Homotopy Theory

Tim Campion

02:39

This seems like an appropriate time to bring up one of my favorite theorems to cite: Freyd's theorem that there is no faithful functor from the homotopy category to $Set$. This implies that there is no faithful functor from the homotopy category to $Grp$ or to any other locally presentable category, or even to any category with a generator. For example there's no faithful functor from the homotopy category to $Top$.

Jonathan Beardsley

How unpleasant.

Tim Campion

As for faithful functors on $Top$, it seems kind of hopeless to get a faithful functor from $Top$ itself to any algebraic category. But if you work with the category $Top_\Delta$ of delta-generated spaces, then $Top_\Delta$ embeds fully faithfully into the category of $M$-sets, where $M$ is the monoid of continuous maps $[0,1] \to [0,1]$. This also works with self-maps of $\mathbb R$.

Actually, according to Trnkova, under set-theoretical assumptions there exist certain presheaf categories $C$ with the property that every category (locally small, with at most universe-many objects, I assume...) admits a fully faithful functor into $C$. I find this really surprising.

In particular, under set-theoretical assumptions, there is a fully faithful functor from $Top$ to a presheaf category.

Wait -- this would also apply to the homotopy category. There must be a wire crossed somewhere...

Oh nevermind

The theorem is that every concrete category embeds into $C$

There is also a category $D$ such that every category embeds into $D$, but $D$ is not a presheaf category apparently.

Of course, presumably these functors are completely useless for actually distinguishing nonhomeomorphic spaces for example.

Also when I said $M$-sets before, I should have said $M^{op}$-sets.

@JonathanBeardsley It sounds like you resolved your question, and I think I agree with the resolution.

Jonathan Beardsley

02:59

@TimCampion yeah, oddly this was actually sort of answered in a question you answered a while back

Tim Campion

cool

Jonathan Beardsley

I think the main issue is that Lurie never explicitly says "here's the model structure on $sSet_+$ I'm using," so one kind of has to guess about what he means by "covariant model structure on $(sSet^+)^C$" and it seems like this should, a priori, vary depending on whether or not one wants to deal with Cartesian fibrations or coCartesian fibrations.

However if one does the obvious thing, it seems not to matter.

As I mentioned above. Because over $\ast$, the Cartesian and coCartesian fibrations are precisely the quasicategories (with the "natural" markings).

So really all that $(sSet^+)^C$ is modeling is "a diagram of quasicategories," which is good, and clarifies why it's showing up in the Grothendieck construction.

Jonathan Beardsley

04:52

Hey @TimCampion, it seems like the straightening functor (over the nerve of a discrete category) $\mathfrak{F}_\bullet(C):sSet^+_{/N(C)}\to (sSet^+)^{C}$ should preserve fibrant objects, does that seem true or untrue to you?

Ah sorry... maybe that's not necessarily true UNLESS the the thing in the domain, $f:X^\natural\to N(C)$ is such that $X^\natural$ is the natural marking for $X$ a quasicategory?

It seems like a fibrant object in the domain is something of the form $X^\natural\to N(C)^\sharp$ where $X\to N(C)$ is a coCartesian fibration, and something fibrant in the target is a functor that takes values in quasicategories with the natural marking. So as long as the the object $X^\natural\to N(C)^\sharp$ arises from a coCartesian fibration of quasicategories, we're good to go.

Jonathan Beardsley

05:18

Hm... but that doesn't seem quite right, because (presumably) $\mathfrak{F}_\bullet(C)(c)$ produces the fiber of $X^\natural$ over $c$... which should always be a quasicategory I think...?

Jonathan Beardsley

05:38

Jeez what am I missing here? A cocartesian fibration has quasicategories as fibers, and the straightening functor produces a functor which takes $c\in C$ to $fib(c)$, which is a quasicategory. So it should induce a fibrant object of $(sSet^+)^C$...?

