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12:02 AM
@SaalHardali Note that the class $\mathcal E$ of morphisms $f$ such that $\Sigma^\infty_+(f)$ is $L$-local (for some fixed Bousfield localization $L$) is closed under 2/3. I'm pretty sure it follows that the induced factorization system $(\mathcal E, \mathcal R)$ corresponds to a localization of the category of spaces, in the sense that $\mathcal R$ is generated as a right class by its maps $X \to 1$.
So if such a factorization system is a modality, then it corresponds to a left exact localization of spaces and must be trivial.
But there should be tweaks of this giving interesting modalities. One thing you might do is look at the class of maps $f$ with $L$-acyclic fiber. Then you manifestly have a class of maps closed under base change, but it will then be delicate trying to ensure you actually have the left half of a factorization system.
I think Anel, Biedermann, Finster, and Joyal said in one of their papers on modalities that they planned to show any set of morphisms in a topos generates a minimal modality, though I don't think a proof has appeared yet.
I'm really curious what modality is generated by a type-$n$ space.
 
@TimCampion You are right of course. When I was thinking about this I indeed recall defining the left class as those morphisms whose fibers have suspension spectra in the bousfield class. I forgot about that point so somehow I thought defining directly in terms of equivalence could work.
Do we know a nom-trivial modality which can be described as maps whose fibers are stably in some fixed bousfield class?
I think rationalization is a good example.
 
It might be easier if one additionally puts some kind of connectivity condition on the fibers?
Like if you look at maps whose fibers are simply-connected and with trivial rationalization, are these the left half of a modality?
 
I don't have any other example though...
I was thinking just that now.
 
Wait, are you saying that rationalization definitely works?
 
For example $L_n$ could be examples
If we only ask for it to be stable under base change with some bound on connectivity.
I think rationalization works for serre class reasons. But I might be missing something...
 
12:15 AM
Would you argue by explicitly constructing factorizations? Or rather by verifying that the class of maps is closed under colimits in the arrow category + composition + cobase change?
Or something else?
 
Composition cobase changes are obvious.
For rational I mean.
Colimits in the arrow category are actually also obvious.
-
 
I think I see how composition follows from Serre class reasons.
How does cobase change work?
I agree cobase change is obvious if you're just applying $\Sigma^\infty_+$ but we switched to looking at fibers, and their interaction with cobase change seems hard to me?
 
Yeah it's not obvious sorry
I got ahead of myself.
 
ah ok
But the condition that the fiber of $f$ be rationally trivial is almost the same as $\Sigma^\infty_+ f$ being a rational equivalence -- it's really just a $\pi_1$ issue.
Which makes it kind of tantalizing...
 
In general I think this might be useful for these kind of questions: mathoverflow.net/q/330711/22810
Anyway yeah for HZ-local spaces being rationally contractible is the same as being stably rationally trivial.
So the only problem is pi_1 issues like you said.
 
12:32 AM
Ok, I think it works, at least if you require the fibers to be $n$-connected for some $n \geq 1$.
 
Interestingly if X is any connected space then the rational homology of OmegaX controls the homotopy of X which means that the map from a point to X is in our class iff X rationally contractible.
 
How so?
 
What I mean is $\pi_{\ast}(X) \otimes \mathbb{Q}$ is a function of $C^{\ast}(\OmegaX, \mathbb{Q})$ (by basically taking the primitive elements).
And $\Omega X$ is the fiber of $pt \to X$
This means that our left class identifies all rationaly contractible spaces. But it might not identify all rationally equivalent spaces...
 
(Here's the argument I had in mind. Let $\mathcal E$ be the class of maps of spaces whose fiber is at least $n$-connected with trivial rational homology, where $n\geq 1$ is fixed: I claim that $\mathcal E$ is the left half of a modality. It's clearly closed under base change and we agree it's closed under composition.
 
And with that i'll go to sleep now hoping I didn't say terrible nonsense...
 
12:40 AM
To see that it's closed under cobase change, by taking fibers we can reduce to the case where the pushout of $B \leftarrow A \to C$ is contractible, where $A \to B \in \mathcal E$. Since maps with 1-connected fiber are closed under cobase change, $C$ is 1-connected, so it suffices to show that $C$ has trivial rational homology.
But in this case there is an equivalence $\Sigma A = \Sigma B \vee \Sigma C$, so this follows from the fact that $A \to B$ must be a rational homology equivalence. For colimits in the arrow category, maybe some more argument is needed...)
Ok, for colimits in the arrow category, we can break it down into coproducts (obvious) and contractible colimits. For the latter, it suffices to show that the class of n-connected spaces with trivial rational homology is closed under contractible colimits. This follows because n-connected spaces and spaces with trivial rational homology are each closed under contractible colimits.
Hmm... is it actually the case that a map with 1-connected rationally trivial fiber is a rational homology equivalence?
Yeah, the coefficients in the Serre spectral sequence can't really get twisted in this case. Right?
 
I would like to point out that my argument for the rationally contractible case assumes Quillen rational homotopy works for which I would need some kind of nilpotence/connectedness assumption...
 
1:10 AM
Rational trivial fiber implies equivalence on rational homotopy groups via the long exact sequence on homotopy right?
And now i'm really going to sleep before making anymore outlandish statements.
 
