Mathematics - 2017-06-23 (page 4 of 5)

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11:00 AM

@Secret what do you think of this solution ?

Still computing, there seemed to be a lot of nested operations whenever i=/=j
s=accdbdab
12345678

Compute
T[1,8]
Check T[1,8]
s[1]=a,s[8]=b=/=s[1]
T[1,8]=min(1+T[1,7],1+T[2,8])
Check T[1,7]
s[1]=s[7]
thus
T[1,7] -> T[2,6]
check T[2,6]
s[2]=/=s[6]
T[2,6] = min(1+T[3,6],1+T[2,5])
Check T[3,6]
s[3]=/=s[6]
T[3,6] = min(1+T[4,6],1+T[3,5])
Check T[4,6]
s[4]=s[6]
T[4,6]=T[5,5]=0
Check T[3,5]

yeah it is normal because we have to check all cases

and it seems the number of checks needed grows roughly in powers of 2

exactly

summer has come here so i can care my hobbies. i am setting my dev. env. up.

11:14 AM

check T[3,5]
s[3]=/=s[5]
T[3,5]=min(1+T[4,5],1+T[3,4])
Check T[4,5]
s[4]=/=s[5]
T[4,5]=min(1+T[5,5],1+T[4,4])=min(1,1)=1
Check T[3,4]
s[3]=/=s[4]
T[3,4]=min(1+T[4,4],1+T[3,3])=min(1,1)=1
T[3,5]=min(1+T[4,5],1+T[3,4])=min(2,2)=2

T[3,6]=min(1+0,1+2)=min(1,3)=1

check T[2,5]
s[2]=/=s[5]
T[2,5]=min(1+T[3,5],1+T[2,4])=min(1+2,1+T[2,4])

check T[2,4]
s[2]=/=s[4]
T[2,4]=min(1+T[3,4],1+T[2,3])

check T[3,4]
s[3]=/=s[4]

T[3,4]=min(1+T[4,4],1+T[3,3])=min(1,1)=1
check T[2,3]
s[2]=s[3]

T[2,3]=T[3,2]=0

yeah it is normal,but I think that is the right way :P

How will you deal with odd pallidromes like abdbdba or abdcdba?, or do you also have a check on the input string on whether it can form a palidrome before the T[i,j] start checking (e.g. a string with all letters distinct cannot from a palidrome for example)?

I just seek for the number of moves of deleting , I am not interested of the length of the string

ok I see

and in such a nice cases where it is palindrome the first thing ofcourse it to chech the start and end character if they are the same we go in depth else we look for the string without the first letter and the string without the last letter

11:21 AM

Ah so you want to count the minimum number of deletions of an input string such that the leftover is a pallidrome

yes

How will your program respond if the user puts in a string that contains no pallidrome e.g. abc, I am guessing T[i,j] will eventually return 1?

but in fact there cannot be any deletion on it that will give you a pallidrome, unless you consider a single letter as a "trivial pallidrome"
or is it irrelevant because of whatever underlying background that the strings came from will guarenteee it contains at least one pallidrome substring?

@SohamC ping back

my solution guarantees because for "abc" it will say 1+1+0 = 2 meaning it will leave a single character which can be considered palindrome

ok that makes sense

Meanwhile this program does makes me wonder about something: If we are given the following string:

abdcfghrkdfsbcgdjeabdbacdhnfjjfcbdbdbss

we as humans can rather quickly find out where the 3 nontrival pallidromes are, yet a computer will take $2^{|s|}$ steps. I wonder how are we able to achieve that...

11:31 AM

because we can scan large interval which seem correct and we can even better seek for palindrome in smaller interval of correct interval

computers don't have that sense

Soham Chowdhury

@Balarka email reminder

@Soham I do remember :) I just got back from school

we can find some solutions but it is not guaranteed we will find all while the computer guarantuees

Soham Chowdhury

@BalarkaSen sure.

I see

Soham Chowdhury

11:34 AM

I cannot believe I am going to be off to ... college ... in a little over a week

Neat!

Soham Chowdhury

:(

I mean, more like :/

Hey @Balarka

Soham Chowdhury

In any case, colleges have libraries, which is a nice thing.

I really want to look at Eisenbud's comm-alg book

@Soham true that, which is convenient

And hope you have a good time in college!

11:39 AM

Eisenbud is too huge 4me

(I am writing the email now)

(For a shorter one, I've heard Atiyah-Macdonald is good)

@Daminark hi

Soham Chowdhury

look. I don't do well with physical math books, it's not how I was brought up :P

I suppose you're the same way to some extent, Balarka.