Harry Gindi

@JonathanBeardsley I think what you're missing is that the straightening of a marked cartesian fibration over a point is not isomorphic to the fibre

Consider the trivial example of the naturally marked n-simplex over a point (i.e. no nontrivial markings). Its image under straightening is not isomorphic to Δ[n] viewed as a simplicial category

Jonathan Beardsley

@HarryGindi Aha, so it doesn't need to actually be a quasicategory.

Harry Gindi

yeah

Jonathan Beardsley

Well shucks.

Harry Gindi

Lurie says something like

"straightening is not a simplicial functor, but unstraightening is"

Jonathan Beardsley

05:49

Alright. Well that's okay I guess. Thought I was going to be able to skate by without having to take a fibrant replacement. Rarely am I so lucky.

Harry Gindi

@JonathanBeardsley In the unmarked case, Lurie defines a cosimplicial object that gives rise to a nerve-realization pair defined by straightening and unstraightening over a point, and it's pretty obvious that the left adjoint doesn't send Kan complexes to Kan complexes

You could probably compute by hand what it looks like for the marked version, some more elaborate nerve realization pairing, but the moral of the story is that the left adjoint is pretty much uncontrolled

section 2.2.2

Jonathan Beardsley

Yeah fair enough. I mean, I'm far enough away from being able to actually check any of this that I'll take your word for it. I was hoping it would work out for formal reasons, but as you say, there's no reason to believe the value of the straightened fibration on an object will actually be a quasicategory.

Harry Gindi

It's actually good enough to show that the unmarked version doesn't send Kan complexes to Kan complexes, since any quasicategory that is categorically equivalent to a Kan complex is a Kan complex

and then it's just a matter of computing the realization of some very simple Kan complex like BZ and showing by hand that it's not a Kan complex

that's how I'd do it

Jonathan Beardsley

06:08

@HarryGindi is there any chance that the relative nerve construction avoids this issue? i.e. when we're working over a discrete category?

Hm probably not actually... I had an idea, but now I see that it won't work.

Harry Gindi

No idea; I know of very few left Quillen functors that also preserve fibrant objects. The only ones I can think of offhand are the embeddings of (∞,n)-categories into (∞,n+1)-categories in several models, but those arise from right-adjoint fully faithful embeddings of Reedy categories

and the anodynes in those models are generated by subsets of the anodynes in the bigger category, so it's very tightly controlled

the relative nerve does look a lot nicer though! If it is true that the left adjoint to the relative nerve preserves fibrant objects, it should be almost immediately obvious

Jonathan Beardsley

@HarryGindi Well, actually my question doesn't even make sense because the relative nerve plays the role of the unstraightening functor, not straightening, lol

Or, perhaps to put it more explicitly, the relative nerve definitely preserves fibrant objects, because it's the right adjoint... =\

Harry Gindi

Yeah, I was talking about its left adjoint

Lurie calls it like \frak{F}_\cdot(C)

Jonathan Beardsley

Yeah.

user1732

random comment, are you enjoying this heat wave we're in?

Harry Gindi

06:23

Where?

Jonathan Beardsley

@HarryGindi oh i see, yeah you said that.

user1732

@HarryGindi pacific northwest

Harry Gindi

ah, I don't think we're having much of a heat wave in south Germany

This is the first summer of my life without air conditioning, and I have to say it's unpleasant

user1732

wow!

Harry Gindi

but it doesn't seem to get nearly as hot here as it does in, say, New Jersey

or even michigan

@JonathanBeardsley By the way, what is the application you were hoping for?

There is a very nice Quillen equivalence between Set^+/* and (Set^+)^* called the identity functor

or the obvious equivalence of categories if you want

so I'm guessing that you care about a more general base?

