Homotopy Theory

Tyler Lawson

04:10

@MingcongZeng have you seen Behrens-Hopkins, something like secs 3-4? arxiv.org/abs/0910.0617

Julian Rachman

04:25

May I get help with the factorization of categories?

Julian Rachman

05:06

For instance,

Let [n] be the finite chain poset [0-->1-->...-->n]. Then define an infinite sequence
F: Seq-->C
in C to be essentially finite if F factors through some [n], i.e., as the composite
Seq -->[n] -->C

What does it mean to "factor through?"

Julian Rachman

06:09

If there is no answer to this, I will move on then.

Julian Rachman

06:37

Then lets take this example: Categorically, you can represent an infinite sequence in C as a functor F: Seq-->C, where Seq is the posetal category [0-->1-->2-->...]. Would you agree that F is equivalent to an infinite decreasing sequence iff it does not factor through a finite chain category [0-->1-->...-->n]?
Does this mean that we can show properties or characteristics from one category to another using this method?

lentic catachresis

09:47

I found a cute question at math.SE: math.stackexchange.com/questions/1542801/… I doubt there is a positive answer to his question, but I also doubt my competence so I put it here in case anybody has a better idea

Jon Beardsley

11:27

@JulianRachman "factor through" is kind of general category theoretic term, where F:C-->D factors through G:C-->B if there's a functor H:B-->D such that H(G):C-->D is equal to F

visually this looks like a commutative triangle of categories

@JulianRachman i'm not entirely sure what a "descreasing sequence" would be in an arbitrary category that doesn't have any kind of ordering on its objects. if a functor F:Seq-->C factors through [n] then your sequence should be constant at [n] and take every number greater than or equal to n in Seq to the same object, and every morphism after n to the identity

Urs Schreiber

12:01

Hi @JonBeardsley , I do come by every now and then. Great that you helped establish this leisurely place here.

Eric Peterson

14:29

@TylerLawson this is nice, i hadn't seen this either

Mingcong Zeng

@TylerLawson I have never heard of it before. It looks like something can somehow solve the mystery in my mind.

Jon Beardsley

@lenticcatachresis can something be done if one homotopes this map to a cellular map?

Julian Rachman

@JonBeardsley an infinite decreasing sequence is one that is like $a_{n+1}\to a_n$. Can you give me an example of your explanation?

Jon Beardsley

14:45

@JulianRachman I'm sorry Julian, I think I still don't understand. Can you define an infinite decreasing sequence rigorously for me?

In particular, can you give me an example of an infinite decreasing sequence in, say, the category of sets, or topological spaces?

lentic catachresis

@JonBeardsley I don't understand, which map? The question assumes you're given a map which is already cellular

Jon Beardsley

Oh, I see that now.

@lenticcatachresis yeah nevermind. I was thinking of things incorrectly.

lentic catachresis

no worries

@JonBeardsley I actually looked at Switzer again now and I think that proposition 10.13 gives the answer. The formula is a bit nasty and it involves the degrees of maps induced by $f$ on the n-cells

Jon Beardsley

15:00

@JulianRachman For the case of factoring through [n], think about what it means to factor through. Let's say, for simplicity, I've got a functor F:Seq-->{Abelian Groups} that takes the object k to the free group on k generators, which we'll denote F(k), so long as k<11. It takes everything from 11 up to F(10). It takes every map k-->j to the identity on F(10) for each k>9.

Julian Rachman

@JonBeardsley An infinite decreasing sequence is, denoted $a_i\sqsupset{a}_{i+1}$, iff ${a_i}\sqsupseteq{a_{i+1}} \ \text{and} \ a_{i+1}\not\sqsupseteq{a_i}$ for all ${i\geq1}$

Ah! I see!

So how does the factoring come into play?

Jon Beardsley

Then let G:Seq-->[10] be the functor that takes k-->k for k less than or equal to 10, and takes every higher number to 10.

Which functor, then, goes from [10]-->{abelian groups} so that we can factor our original functor?

Have to run, sorry can't say more. Just play with it.