@Daminark, I've worked through the first couple chapters of that at one point.

Nice

I can't read pdf's as well as I read physical books actually

11:40 AM

How's it going @Balarka?

Soham Chowdhury

@BalarkaSen I like having physical books for the simple reason that I don't have two screens and typing notes is easier without switching back and forth. But I'm more comfortable with pdfs.

Force of habit, I suppose.

you techno

i am thinking of refuting a lot of modern technology

cyberpunk is not the greatest post-apocalyptic world to be in.

@Soham I've got that Macintosh computer so multiple desktops

Soham Chowdhury

I mean physical screens.

I just sorta like the idea of having all my books with me at most times

Soham Chowdhury

11:43 AM

Unless you meant that too?

I mean, it's easy to switch between desktops on a Mac is the point, so you may as well have multiple screens because you just slide 4 fingers to the left or right, which is faster

(Also note that I don't really take notes)

@Balarka o lawd

Soham Chowdhury

Virtual desktop-y setups are extremely inadequate. I have an excellent tiling window manager setup (completely keyboard-controlled), but still.

I'm talking about being able to look at something while typing on another screen. You need multiple monitors for that, right?

@Daminark there are better options for a post-apocalyptic world.

Soham Chowdhury

@Daminark ahahaha.

get used to daminark's puns

and try not to laugh

11:45 AM

Oh, yeah likely, they're not simultaneous, I do alternate

Don't listen to @Balarka's advice, let the laughter flow, it's good for the soul

Speaking of souls, when you're done with your email shtick, want to talk a little about algebraic topology?

Soham Chowdhury

@Daminark I hope you don't mean me, I don't know the first thing about topology

I guess you didn't mean me. :)

people.maths.ox.ac.uk/harrington

I was talking about Balarka lmao

Sorry for some reason, I missread algebraic topology as algebraic biology

But nah I demand YOU to help me with this! :P

Soham Chowdhury

11:47 AM

math.berkeley.edu/~lpachter

@Daminark Sure, but give me a few minutes

Sure

Kek @Secret

westernsydney.edu.au/__data/assets/pdf_file/0006/513906/…

Turns out that subject does exists in some form...

and there's a springer book

Good lord

Pure mathematics

Broadly speaking, pure mathematics is mathematics that studies entirely abstract concepts. This was a recognizable category of mathematical activity from the 19th century onwards, at variance with the trend towards meeting the needs of navigation, astronomy, physics, economics, engineering, and so on. Another view is that pure mathematics is not necessarily applied mathematics: it is possible to study abstract entities with respect to their intrinsic nature and not be concerned with how they manifest in the real world. Even though the pure and applied viewpoints are distinct philosophical positions...

Looks like nothing is pure nowadays, be it maths, arts etc.

11:53 AM

$\infty$-topoi tho

Jk but I mean, I think the point is more, the aim of the study is the abstract object

So many things end up being applicable as a matter of course, but the people working on pure _____ probably ought be defined via taking _____ as an end in itself, as opposed to a means to an other end

What?

@Soham Sent.

I see. For me, I have interest in both fundamental and applied side of ____, which means I can be a pure researcher or an applied researcher.

For example, the recent integral and ordinal mumblings of mine are really more pure driven and applications came as a side product

Whereas for things like dynamical systems, more often there's an application in mind

Due to how I tend to conflate relational existence with physical existence, most abstract entities are tangible to me after I understood them, and I can imagine tossing them around as if they are shapes

Hmm

@Daminark So, algebraic topology?

11:58 AM

I tend to look at subjects with use math more in terms of, wow that's a cool fact about this subject that it involves math somehow. shrugs

Ah, yes, perfect!

Okay so I'm trying to understand what I've now identified to be the Eckmann-Hilton duality

Namely $[\Sigma X,Y] = [X,\Omega Y]$

(equal meaning homeomorphic)

Mhm

Now, May just sorta wrote that there and possibly some other stuff related to it but I was completely lost

So you just want to understand that homeomorphism?

For now that's my main goal

So, I remember he wrote something to the effect of how the compact-open topology made one thing work nicely

Don't worry about that for now. Do you see how to get a map $X \to \Omega Y$ from $\Sigma X \to Y$?

Let's just try to work out a bijection for now.

12:03 PM

Like, that $C(X,C(Y,Z))$ bijects to $C(X\times Y, Z)$, and that in the compact open topology this is a homeomorphism

Correct.

