Homotopy Theory

3:31 AM

@TimCampion I think because the lim^1 term in the milnor exact sequence automatically vanishes in this case (since any sequence of finite dimensional vector spaces has vanishing lim^1)

Jonathan Beardsley

9:49 AM

@AlexanderCampbell ah okay, yeah that makes sense. thanks for pointing that out.

I suppose... so here's the thing I'm trying to prove: suppose I have a Cartesian fibration $p:E\to B$. Then I can take the fiberwise arrow categories as follows: consider the inverse Grothendieck construction $I(p):B^{op}\to Cat_\infty$, then compose this with $Fun(\Delta^1,-):Cat_\infty\to Cat_\infty$, then take the Grothendieck construction of $Fun(\Delta^1,-)\circ I(p):B^{op}\to Cat_\infty$. Note that there is a natural transformation $ev_1:Fun(\Delta^1,-)\to Fun(\Delta^0,-)$ which gives...

...a map of Cartesian fibrations $Gr(Fun(\Delta^1,-)\circ I(p))\to p$, which corresponds to, I think, evaluating at 1 on fibers. I think that this is a Cartesian fibration... but I'm not exactly sure how to prove it.

I suppose $p^{\Delta^1}$ is probably a decent notation for $Gr(Fun(\Delta^1,-)\circ I(p))$

Alexander Campbell

10:06 AM

@JonathanBeardsley I think the result of your Grothendieck construction ought to be equivalent to the pullback of Fun(Delta[1],p) : Fun(Delta[1],E) --> Fun(Delta[1],B) along the "diagonal" B --> Fun(Delta[1],B).

Jonathan Beardsley

In fact I'm like 99% sure I know how to check this in the discrete case, for 1-categories, i.e. I know what the Cartesian lifts are.

Ah, so is that first fibration Fun(Delta[1],p) still Cartesian?

Alexander Campbell

That's right. (If you like, Fun(Delta[1],-) is a cosmological endofunctor of the infinity-cosmos of quasi-categories, and cosmological functors preserve cartesian fibrations.)

Jonathan Beardsley

Ah nice, okay.

So because Cartesian fibrations are preserved under pullback, that'd give the result.

(In case anyone else is reading this...)

Alexander Campbell

Well, that tells us that the pullback I described is a cartesian fibration. It might give your result too, but I would have to draw some diagrams to see that.

Jonathan Beardsley

Oh sorry. Wait right. That's just giving me the Cartesian fibration over B.

That's not telling me that the induced $ev_1$ functor of Cartesian fibrations is itself a Cartesian fibration.

Alexander Campbell

10:19 AM

Note that the pullback I described is the cotensor by Delta[1] of p : E --> B in the infinity-cosmos of cartesian fibrations over B. Your functor of interest is given by cotensoring p with the morphism {1} : Delta[0] --> Delta[1].

That seems like it should help, but I'm not sure how to continue the argument at the moment.

Jonathan Beardsley

No worries! Thanks. :)

Ah I mean one thing that's clear is that all of the fibers need to admit pullbacks for this to be true. I forgot to mention that.

Alexander Campbell

Right, of course. I reckon there's probably an infinity-cosmos way to prove this. Probably it's true in any infinity-cosmos that for an infinity-category C with pullbacks, the codomain projection C^{Delta[1]} --> C is a cartesian fibration. Your desired result might be the specialisation of this result to the infinity-cosmos of cartesian fibrations over B. But I'm not sure.

Jonathan Beardsley

Yeah I'm trying to prove this this all sort of by hand, but I'm having trouble even showing it's an inner fibration. So maybe that's not the right way to do it.

Alexander Campbell

10:42 AM

Does your B have pullbacks?

Jonathan Beardsley

Yes.

(I'm actually just doing all of this for the tangent category T(Spaces)-->Spaces)

Alexander Campbell

Is T(Spaces) = parametrised spectra?

Jonathan Beardsley

Sorry, that's right. The fiber over X is Fun(X,Spectra)

(equivalently take the codomain fibration and tensor with Spectra fiberwise)

Alexander Campbell

Cool.

Jonathan Beardsley

11:18 AM

Not sure I'm understanding definitions here... but is {1}:Delta[0]-->Delta[1] inner anodyne?

Alexander Campbell

No, since inner anodyne maps are bijective on objects.

But cotensoring an infinity-category by any mono yields an isofibration.

Jonathan Beardsley

11:58 AM

Ah, oh I see, right, since inner anodyne maps are always categorical equivalences, is that right?

Lol I swear to god if I ever finish this project I'm never touching a quasicategory ever again...

But that's a big "if." This dorky stuff might just be what I try to figure out for the rest of my life, lol.

