Homotopy Theory

12:00 AM

@PiotrPstrągowski a map $A \to B$ is an epimorphism iff the square $(A \to B) \to (B \to B)$ is a pushout, and pushouts in a functor ($\infty$-)category are determined componentwise. so a componentwise epimorphism is an epimorphism.

Piotr Pstrągowski

@AaronMazel-Gee That's great, thank you! Any idea about the latter part..?

Aaron Mazel-Gee

12:12 AM

i'm not exactly sure how to prove it, but i certainly believe it -- it feels like in the end formula that marc referenced you should end up taking a limit of homotopy (-1)-types, which will again be a (-1)-type since these are reflective inside of Spaces

Tom Bachmann

12:34 PM

Why does what Aaron said not constitute a proof?

Piotr Pstrągowski

12:45 PM

@TomBachmann, It does give a very nice characterization of epimorphisms in the functor category, but the question is when does a natural transformation of functors factor through an epimorphism.

Tom Bachmann

yes but you just do what he said? You have f: X -> Y epi and some g: X -> Z. You want to know if g factors through Y. You know that Map(Y, Z) -> Map(X, Z) is an inclusion of components. Let T be the homotopy fibre of this map at the point corresponding to g (this is a certain homotopy pullback).

You know (because f is epi) that T is either empty or contractible depending on wether or not g factors. Now you use the end formula to write Map(X, Z) and Map(Y, Z) as certain homotopy limits and commute this with the homotopy pullback. Then, because each component is epi, you write T as a certain homotopy limit where each space is either empty or contractible.

This is going to be again empty or contractible, and it seems clear that it is empty if and only if there is one empty space in the diagram

(although I suspect there is something missing here, because otherwise Aaron probably would not have said "I'm not sure I can prove it" - hence my question ^^)

Piotr Pstrągowski

i see, you meant the end formula! It certainly feels like this should work, I'll take a closer look at Saul's paper and try to write this down in detail. Thanks everyone!

Riccardo

1:25 PM

Hi all! do you know any reference about existence of rational homology 4-sphere with fundamental group the Dihedral group?

Mike Miller

2:20 PM

@Riccardo Construct an arbitrary 4-manifold with such fundamental group, and your only problem is $H_2$. Kill off the image of the Hurewicz map by surgery, then all that's left is $H_2(K(D_n,1);\Bbb Z)$ which is finite as desired

Riccardo

4:11 PM

@MikeMiller Thanks for the answer. There is something I'm not able to understand, on the reference I'm reading (P. Teichner PhD thesis) it's said that it's even possible for a given fundamental group that there exists no rational homology 4-sphere. So an obstruction (only one?) would be that H_2(\pi_1M,Z) begin finite?

Mike Miller

Yup. It's possible iff the fundamental group itself has $b_1=b_2=0$. Obviously you can't fix $b_1$, and in general just do the above argument. The essential point is that $H_2(M)/\pi_2(M) = H_2(\pi_1 M)$.

Sean Tilson

@ArunDebray there seems to be a discrepancy on page 5 in terms of the notation for he different mapping objects.

Piotr Pstrągowski

4:35 PM

I have another question. If $C$ is a small $\infty$-category, let $P(C) = Fun(C^{op}, S)$ be its presheaf category and $y: C \rightarrow P(C)$ the Yoneda embedding.

I expect that for any $c \in C$ there should be an equivalence $P(C) _{/y(c)} \simeq P(C_{/c})$, where on the left I have the overcategory of presheaves and on the right I have presheaves on the overcategory. Is there any reference for this?

Piotr Pstrągowski

4:47 PM

I think there's a recognition principle for overcategories of $\infty$-topoi in Lurie's book (6.3.5.11), but perhaps it's not immediate for me how to use it. In any case, I'd be really happy just to reference the above fact, assuming it's true.

Dylan Wilson

5:07 PM

Well you have a functor C_{/c}---> P(C)_{/yc} which is fully faithful and it generates the target under colimits (bc you can write F as colimits of representables and then F--->yc is colimit of those representables and the composed map to yc... which is the same as a map to c). So what else could it be?