Soham Chowdhury

@BalarkaSen Thank you. Do you have any idea how long he takes to reply, usually?

I see you're working with @Daminark, so I'll leave y'all alone for now and go do something useful.

@Soham We are talking now. He's on email for 24/7 generically. :P

So I think the idea is to use the natural transformation, right?

@Daminark do NOT think categorically.

What is $\Sigma X$?

Soham Chowdhury

12:08 PM

@BalarkaSen Oh, I see.

Back again

$(X\times I)/(X\times \{0\} \cup X\times \{1\} \cup \{x_0\}\times I)$

He agreed to talk, @Soham. I am right now asking whether you should shoot him an email, or vice versa.

Soham Chowdhury

Okay. (He remembers me?)

I am not sure. Should he?

I didn't think you knew him.

Soham Chowdhury

12:10 PM

He connected me to SB.

Don't really know him per se.

Just sent him an email, and he said "talk to this prof".

@Daminark Right, OK. So what's a way to get a 1-parameter family of maps $X \to Y$ from a map $\Sigma X \to Y$?

Soham Chowdhury

But, no, I think it's likely he doesn't. Ignore that.

@SohamC OH. I see. You should mention that to him when you send an email to him.

(He's very forgetful, so it's very possible he does not remember)

Soham Chowdhury

So, what is he saying?

Okay, I'll mention it.

hasn't replied to the last email yet. i'll let you know if he wants you to send an email to him

12:14 PM

I mean, I think it should be $F_t(x) = F(x,t)$, somehow being careful about the collapsing of $\{x_0\}\times I$

@Daminark Good, good. Yes.

(Kek Peter May's influence is taking hold on me)

Oh wait hold on, $(x_0,t_1)$ and $(x_0,t_2)$ are the same point so $F$ should basically take the same value on each, meaning that this one parameter family of maps is based properly, I think

:)

So, yes, if $F : \Sigma X \to Y$ is a map, then you get a 1-parameter family $f_t : X \to Y$ given by $f_t(x) = F(t, x)$. This one parameter family, as you said, sends $f_t(x_0) = y_0$ for all $t$, where $x_0$ is the "base" of $\Sigma X$, and $y_0 = F(x_0)$

But more importantly, this 1-parameter family $f_t$ "starts" ($f_0$) and "ends" ($f_1$) at the constant map at $y_0$

Hmm, is this not starting to look like a loop of based maps? :)

So this should generate a loop somehow, and I imagine this might have something to do with our currying?

Internet...

But yeah awesome, I'll think a bit and make sure to formalize this

Ok, let me know if you want me to spell more details out. Remember you need to construct a map $X \to \Omega Y$ out of this.

It's helpful to understand what $\Omega Y$ looks like.

(Not literally, just interpret the definition visually)

12:23 PM

Eww, math.

In the end it's just $C(X \times I, Y) \cong C(X, C(I, Y))$ with appropriate basing considerations.

(the bijection for that is super-obvious; any map $f : X \times I \to Y$ gives rise to an $X$'s worth of maps $f(x, -) : I \to Y$)

So you take that homeomorphism, call it $\phi$, and then you're looking at a precomposition $\phi\circ f$

Or wait hold on not that

$\phi(f)$

@SohamC Shoot.

Tell him what you know, what you do, and where you're at.

among other things maybe

Well, you're having $f_t(x)$ which we can revert to thinking about as $F(x,t)$ for a bit, then it looks like $\phi(F)$

@Fargle kek

And it works basedly because each $f_t(x_0) = y_0$, so that when you pull it through, you get that the loop fixes $y_0$

Soham Chowdhury

@BalarkaSen I'm supposed to email him?

12:29 PM

Yes, you are.

Soham Chowdhury

okay.

Okay I think I better understand this, and you were right that it's helpful to not just say that suspension and loop are functors, so then you have natural transformations and then just compose them

At least now I see it, for real

Soham Chowdhury

rkmvu email or tifr one (or something else)?

Those are official ones. Do you know the gmail one? If not, send me an email and I'll send it to you.

Soham Chowdhury

Copy it and I'll delete the message.

12:32 PM

And now I just need to remember exactly how we linked it to Eilenberg-MacLane

delete

@Daminark Give me a few more minutes, I'll see what you have to say.

Yeah it's fine, I'll just type it out as you talk

@Soham Done.