@JonathanBeardsley Actually, it says right here that the relative nerve over a point is the identity, right at the top of the section on the relative nerve

" For every vertex s ∈ S, the fiber X s is an ∞-category
which is equivalent to F(φ(s)) but usually not isomorphic to F(φ(s)). In the
special case where C is an ordinary category and φ : C[N(C) op ] → C op is the
counit map, there is another version of unstraightening construction Un +
φ which does not share this defect."

so the fibres are isomorphic, i.e. over a point it works

It's in the introduction to 3.2.5

I'm not sure if the left adjoint to the relative nerve sends cartesian fibrations to projectively fibrant diagrams, but it does at least send them to injectively fibrant ones

so there's a satisfying answer to this question you've asked!

Jonathan Beardsley

06:49

@HarryGindi trying to show that the operadic nerve commutes with "op" in a suitable setting

@HarryGindi hm so I'm not sure I'm following. The goal, for me, was to show that straightening preserved fibrant objects, where fibrant in the target meant "in the projective model structure" but I can't tell if you're saying that's true or not

Harry Gindi

I don't think it is true.

even with the relative nerve

It only gives you something where the left adjoint to the relative nerve sends a cartesian fibration to a diagram whose value at an object in the base is isomorphic to the fibre over that vertex in the base

Jonathan Beardsley

Yeah ok. I agree.

Harry Gindi

wait, have I screwed something up?

Jonathan Beardsley

¯_(ツ)_/¯

Harry Gindi

projectively fibrant is pointwise fibrant

Jonathan Beardsley

06:59

Haha I'm sort of hanging on by my fingernails here.

Yes that's right.

Harry Gindi

you're good to go, lad!

user1732

\o @DenisNardin

Jonathan Beardsley

Hmmmm... Ok I'll have to think about that. I saw the bit from the intro you mentioned before actually... But this didn't seem to be telling me anything about straightening, only unatraightening

Ugh in case you can't tell I'm now on my phone, so can no longer type.

Harry Gindi

the fraktur F is the left adjoint?

Jonathan Beardsley

Yes

Harry Gindi

07:02

I'm on my phone because I can't sit in a chair for more than 2 hours

my shoulders are all messed up lol

Jonathan Beardsley

Well you're doing quite well for typing on a phone.

I probably shouldn't sit in a chair for that long, but I do anyway. My body is crumbling.

Harry Gindi

I'm going to spend the summer swimming, I hear it's good for shoulders and upper back haha

user1732

don't you lift weights any more Professor?

Harry Gindi

but if I am reading that intro part correctly, it says start with a fibration X/S

apply F

Jonathan Beardsley

@user1732 hi you're giving me the creeps please stop

Harry Gindi

07:06

then when you evaluate the phi you get from this

it is isomorphic to the fibre

which would imply that F sends cartesian fibrations to projectively fibrant diagrams

I can't find the statement in the chapter though

Jonathan Beardsley

Hm. No I don't think that's it. Are you maybe confusing the fraktur F with the script F?

Harry Gindi

ah, that's the problem

Jonathan Beardsley

To me it seems to say that you start with a functor, apply the right adjoint, and examine the fibers of the fibration you get. They're isomorphic to the values of the functor you started with

Harry Gindi

Yeah, that is what it says

Jonathan Beardsley

But this doesn't actually tell me much about frak(F) I don't think

Harry Gindi

07:10

Yeah, I guess not =[

Jonathan Beardsley

Anyway I think it's okay. I have large, hideous diagram that I want to commute but I'm okay if it only commutes on "underlying infinite categories" and I think that means I can replace functors with derived functors.

Or, like, it's okay to fibrantly replace crap.

I'm not particularly well versed in the yoga of ∞-categories.

Harry Gindi

Yeah, someone told me he thought you could use the yoga of infinity cats to avoid some horrible combinatorial thing, but I can't see how

(Moerdijk's calculation of the homotopy-monoidal structure on dendroidal sets, specifically)

Jonathan Beardsley

One thing in particular is that I want control over the unit and counit of straightening and unstraightening and I guess I only have it "up to homotopy" because they're not actual equivalences.