Matt Feller

2:06 PM

There are two ways to see why {1}:Delta[0]-->Delta[1] is not inner anodyne: inner anodyne maps are (a) bijective on 0-simplices and (b) acyclic cofibrations in the Joyal model structure, i.e., monic weak categorical equivalences. One has to be a bit careful with these things, because it turns out there are maps satisfying (a) and (b) which are not inner anodyne. Alexander has a short paper which explains a simple counterexample: https://arxiv.org/abs/1904.04965

But yeah, the map {1}:Delta[0]-->Delta[1] satisfies neither (a) nor (b), so is not an inner anodyne map.

Jonathan Beardsley

2:29 PM

@MattFeller thanks!

I guess the thing also is that inner anodyne maps are generated by inner horn inclusions under... pushouts? and something else (I can't remember). and maybe it's kind of clear that that inclusion can't be built from inner horns.

Dylan Wilson

2:46 PM

@JonathanBeardsley if you have any natural transformation of functors B^{op}—>Cat_infty which is pointwise a Cartesian fibration and such that the naturality square for a map in B is a map of Cartesian fibrations, then the induced map on Gr will be a Cartesian fibration.

This follows from HTT.2.4.2.11 + checking the condition I gave implies that the locally Cartesian arrows are closed under composition

So in your case you should be okay because ‘restriction’ functors Sp^X —>Sp^Y preserve pullbacks

(Of course, they preserve all limits and colimits, but pullbacks should be enough)

Jonathan Beardsley

i see, so you basically need to check that E'(b)-->E(b) is a Cartesian fibration for each b\in B, and then check that the morphisms E'(b)-->E'(c) over E(b)-->E(c) preserves Cartesian fibrations.

er,preserves cartesian lifts

Dylan Wilson

**morphisms

Yeah

Jonathan Beardsley

Yes, right.

well that's pretty nice.

Dylan Wilson

:)

Jonathan Beardsley

3:04 PM

@DylanWilson sorry just checking back on this. the point here being that the cartesian morphisms are precisely given by pullbacks?

(at least fiberwise, since they're just the arrow categories)

Matt Feller

3:25 PM

@JonathanBeardsley Yeah, the class of inner anodyne maps is the closure of the set of inner horn inclusions under pushouts, transfinite composition, and retracts. I think it's intuitive that you couldn't get {1}:Delta[0]-->Delta[1] through those operations, but probably the easiest way to make that intuition precise is the bijective-on-0-simplices argument: all inner horn inclusions are bijective on 0-simplices, and that property is stable under the generating operations.

Jonathan Beardsley

@MattFeller yeah that's a very good fact to know, that they're bijective on 0-simplices

asdq

3:37 PM

Is the mapping space between two \pi-finite \infty-groupoids (i.e. truncated \infty-groupoids with finite homotopy groups resp. connected component sets) again a \pi-finite \infty-groupoid?

Dylan Wilson

3:56 PM

@JonathanBeardsley right. Let's say you've got $F: B^{op}\to\mathsf{Cat}_{\infty}$. as you say, an arrow in $F(b)^{[1]}$ is ev_1-cartesian exactly when it's a pullback square. Given $f: b'\to b$, you want $F(b)^{[1]} \to F(b')^{[1]}$ to preserve cartesian morphisms, which amounts to asking that $f^*: F(b)\to F(b')$ preserves pullbacks.

Jonathan Beardsley

Right. And then the fact that composability of locally Cartesian morphisms gives you a Cartesian fibration is basically Remark 2.4.2.9 of HTT

I mean, again I sort of "know" why that gives you a Cartesian fibration, but it's nice to have a concrete reference for the \infty stuff.

Dylan Wilson

definitely- I would probably cite Proposition 2.4.2.8 though, the equivalence between (1) and (2)

Jonathan Beardsley

Oh and the requirement that $ev_1$ be an inner fibration follows from something Denis showed me earlier where you just lift along inner horns.

Dylan Wilson

in fact it's always a cocartesian fibration (as long as B is an $\infty$-category and not some random simplicial set)

Jonathan Beardsley

Oh, why is that?? That's useful.

Dylan Wilson

4:07 PM

you can prove that by hand or else use the overkill 2.4.7.12 when f is the identity map

intuitively the pushforward is given by composition: $\mathrm{ev}_1: \mathcal{C}^{[1]}\to \mathcal{C}$ is classified by $c \mapsto \mathcal{C}_{/c}$ and given $c\to c'$ a choice of composition gives $\mathcal{C}_{/c} \to \mathcal{C}_{/c'}$

Jonathan Beardsley

Er, wait, so 2.4.7.12 is showing something is a Cartesian fibration. Is it obviously the opposite of the thing we're looking at?