Oh wait I think I see it

Soham Chowdhury

@BalarkaSen got it

12:36 PM

So we know that $\Sigma S^{n-1} = S^n$

also we should move further discussions about this to email

in my opinion

So then $G = [S^n, K(G,n)] = [S^{n-1},\Omega K(G,n)]$, which is suggestive of the identity $\Omega K(G,n) = K(G,n-1)$

Oh wait you could get that identity by directly thinking about loops, I imagine

Bingo, @Daminark.

Soham Chowdhury

@BalarkaSen okay.

Okay need to make this precise but the pieces are starting to fit together

12:39 PM

So just to understand where we were, we understood the loopspace-suspension adjunction?

Yeah, I think that part checks out alright for me now

Great.

So you want to prove $\Omega K(G, n) = K(G, n -1)$ now?

Yup

I like your observation. I think that means $\pi_n(K(G, n)) \cong \pi_{n-1}(\Omega K(G, n))$?

Actually, you don't need $S^n$. You can work with $S^k$

Which would tell you $\pi_k(K(G, n)) \cong \pi_{k-1}(\Omega K(G, n))$

Well I think that is something that just holds more generally because the suspension of $S^n$ is $S^{n-1}$

Wait what? What's the difference between $S^n$ and $S^k$ here?

12:41 PM

you used $n$ in both the dimension of the sphere and K(G, n)

that is not necessary

Oh I see

Soham Chowdhury

@BalarkaSen because the groups are 0 in every other dimension, right?

if I remember what K(G,n) means

Oh that's true. I was more like, saying, write it out for all homotopy groups than one :)

What I was in fact getting at was $\Omega K(G, n)$ has the same homotopy groups in all dimensions as $K(G, n-1)$

which your comment + Daminark's observation proves

Right, nice

Perfect!

Okay now I'm much happier with that

Does isomorphic homotopy groups in every dimension imply homotopy equivalent spaces?

12:44 PM

And we just need to wait for Peter May to actually construct said spaces and show that their homotopy stuff form groups :P

Ponder on that while I go away from keyboard for a while.

I know this should give a weak homotopy equivalence

Or at least I think

Hey everyone!

My immediate hope for a counterexample would be something which is not contractible but with vanishing homotopy groups

seems like my first bounty is going to expire without an answer :(

did manage to bring the question from ~2 to 35 upvotes though and my own work has at least convinced me the question is "hard"

Akiva Weinberger

12:53 PM

Whoa your avatar changed

(Did you get a haircut?)

Hey @Akiva and @GPhys!

Akiva Weinberger

@GPhys What's the question?

And hejhej

35

Q: Prove that $\sum_{n=1}^\infty \frac{n!}{n^n}$is irrational.

Let $$S= \sum_{n=1}^\infty \frac{n!}{n^n}$$ (Does anybody know of a closed form expression for $S$?) It is easy to show that the series converges. Prove that $S$ is irrational. I tried the sort of technique that works to prove $e$ is irrational, but got bogged down.

sequences-and-series rationality-testing

@Daminark Do you know Whitehead's theorem?

I went through the full notes for a lower level grad course on irrationality and transcendence @AkivaWeinberger

I made no progress on this question, but I learned a lot

12:56 PM

Nope, that hasn't happened yet

René Descartes

Is Science the knowledge of what is around us? Please give your opinion. I request you.

and I proved or empirically showed several of the methods wouldn't work for the series as given

It says for CW complexes, a weak homotopy equivalence (a map $f : X \to Y$ such that $\pi_n(f)$ is an isomorphism for all $n$) is a homotopy equivalence.

So the main atop course is going a manifoldsy route, but so far what we've done is briefly review the basic topology stuff, then do homotopy, induced homomorphism, weak homotopy equivalence, fundamental group, a bit on higher homotopy groups (including a proof I'm not yet sure about that they're abelian), categories/functors/natural transformations, Eilenberg-MacLane, "the nth cohomology group", and the loop/suspension stuff

If what you guess is true, that is, $\pi_n(X) \cong \pi_n(Y)$ for all $n$ implies there is a weak homotopy equivalence $f : X \to Y$, then $X$ and $Y$ in fact should be homotopy equivalent. (assuming everything is a CW complexes). But is it?

12:57 PM

(starting from higher homotopy groups, it was the spinoff seminar by May)

We haven't quite defined CW complexes yet

I am not liking May's course.

@Daminark It's just "the space is nice enough". Don't worry about it.

Alright

So, based on this statement, are weak homotopy equivalence and homotopy equivalence the same in CW complexes?

Yes: that is precisely the Whitehead theorem.