Harry Gindi

Whatever it is, if you can get away with the relative nerve, it's much better

Jonathan Beardsley

I.e. c-->Un(St(c)) is not an equivalence but c-->Un(St(c))-->Un(St(c)^f) is

Harry Gindi

07:16

yep

Jonathan Beardsley

Well yeah I'm using the relative nerve because I only am working over a discrete category in every case I'm interested in.

Harry Gindi

I'm still working on generalizing straightening. One neat thing I realized a while ago is that when you are taking the cone over X, you're really taking a kind of lax cone

it's all sort of hidden in the combinatorics and construction, but it makes it clear what you should be doing in infinity,2 categories

(Lurie's scaled straightening isn't analogous to straightening fibrations but to straightening groupoidal fibrations)

anyway, seeya later!

user1732

cya

Girish

09:15

If X is reedy fibrant cosimplicial diagram of simplicial sets, is it injective fibrant ?

Harry Gindi

13:32

@Girish Don't think so

that sounds backwards

Jonathan Beardsley

Yeah that's not true I don't think, see here ncatlab.org/nlab/show/…

Reedy fibrant implies projectively fibrant, and reedy cofibrantly implies injectively cofibrant

Harry Gindi

Reedy fibrant simplicial diagrams are "injective resolutions" and Reedy cofibrant cosimplicial diagrams are "projective resolutions"

yeah

Jonathan, nope

Jonathan Beardsley

Anyone know if Lurie is taking "functorial fibrant replacements" as part of his definitions of model structures in HTT? Seems like the answer is no.

Harry Gindi

It depends on the variance of the Reedy category

Tim Campion

@JonathanBeardsley Def A.2.1.1 doesn't ask for them. But I don't think he ever considers a non-combinatorial model category. So he always has functorial factorizations by the small object argument

Harry Gindi

13:38

@JonathanBeardsley He doesn't, but he constructs them by SOO because he only works with LocPres so it doesn't matter

yeah exactly

Tim Campion

:)

Jonathan Beardsley

Oh ok nice. That makes my life easier.

Tim Campion

Does anybody have a table of contents for HTT that includes all the subsections? I really should have known by this point that there are bits about straightening / unstraightening in both chapter 2 and chapter 3, though you'd never guess it from the section titles.

Jonathan Beardsley

@TimCampion ugh I wish

Or with hyperlinks

Tim Campion

Isn't there a hyperlinked version?

Harry Gindi

13:42

There is, but irs TOC is incomplete

@TimCampion I just constructed straightening/unstraightening for cellular sets

Tim Campion

sweet

Harry Gindi

I can show that unstraightening a diagram of omega-qcats is a cartesian fibration, up to a hard combinatorial calculation of lax cones

No decomposition, so a bit stuck rn

I e-mailed Dimitri Ara for help

Jonathan Beardsley

It seems like, again according to the nlab, there is some issue with functorial fibrant replacement being simplicially enriched... Unless our category has all objects cofibrant?

Where the hell is all this stuff written down?

Harry Gindi

Appendix 3 of higher topos theory

it's the most horribly technical section of the book except for straightening/unstraightening

Another reference IIRC is Clark's paper "on enriched left Bousfield localization"

IIRC he gives a bunch of cleaner proofs of the stuff in Lurie Appendix 3

and more general statements

I last read it in 2010 though so I'd have to check if that is the right paper

Jonathan Beardsley

14:05

Hrm... Do you mean the stuff about chunks? In HTT?

Ah actually Riehl's book, Categorical Homotopy Theory, Section 13, does this really well!

Girish

14:21

@Har

@HarryGindi @JonathanBeardsley Thank you. Is there any other way to find homotopy limit of cosimplicial simplicial sets other than taking injective fibrant replacement. I have Reedy fibrant cosimplicial simplicial set of which I am taking homotopy limit.

Greebo

14:50

do homology count as close enough to homotopy theory that I can ask questions here? Could not find an active chat room for homology

(specifically locally finite / Borel-Moore homology)

user2236

Askaway.