(Admittedly you'll have to go cite all the lemmas characterizing mapping spaces in terms of other mapping spaces, but that shouldn't be too bad)

Espen Nielsen

@DylanWilson That shows that the induced functor $P(C_{/c})\rightarrow P(C)_{/y(c)}$ is essentially surjective. You still need an argument for fully faithfulness, right?

Dylan Wilson

That should follow from fully faithfulness of the first functor I wrote down... maybe combined with the slightly stronger assertion that elements of the target are colimits of the canonical diagram of representables, instead of some arbitrary diagram?

We wait... hmmm

**er

maybe just wrote down the inverse then, and check that the unit and counit are equivalences by reducing to representables?

**write. iPhone's suck.

Huh... what is the inverse actually? Are we proving a true statement?

Oh duh nvm

Wait unduh. Now I'm confused. But also hungry. Lunch!

I guess the inverse is annoying to write down so maybe just check directly that the functor is fully faithful. We can reduce to the case where the domain is representable and then use the better form of the Yoneda lemma which tells you how to map out of representables :)

And you need that lil fiber square telling you how to compute mapping spaces in overcategories

Omar Antolín-Camarena

5:43 PM

@PiotrPstrągowski I think both $P(C_{/c})$ and $P(C)_{/y(c)}$ should have the following universal property for an $\infty$-category $E$: given a presentable $\infty$-category $D$, cocontinuous, terminal object preserving functors $E \to D$ are equivalent to functors $C \to D$ that send $c$ to the terminal object of $D$.

Omar Antolín-Camarena

6:14 PM

If $Pr^L$ denotes the $\infty$-category of presentable $\infty$-categories and cocontinuous functors, $Cat^t$ denotes the $\infty$-category of $\infty$-categories possessing terminal objects and terminal object preserving functors, and let the subscript $\ast$ denote pointing categories by picking an object. Then I'd hope you have a commuting square of forgetful functors (the vertical ones point at the terminal object):
$$\begin{CD} Cat_ast @<<< Pr^L_\ast \\
@AAA @AAA \\ Cat^t @<<< Pr^L
\end{CD}$$

Arun Debray

6:49 PM

@SeanTilson Thanks! Looking at it again, there are inconsistencies elsewhere in the notes too. I'll go fix them.

Re: yesterday's discussion, I asked Andrew Blumberg and Ernie Fontes about it. They said that it would be nice to have an RO(G)-graded triangulated structure, but that it's not at all clear how to produce triangles of the form X -> Y -> C(f) -> X[V] (which is what people were curious about), so in general you can probably only get an action by Pic(Sp^G).

In the case G = C_2, you can apparently say something, but it's not as nice as one would expect, and it doesn't extend to other groups

So what I wrote in the notes was misleading. I'll fix it later today.

Dylan Wilson

7:50 PM

Well for C_2, it's magically true that the cofiber of C_{2+}--->S^0 is a representation sphere. This lets you build something in the triangulated category. In general, permutation rep spheres and reduced permutation rep spheres have descriptions as hocolims over n-cubes... but triangulated cats have trouble with n-cubes as soon as n>2

(of course this is all a shadow of the dream-team's 'G-stability', and G-(co)limits etc... see Denis's paper)

Piotr Pstrągowski

9:53 PM

@OmarAntolín-Camarena Maybe I'm misunderstanding something, but it seems to me that the argument is that $C_{/c}$ is the free $\infty$-category with a terminal object on the pointed $\infty$-category $(C, c)$. Is this true? Say if $C = \{ c, c^\prime \}$ is the discrete category with two-points, then the slice $C_{/c}$ is trivial, but there can be plenty of functors $C \rightarrow D$ that take $c$ to the terminal object of $D$.

I just found the needed statement! It is, in fact, 5.1.6.12 in Lurie's book. Thanks everyone for helping!

Omar Antolín-Camarena

10:32 PM

@PiotrPstrągowski You didn't misunderstand what I was saying, I was just plain wrong!

Omar Antolín-Camarena

10:56 PM

I don't see any easy way to fix it. :(

Espen Nielsen

11:52 PM

@DylanWilson Fully faithfulness of the yoneda embedding $C\rightarrow P(C)$ should allow us to conclude that the functor is fully faithful on representables at least...

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