Oh wait sorry I missed that message

Yeah okay we're good

But what I am asking is more like, does $\pi_n(X) \cong \pi_n(Y)$ for all $n$ really mean $X$ and $Y$ are weak homotopy equivalent?

@Daminark It's actually not entirely trivial.

Say, you don't know about covering spaces yet, do you?

1:07 PM

Nope

All that's happened is the syllabus you're iffy about above

Mmkay.

Ask me the question I asked you after you learn covering spaces and homotopy lifting.

Note it down somewhere meanwhile.

How do homotopy groups behave with Cartesian product?

Akiva Weinberger

Every space is weakly homotopy equivalent to a finite space, fun fact

For example, for spheres, you quotient out by equators and stuff

Ah, yeah Peter May mentioned that

@AkivaWeinberger I don't think this is true for every space.

1:09 PM

Maybe every CW complex?

Akiva Weinberger

Like for a 2-sphere, you quotient out the northern hemisphere, the southern hemisphere, the eastern… hemiequator, and the western hemiequator

and you have two points left

so it's a six-point space

Should be true for all simplicial complexes.

And @Balarka will do

Akiva Weinberger

@Daminark Oh, probably.

@BalarkaSen *Oh, probably.

Actually I think something stronger than that is needed.

1:11 PM

@AkivaWeinberger as for the question I bounty'd, I suspect you need some clever integral representation

@Daminark The answer is "false".

Also @Balarka what in particular are you iffy about wrt Peter May's shtick

Steamy was not happy about Peter May's defining $\widetilde{H}^n(X,\pi) = [X,K(\pi,n)]$

Ok, no, apparently this is true for all finite simplicial complexes.

@Daminark I am totally not. UGH.

It was rather funny when he did it though

That is a fantastic theorem I appreciated after I learnt individual definitions of all those.

1:13 PM

He was like "So we've got $[X,K(\pi,n)]$, and I wish to give it a name. Let's call it the H stuff!"

(He didn't say the H stuff but I don't feel like TeXing it again)

So that was amusing. But yeah he defined that and said something like "So we're gonna see that this works when we construct E-M spaces, and then we'll see it's dual to something, hence homology exists!"

I don't like it.

But I wonder; it might be a fun pedagogy.

lol i can't decide whether i like it or not

He's apparently never done it quite this way before, he doesn't like the whole chains stuff used to introduce homology, apparently, and thinks that this is closer to how modern algebraic topologists think about it. Also he likes that you can get other homology theories out of different types of spaces

Yes.

He wants to introduce the Eilenberg-Steenrod axioms at some point, and that'll explain this

$K(G, n)$ is the "Eilenberg-Maclane spectrum" to the homotopy theorists nowadays

You can use different spectrum to get different homologies

1:17 PM

I find it rather amusing that he's one of the more modern people around and yet he's one of the oldest people at this school, if not the oldest

I don't think any of this is modern. It's more useful from the homotopy theory perspective.

No I meant like, his research

The chain complex definition is more useful from the geometry perspective.

@Daminark Ah, ok.

I think he remarked once that he does research in infinite loop space theory and higher category theory (model categories or smth?)

vOv

1:21 PM

So I guess you can see why he wants to put a lot of stuff in categorical language

I know why. I just don't appreciate it :P

Peter May is a pretty famous name.

Hi chat

Hi semi

With respect to the homeomorphism between looped and suspended space, he basically said that the two are functors and that this would basically have to do with a natural transformation between them

classics

semiclassics

1:22 PM

He actually said that you straight up can't do math without categorical language, though if you're in analysis you can avoid category theory :P

Hey you know this one $\dot{x} = 1(mod 1) , \dot{y} = \omega (mod 1)$

not really

those are independent ODEs, though

actually I am searching what this type of system is known as ?

so wouldn't seem like there's anything terribly interesting there

@Daminark $[\Sigma X, Y] \cong [X, \Omega Y]$ is a categorical relationship, namely, "Loopspace and suspension are adjoint functors". But that doesn't explain shit about where that isomorphism comes from, unfortunately. :P

1:23 PM

it is when you plot x vs y

also, though, what is exactly being taken mod 1 here?

I don't believe you... You're a liar!!!

If it's 1, well---1 mod 1 = 0

and omega mod 1 is a constant.

Lel @Balarka

I love some of the categorical formalism and results, but not the pedagogy.

$H^n(X; G) = [X, K(G, n)]$ is proved (from the Eilenberg-Steenrod axioms of $H^*$) from a categorical theorem, actually.