Jonathan Beardsley

@Greebo Anyone can basically ask a question about anything. There's just no guarantee that anyone will answer it!

@Girish I don't know off the top of my head, unfortunately.

Tim Campion

As a terminological point, "homotopy theory" doesn't mean "the study of homotopy groups" -- it means something more like "the study of homotopy-invariant structures". cf. the MO tag description. In practice, that encompases all of algebraic topology, as well higher category theory, and parts of algebraic geometry / number theory.

and other things, like parts of symplectic geometry and other differential geometry areas, parts of field theory, etc.

Greebo

ok testing latex first $f\colon \mathbb{R} \to X$ this should work right?

nice

I am looking at the long exact sequence $\ldots \to H^\infty_n(Z) \to H^\infty_n(X) \to H^\infty_n(U) \to \ldots$ for Borel-Moore/locally finite homology ($Z$ closed, $U = X\setminus Z$), and as I feel more comfortable with the locally finite variant, I wanted to find a proof going through locally finite singular chains.

I know a short exact sequence $0 \to C^{BM}_*(Z) \to C^{BM}_*(X) \to C^{BM}_*(U)\to 0$ exist for the Borel-Moore version, which gives the long exact sequence, but I have problems proving the existence of a similar short exact sequence for locally fininte simplicial chains.

user2236

@TimCampion is there any relation to Langlands project?

Greebo

15:04

($X$ locally compact also a condition btw)

ofc the short exact sequence might not exist for lf-homology, but information about lf-homology seems scattered overall, so I wondered if anyone knew anything about it

Tim Campion

@user2336 I don't know much about it, but people like Gaitsgory definitely use higher-categorical language in their approach to Langlands stuff (geometric Langlands, I think?) and to me that qualifies as homotopy theory.

Greebo

mostly what I've learned about lf-homology has been from ppl saying "it is well known that" in comments on stackexchange :P

(technically BM-homology isn't homotopy invariant, so even less relevant to the chat name)

Eric Peterson

is borel-moore the one you get by taking relative homology of the one-point compactification against that new point?

Greebo

basically

the actual definition is something to do with cosheaves or something

(haven't had much luck finding a proof that it's equivalent to taking one-point-compactification either, for that matter)

ncatlab tells me it's "based on results in On the Steenrod homology theory" by Milnor

which I haven't had time to read yet

but for the locally finite variant you just look at standard chains except you allow them to be formally infinite sums instead of finite sums

but they have to be locally finite instead

problem with the short exact sequence is I'm guessing the map $C^{lf}(Z) \to C^{lf}(X)$ should be inclusion, but have problems with what the map $C^{lf}(X)\to C^{lf}(U)$ should look like. Chains that are locally finite in $U$ might not be locally finite in $X$, creating problems.

Tim Campion

Presumably that map only exists if $U$ is an open subspace of $X$, in which case the problem disappears, right?

Greebo

15:20

$U$ is open, not sure why that makes the problem disappear?

I need $j\colon C^{lf}(X) \to C^{lf}(U)$ to be surjective, I have problems hitting all locally finite chains in $U$ from $X$

my earlier best bet for $j$ was barycentrically splitting simplexes that partly hit $U$ an infinite amount of times, removing the pieces that ended up in $Z$ and just keeping the pieces that ended up wholly in $U$ at each step

which I think gives me a locally finite chain, and is a chain map, not 100% sure

but I don't think it's surjective

Tim Campion

I see, I thought you were struggling with constructing the map. I agree now that you probably don't get surjectivity on the nose.

Greebo

e.g. if $X$ is $\mathbb{R}$ and $Z = (-\infty, 0]$, the chain $\sum_{n=1}^\infty n \cdot [\frac{1}{2^n}, \frac{1}{2^{n-1}}]$ is locally finite in $U$ but not in $X$. And as far as I can see will not be hit by my constructed $j$

Denis Nardin