Dylan Wilson

yep

notice that, while the fiber product involves ev_1, the projection involves ev_0. Dualizing reverses those choices, and we had f=id, so we get the right thing

Jonathan Beardsley

Oh... okay yeah I missed that {0} there... I was thinking we were just getting the codomain fibration again.

Dylan Wilson

incidentally, 2.4.7.12 and its dual are really useful. this construction is actually the free (co)cartesian fibration on the arrow f, in the sense that it gives an explicit left adjoint to the inclusion of the (non-full) subcategory (co)Cart(D)-->Cat_{/D}

Jonathan Beardsley

right, this is the thing from Gepner-Haugseng-Nikolaus right?

I was just banging my head against that today too...

Dylan Wilson

4:14 PM

yeah that does show up in that paper, good call!

maybe it's easier to think about the classifying map: given an arrow f: C \to D, the free cocartesian fibration is classified by the functor $d \mapsto C\vert_{D_{/d}}$. In other words, we don't have functoriality on literal fibers unless we're special (i.e. already cocartesian), but we can use the functoriality of $D_{/(-)}$ and pullback to say we always have functoriality on these sort of 'lax fibers'

Jonathan Beardsley

Sorry, I'm really not seeing how the fibration in 2.4.7.12, with f the identity arrow (on B?) is the same as the fiberwise evaluation E'-->E of the fiberwise arrow category.

Dylan Wilson

sorry, perhaps I didn't explain clearly: I was using 2.4.7.12 to explain why $\mathrm{ev}_1: F(b)^{[1]}\to F(b)$ is a cocartesian fibration

(always) and hence in particular inner, which I thought you were worried about. Then, assuming the existence of pullbacks, one can also show it's cartesian

Jonathan Beardsley

Yeah perhaps I'm confused about some things here actually. So... your use of 2.4.2.11 is to show that what exactly is locally Cartesian?

Dylan Wilson

Let's start over. Here is my claim: suppose $F, G: B^{op} \to \mathsf{Cat}_{\infty}$ are functors, where $B$ is an $\infty$-category, and $F \to G$ is a natural transformation. Suppose that (i) $F(b) \to G(b)$ is a cartesian fibration for every $b$ and, (ii) given $b' \to b$, the map $F(b') \to F(b)$ takes morphisms which are cartesian over G(b') to morphisms which are cartesian over G(b). Then I claim that $Gr(F)\to Gr(G)$ is a cartesian fibration.

Jonathan Beardsley

Right. Yes. And I can see why this particular example satisfies those conditions.

Dylan Wilson

4:28 PM

ah, I see- you are wondering why Gr(F)-->Gr(G) is an inner fibration. that's fair. I guess I belong to the camp of people who replace Lurie's definition of (co)cartesian with the homotopy invariant one, in which case you drop the requirement that you have an inner fibration... because now you'd have choose some specific unstraightening functor or model of Gr(F)-->Gr(G) which happens to be an inner fibration, I see

that's fine, in this case I guess there is probably a concrete construction of the 'fiberwise arrow category'... (maybe y'all did this above already?)

Jonathan Beardsley

There is a concrete construction. And Denis showed me earlier how proving it's inner is basically checking a relatively simple diagram.

Dylan Wilson

I guess in general, given $E,E' \to B$ cartesian, you can always form a new cartesian fibration $\underline{\mathsf{Fun}}(E,E') \to B$ concretely like the parameterized folks do for cocartesian things

great

then you're all set

Jonathan Beardsley

Well, actually, so I think I was still trying to understand how exactly 2.4.2.11 is coming in here. Are you applying it to the triangle $E'\to E$ over $B$?

Dylan Wilson

yes

Jonathan Beardsley

Okay great, so once we know this thing is inner, we get that it's locally Cartesian, by checking it on fibers, as you said.

Wonderful.

Dylan Wilson

4:33 PM

(1) is ok, (2) is essentially the statement that ev_1: Fun([1], C)-->C is natural in C, and (3) is because of the existence of pullbacks; then we check locally cartesian arrows are closed under composition using the fact that the restriction maps preserve pullbacks

mhm

Jonathan Beardsley

I know those were two different messages but I like the idea of you ending all of your messages, no matter what, with "mhm."

