric monoidal functors (although in any case, I had the additional condition at my disposal) $\endgroup$ Commented Sep 29, 2020 at 11:51
1
$\begingroup$

I have accepted Yonatan's answer because ultimately his proof is what's appearing in the reference, but for future reference, we included his proof in this preprint, as Appendix B (rather than the one I had in mind at first, which was longer and used more hypotheses). So now, there is a reference.

$\endgroup$

You must log in to answer this question.

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