Julian Rachman

@Jon will do. Thank you.

lentic catachresis

16:08

@JonBeardsley I'm actually not getting how Switzer defines his map on the bottom of page 178. I mean, out of an n-cell from X and a completely unrelated n-cell from Y he gets a map $S^n\to S^n$ out of the given map $f:X\to Y$

Jon Beardsley

16:25

@JulianRachman this statement requires an ordering on your target category. there are obvious ones you could pick, like inclusion, or cardinality, or whatever, but an arbitrary category doesn't, assuming i'm not making a mistake, have any way of saying one object is less than another object.

lentic catachresis

16:36

I actually think that Switzer's formula is wrong and that the thing is much more complicated than it seems. I'm looking at Lundell and Weingram's "The topology of CW complexes" and def. 3.7 p. 165 is quite convoluted, with some weird incidence numbers stuff that I haven't looked in detail

Bertram Arnold

16:55

These "incidence numbers" should just be the degree of the map constructed by Switzer. The n-cells of X and Y are related by f: X^n/X^{n-1} -> Y^n/Y^{n-1}, and you use the characteristic maps of the n-cells to map from and into these bouquets of spheres to/from S^n

lentic catachresis

But the map constructed by Switzer is ill-defined. I've just answered the question there, what do you think?

I mean, I understand what you say, but you will get a zig-zag (well, a roof)

a map $S^n\to Y^n/Y^{n-1} \leftarrow S^n$

Bertram Arnold

17:12

But this admits a retraction - it's just the inclusion of a wedge summand

Essentially because the category of pointed spaces has a zero object

Bertram Arnold

17:34

And this retraction is given by also collapsing all $n$-cells except the one you are considering. In your writeup it looks as if you want to collapse only this $n$-cell (which won't work). Look also at the definition of $X^{n-1}_\beta$ after Definition 10.10

lentic catachresis

Oh. I thought $X^{n-1}_\beta$ was his notation for $X^{n-1}$ with beta attached.

ok, you're right, I was hasty. I deleted the answer

Julian Rachman

19:20

@JonBeardsley I see you are getting at something. Go on.

Do you have a formal statement that you can make out of what you have said?

Saul Glasman

19:42

@JulianRachman I don't think there's much of a formal statement to be made beyond what Jon's already said. It's just that a general category doesn't come with a natural ordering on the objects.

Julian Rachman

19:54

@Saul So basically my target category $C$ does not originally have any order in its objects and thus I must define an ordering whether it is inclusion, etc. in order for me to show greater than or less than.

Saul Glasman

I think you might be confusing putting an ordering on objects in $C$ with giving a functor from a totally ordered set to $C$

to do the latter, you just need to give a string of composable morphisms in $C$

Julian Rachman

Oh. I definitely am because I looked above and define $F:\text{Seq}\to C$ where $\text{Seq}$ is a posetal category.

user105491

the latter (sequences of composable morphisms in $C$) is useful because a string of $n$ composable morphisms in $C$ is just an element of $N(C)_n$ where $N(C)$ is the nerve of $C$, which is an $\infty$-category

Julian Rachman

@SanathDevalapurkar so how may I use it?

user105491

It depends on what it is that you are trying to do

Julian Rachman

20:00

And we are talking comparable like hom(A,B)\circ hom(B,C)\to hom(A,C)?

Well what I said above is what I am trying to accomplish.

user105491

how do you compose hom-sets?

Julian Rachman

I said it above.

I am sorry if I have no knowledge of \infty-categories but I am just not there yet. However I believe that your statement above is understandable.

Here is the question I asked recently: math.stackexchange.com/questions/1632880/… it is poorly written in my opinion but it might help a little.

user105491

no, no, what I'm asking is how you compose hom-sets? you can compose morphisms like $A\to B$ and $B\to X$ to get a morphism $A\to X$. Sequences of morphisms of the form $A\to B\to X$ form elements of $N(C)_2$.

user105491

maps $[n]\to C$ are just strings consisting of $n$ composable morphisms of $C$.