1:34 AM
I think this generalizes -- and the connectivity is actually a distraction. If you have any Bousfield localization $L$ of spectra, then the class $\mathcal E$ of maps of spaces with $L$-acyclic fibers is the left half of a modality. You can also intersect $\mathcal E$ with the left half of any other modality $\mathcal E'$ to get another modality. At least, I think I've checked everything up to accessibility issues.
 
 
6 hours later…
7:35 AM
@TimCampion That's a bit surprising. Did you use anywhere that it is a bousfield localization instead of some abitrary localization?
 
 
8 hours later…
3:42 PM
@SaalHardali Most of the argument is the same. For closure under composition, it suffices to show that if $A,B$ are $L$-acyclic, and $A \to C \to B$ is a fiber sequence, then $C$ is $L$-acyclic. But from the fiber sequqence we get $C = \varinjlim_{b \in B} A$, so $C \to B$ is a colimit of $A \to 1$'s, so it's an $L$-equivalence. Since $L$-equivalences are stable under composition $C \to B \to 1$ is also an $L$-equivalence as desired.
For closure under cobase change, again reduce to the case where the pushout of $B \leftarrow A \to C$ is contractible. Use the fact that in this case $\Sigma A = \Sigma B \vee \Sigma C$ and (crucially!) $L$-equivalences are preserved and reflected by $\Sigma$. So this part uses that it's a stable localization, and not just an arbitrary localization of spaces.
 
 
1 hour later…
5:01 PM
@TimCampion How do you reduce to the case of contractible pushout?
 
5:11 PM
@SaalHardali Consider a span $B \leftarrow A \to C$ with $A \to B \in \mathcal E$. We want to show that $C \to B \cup_A C \in \mathcal E$. By definition, this means that its fiber $C' \to 1$ is in $\mathcal E$. So we take the fiber. Since colimits are basechange-stable (since Spaces is locally cartesian closed), the whole pushout square in fact base changes to a pushout square $B' \leftarrow A' \to C'$ whose pushout is contractible.
And by base change-stability we have that $A' \to B' \in \mathcal E$. So we've reduced to considering this new pushout square with contractible pushout.
At least I think this works -- I still find it surprising! It's what inspired this question the other day.
 
@TimCampion Oh yeah that actually works. I'm more surprised by the fact that this gives a modality. I guess the counter-intuitive thing here is that there are much more $L$-equivalences than maps with $L$-acyclic fibers.
Even for rationalization this can't be fixed because there's always $pt \to BG$ for $G$ finite group.
So I think the only modality of this form for which the L-equivalences are the same as the left class is the acylic modality.
 
5:43 PM
Let $f(n)$ be the maximal integer s.t. $\pi_{\ast}(\mathbb{S}) \to \pi_{\ast}(L_{E(n)} \mathbb{S})$ is an equivalence in degrees $[0,f(n)]$. Is there a (non-stupid) lower bound for $f(n)$? Perhaps more easy is the same sort of question for $Ext^{\ast, \ast}_{BP_{\ast} BP}(BP_{\ast}) \to Ext^{\ast, \ast}_{E(n)_{\ast} E(n)}(E(n))$.
I have a feeling that at least for the second question there must be an answer somewhere in one of ravanel's books.
 
 
4 hours later…
10:11 PM
Turns out this construction of modalities generalizes much further: Let $\mathcal C$ be any locally cartesian closed presentable $\infty$-category and let $(\mathcal E, \mathcal M)$ be any accessible factorization system on $\mathcal C$. Let $\mathcal E'$ be the class of maps $f \in \mathcal C^{[1]}$ such that every base change of $f$ is in $\mathcal E$. Then $\mathcal E'$ is the left half of an accessible modality on $\mathcal C$. The proof in this formulation is straightforward.
When $\mathcal C$ is spaces, $\mathcal E'$ consists of those maps whose fibers are in $\mathcal E$: one containment is obvious while the other uses the fact that for $ A \to B$ we have $A = \varinjlim_{b \in B} \{b\} \times_B A$. So this recovers the modalities we were talking about.
But it also means that if $L$ is any unstable localization (i.e. a localization of spaces), then there's a modality whose left class consists of maps whose fibers are all $L$-acyclic.
 
 
1 hour later…
11:26 PM
Let $C$ be a presentable $\infty$-category and let $S$ be a small set of morphisms in $C$. Let $S^+$ be the saturated class of morphisms generated by $S$ (i.e. the left-class-of-a-factorization-system generated by $S$). Let $\bar S$ be the strongly saturated class of morphisms generated by $S$ (corresponding to the local morphisms when you localize at $S$). Then $S^+ \subseteq \bar S$.
But it's not clear to me that $\bar S$ admits a small generating set as a saturated class (although of course it's "small-generated as a strongly saturated class") -- this seems to be used implicitly in the first line of the proof of HTT 5.5.4.15. I think I can see that $\bar S$ consists precisely of those morphisms $f$ such that there exists $g \in S^+$ such that $gf \in S^+$. But I don't see how to get a generating set out of this.
Note that strongly saturated = saturated + that last case of 2/3.
 

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