Dylan Wilson

4:56 PM

Lol

Yuri Sulyma

6:01 PM

@AlexanderCampbell it is indeed true that the codomain projection is a bifibration in any ∞-cosmos with pullbacks, reference is Example 4.1.19 of Riehl-Verity IV (arxiv.org/pdf/1506.05500.pdf)

Tim Campion

@DylanWilson Let me make sure I understand. You're saying that $F(HF_p, HF_p) = F(\varinjlim HF_p^{(i)}, HF_p) = \varprojlim F(HF_p^{(i)}, HF_p)$, so that we have a short exact sequence $0 \to \varprojlim^1_i H^{\ast+1}(HF_p^{(i)}) \to \pi_\ast(F(HF_p,HF_p)) \to \varprojlim_i H^\ast(HF_p^{(i)}) \to 0$. We can choose the system $HF_p^{(i)}$ to have finite cohomology in each degree for each $i$, so the $\varprojlim^1$ term vanishes.

If we simultaneously choose the $HF_p^{(i)}$ to be finite, this means that every nonzero map $HF_p \to \Sigma^\ast HF_p$ restricts to something on some (finite) $HF_p^{(i)}$ and so is not phantom. Cool! One nice thing about this argument is that you can make it without actually computing the Steenrod algebra.

Charles Rezk

6:19 PM

To know that you can choose $HF_p^{(i)}$ to be finite, you need to know HF_p is finite type, which seems to me means you need a little computational control over it.

Maybe better to let $HF_p^{(i)}$ be a skeletal filtration for some CW-structure (which you can do because $HF_p$ is connective). Then for each $k$ the tower $\{H^k HF_p^{(i)}\}_i$ is eventually constant, so it is obvious that lim-1 vanishes.

Ted E

Hi. I have a vague question: What is the moral meaning of separatedness. I have read that it should be thought of as a topological notion (where a separated scheme isn't necessarily hausdorff). But separatedness is a relative notion, so you may not be separated over $\text{Spec}(\Bbb Z)$, but still be separated over some scheme $S$, so it doesn't seem like it really is purely topological. (A vague question typically generates a vague answer, which is fine)

Ted E

6:38 PM

(I mean the moral meaning of a scheme $X$ being $S$-separated, I don't mean separatedness abstractly or anything)

Tim Campion

7:01 PM

@CharlesRezk Ah, I see -- if we let $HF_p^{(i)}$ be the $i$-skeleton, then lim^1 vanishes for a different reason than in Dylan's argument. We still need to know that $HF_p^{(i)}$ is finite to complete the argument, though. What I was thinking is that one could just start with the Moore spectrum $HF_p^{(0)} = M(p)$, then glue in $M(p^k)$'s to kill $\pi_1$, and get $HF_p^{(1)}$, then glue in more $M(p^k)$'s to kill $\pi_2$, etc.

Inductively assuming that $HF_p^{(i)}$ has finite and $p$-torsion homology, then $\pi_{i+1} HF_p^{(i)}$ is finite and $p$-torsion (by a simple Serre class argument), so $HF_p^{(i+1)}$ is also finite and $p$-torsion. Then all $HF_p^{(i)}$'s are finite with finite $F_p$ homology and we win. This feels like the minimum of computational input to me.

@TedE I'm confused what you mean by "topological notion". Why shouldn't a "topological" notion be allowed to be relative? "Relative" just means that it depends on the data of a map $X \to S$, and we can talk about maps in topology. On the other hand, the strictest meaning of "topological notion" I can think of in AG is "notion which depends only on the underlying Zariski space of the schemes involved", and separatedness is not "topological" in this sense.

Ted E

@TimCampion Oh, I hadn't thought of it like that. I was thinking about 'topological notions' in that 'strictest' sense you mention, since the topology is independent of whichever structure morphism we wish to associate with the scheme. You're right though, any chosen structure map is continuous, so this is still topological.

Charles Rezk

8:16 PM

What do we know about maps between equivariant classifying spaces: e.g., $\mathrm{Map}_G(B_GH, B_GK)$?

Saal Hardali

9:32 PM

I'm looking for the precise version of the following statement which might or might not exist: "the infinity category of product preserving functors $\mathcal{P}_{\Sigma}(C^{\omega})$ is equivalent to the localization of the simplicial objects in $C$ by ?????".

I think the most you can get out of what's directly stated in HTT is an essentially surjective functor $sC \to \mathcal{P}_{\Sigma}(C^{\omega})$. Oh and I'm assuming a bunch of stuff about $C$ obviously. I rpefer to imagine $C$ is spaces for the moment and add the relevant clauses at the end.

*finite product preserving

Dylan Wilson

10:00 PM

@SaalHardali what’s C^{Omega}? If C is an ordinary category then what you want is 5.5.9.3

If C is simplicial then folklore says you use the Dwyer-Kan-Stover model structure

Saal Hardali

10:26 PM

My $C$ is an infinity category and $\omega$ means compact objects.

@DylanWilson I don't know a lot about thess model structures but from what I did manage to read I found it hard to pinpoint what the weak equivalences are in terms of the infinity category (rather than the simplicial model category presenting it).

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