Julian Rachman

I don't know.

Can you elaborate?

user105491

20:11

not right now, unfortunately. i have to go out somewhere

Julian Rachman

Ok. Any hints that you can leave me with?

@SanathK.Devalapurkar just tag me in your response when you come back.

Julian Rachman

21:41

@SanathK.Devalapurkar Even if you are here or not, I now understand your uss of nerves. However, if you took a look at what I am currently working on, I am concerned with infinite sets not finite sets. So how are we going to construct this around infinite sets (sequences) and how does \infty-categories come into play?

user105491

Unfortunately, I have no idea what you are working on, so I don't know the answer to your question.

Julian Rachman

Ok but I am asking how or if nerves will work for infinite sets (sequences).

And my work is above in my discussion with Jon.

For instance, can we have $N(C)_\infty$? How would this be constructed?

@SanathK.Devalapurkar

user105491

22:14

maybe by considering the inclusions $\Delta^n\subset\Delta^{n+1}$ and taking the colimit of the tower; the resulting thing could be called "$\Delta^\infty$"?

Julian Rachman

22:24

Where \Delta denotes the nerve of C?

Jon Beardsley

@SanathK.Devalapurkar @JulianRachman I think infinity categories are kind of a red herring here.

user105491

Yeah, I agree. I only mentioned infty-categories as an aside.

Jon Beardsley

Julian is not asking for anything having to do with $\Delta$, the category of all finite sequences [n], but rather thinking of [n] itself as a category and mapping it into something.

Julian Rachman

Yes. Factorization.

Jon Beardsley

@JulianRachman did you understand the example I gave above about factorization?

Julian Rachman

22:27

Not 100%.... :/

Jon Beardsley

Think of this: given a number, say, 15, you can factor it as 3*5, given a functor F:C--> D, you can ask "Can I factor F as two functors, say G:C-->B and H:B-->D?"

Julian Rachman

Ok. I understand that much.

Jon Beardsley

Suppose you can, then you say that F "factors through B"

Now, do you recall the example I gave earlier?

Julian Rachman

The one about arbitrary categories? Then yes

Jon Beardsley

We had a functor $F:Seq\to Ab$ defined by $F([n])=\{\mathrm{free~~abelian~group~on~}n\mathrm{~generators}\}$ if $n\leq 10$ and $F([n])=\{\mathrm{free~abelian~group~on~}10\mathrm{~generators}\}$ if $n>10$

The morphism $[n]\to [n+1]$ was taken to either the subgroup inclusion of the free group on n generators into the free group on n+1 generators (in the case that n+1 is less than or equal to 10), and the identity otherwise.

Does that make sense? I've defined a functor $F:Seq\to Ab$ by telling you what each object and each morphism go to (and you can check that $F$ preserves compositions of morphisms).

Oh sorry, I think I'm misremembering the notation.

Above should say $F(n)$ not $F([n])$.

There's another functor, now using the notation you had from earlier, $G:Seq\to [10]$, that takes $\{1\mapsto 1,~2\mapsto 2,\ldots,~9\mapsto 9,~10\mapsto 10,~11\mapsto 10,~12\mapsto 10,\ldots\}$

And then there's a functor $H:[10]\to Ab$ that takes $k\in\{1,\ldots,10\}$ to the free abelian group on $k$ generators.

I'd recommend getting out a piece of paper and writing down what $F:Seq\to Ab$ looks like, versus what $Seq\overset{G}\to[10]\overset{H}\to Ab$ looks like.

You will see that $H\circ G=F$.

Julian Rachman

22:46

I will to go through this for about an hour or two and I will be right back. Thank you, @Jon. You saved me a lot of self-cleaning.

I will be back with my thoughts.

Julian Rachman

23:35

@Jon I have had a spark of knowledge.

G:Seq\to[n] forms my infinite decreasing sequence if I define the morphisms such that $x_j\to x_i$ iff $x_j\sqsubseteq x_i$ where $i<j$.

I which was my original question.

@Jon If there is is more I am not seeing, let me know.

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