1:26 PM

$x ̇ = 1(mod1); y ̇ = ω+αsin(2πx)(mod1)$

a coupled one!

The "Yoneda lemma"

any body can search and can help me in solving this analytically?

that's a bit different, yeah.

I mean if you wanna prove a theorem, Yoneda lemma

Like just a sample of how to do it?

1:27 PM

but, again, I don't know what you mean by mod 1 here. Do you mean you only care about the interval [0,1]?

Soham Chowdhury

Infuriatingly trivial if you don't understand it, and unexplainably trivial if you do.

yeah I think we will be on a unit square

Soham Chowdhury

@Semiclassical often people mean fractional part by that, yeah.

@Semiclassical

@baymax sure, but that'd be $x,y$ mod 1

1:27 PM

chuckles

Alessandro Codenotti

Every time cathegory theory is mentioned the yoneda's lemma pops up, I'll discover what is says sooner or later

like i didn't get that?

Let's see how long before May teaches us that. Next class is on Tuesday

Also lol I wonder what Ted would think

Ted is even more conservative than me about category theory.

(I like it)

I know, I kinda wanna see his reaction sitting in on a lecture

Soham Chowdhury

1:30 PM

Well, doesn't it depend on the math you do?

Mike has unfortunately become an infinity category theorist

too bad

Has he really?

not technically, but he's doing lots of categorical algebra

I mean given that I've got 8 weeks (hopefully no bootcamp conflict), it May happen to me too

Soham Chowdhury

he's grumpy about it, but characteristically full of good cheer

1:32 PM

And ah, well algebra is good

no u

Soham Chowdhury

@Daminark have you done/are you doing the Chicago REU?

The program I'm officially in is called the bootcamp

It's something my analysis professor set up, we'll do complex, probability, dynamics, and diffgeo

I am annoyed. I don't know an entirely elementary proof of $\Omega K(G, n) = K(G, n - 1)$.

I can use ze path space fibration but I don't want to

$\dot{x} = -x + x^3$ , x(0) = a ;$-1<a< 1$ , does solution exist for all times?

1:34 PM

Now, I'm also sitting in on the REU talks, with my top priorities being atop, followed by NT

@Semiclassical any help?

I have done the apprentice last year, which was a 5 week course in linear algebra and a side of graph theory

Akiva Weinberger

What's $[X,Y]$ for spaces?

Homotopy classes of maps $X \to Y$

Also used as based maps sometimes

Akiva Weinberger

Ah, right

Wait, so the homology of a space is the homotopy classes of maps from it to $K(\Bbb Z,n)$?

1:41 PM

K(Z, n)

and *cohomology with Z coefficients

Akiva Weinberger

Oh, cohomology.

How is that at all useful? I thought $K(\Bbb Z,n)$ was in general a horrible, infinite-dimensional thing

(Also I have no idea how one proves this)

Do you know the cellular approximation theorem?

Akiva Weinberger

Vaguely.

Ok. If $X$ is an $n$-dimensional CW complex then any map $X \to K(\Bbb Z, n)$ is homotopic to a map $X \to K(\Bbb Z, n)$ which sends all of $X$ to the $n$-skeleton of $K(\Bbb Z, n)$.

This is because of cellular approximation; you can homotope any map between CW complexes so that it sends $i$-skeleton to $i$-skeleton for all $i$.

$K(\Bbb Z, n)$ is a sad space, but it admits a model (by that I mean another space which is homotopy equivalent to it) so that $n$-skeleton of $K(\Bbb Z,n)$ is $S^n$. (you can prove this)

Soham Chowdhury

@Balarka can you come online on Hangouts?

1:45 PM

So you end up looking at maps $X \to S^n$

Ah, sure, @Soham.

Akiva Weinberger

Right, I think I remember your construction of $K(\Bbb Z,S^n)$ from a while ago which should work

@BalarkaSen Wouldn't that give you cohomotopy or something?

K(Z, n), but yeah.

@Akiva Right, cohomotopy.

Akiva Weinberger

Dammit

That's what I meant

Got to go, be back later

As a particular example of what I said, if $M$ is an $n$-dimensional manifolds (do topological manifolds always admit CW structure? I think so) then $H^n(M; \Bbb Z)$ is in bijection with $[M, S^n]$

So $[M, S^n] = \Bbb Z$

With a little more work you can show Hopf degree theorem from this

(you just have to show that the bijection is given by taking preimage of a point and counting the number of preimages; there's a categorical work involved here)

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