1:11 AM
er maybe that's not so bad.
Anyway, I think there's a bijection between nullification functors (= slocalization at a set of maps with contractible codomain) on the $\infty$-category of spaces and modalities on the $\infty$-category of spaces. The bijection takes a nullification $P$ to the modality whose left class is those maps with $P$-acyclic fiber, and conversely simply takes a modality $(\mathcal E, \mathcal M)$ to localization at $\mathcal E$.

16 hours later…
5:11 PM
There's a theorem saying that if both $\mathcal{C}$ and $\mathcal{C}^{op}$ are (locally-) presentable 1-categories then $\mathcal{C}$ must be a lattice. I wonder if this is true for $\infty$-categories...
I believe it should fail. I would not consider it a completely fleshed out counterexample but here's my thought:
I think verdier duality can be stated as an equivalence $CoSh^{Sp}(\mathfrak{X}) := Fun^L(\mathfrak{X},Sp)^{op} \cong Sh^{Sp}(\mathfrak{X}) = Fun^{R}(\mathfrak{X}^{op},Sp)$ for $\mathfrak{X}$ the $\infty$-topos associated with a sufficiently nice locally compact hausdorff space. But regardless of that $Fun^{L}(\mathcal{C},\mathcal{D})$ when $\mathcal{C}$ and $\mathcal{D}$ are presentable is always presentable. I admit that I do not immediately have a reference for either statemen
I just found a reference for this kind of Verdier duality statement! people.math.harvard.edu/~lurie/282ynotes/LectureXXI-Verdier.pdf
There's no proof there unfortunately.
Still though, if the theorem I mentioned about 1-categories holds for $\infty$-categories it seems like no such statement can hold regardless of how nice $\mathfrak{X}$ is...
To see the relation between my formulations and lurie's: $Sh_{\mathcal{C}}(X) = Sh_{\mathcal{C}^{op}}(X)^{op} = Fun^{R}(\mathfrak{X}^{op},\mathcal{C}^{op})^{op} = Fun^{L}(\mathfrak{X},\mathcal{C})^{op}$
Also the fact that verdier duality involves in a crucial way stability makes this plausible as any stable 1-category must be trivial I believe.
@TimCampion That's really nice to know! Thanks for clearing that up.

5:49 PM
@SaalHardali This cannot be right: for $\mathfrak{X}=\mathrm{Space}$ (i.e. pick your space to be the point) this would say $\mathrm{Sp}=\mathrm{Sp}^{op}$, which is false. Verdier duality usually has finiteness restrictions
I suspect the correct statement of Verdier duality is $\mathrm{Fun}^L(\mathfrak{X},\mathrm{Sp}^{op})^{op}\cong \mathrm{Fun}^R(\mathfrak{X}^{op},\mathrm{Sp})$. Unfortunately for you this does not provide a counterexample to the theorem (which I believe to be true)

@SaalHardali I'm glad it's of some interest!
Regarding the "opposites" question, it's weird
The proof for 1-categories relies on epi-mono-type factorizations
so it might not be possible to generalize the proof to $\infty$-categories

@SaalHardali I believe you have an op too much in the last step ($\mathrm{Fun}^R(\mathfrak{X}^{op},\mathcal{C}^{op})=\mathrm{Fun}^L(\mathfrak{X},\mathcal{C})^{op}$ )

@DenisNardin Yep! That's the one. Just found it with the help of a third party. Thanks.
Great! I'm glad it all comes down to this stupid point. This actually solves the original problem that motivated this question.
So I'm back to being agnostic about the opposite presentability thing.
And of course $Sp=Sp^{op}$ is false (silly me).
And I agree with your statement of Verdier duality as well.
Everything makes a lot more sense now. Thanks.