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09:01
I'm proving that if a sequence $\{x_n\}$ has the property
\[
\lim_{n\rightarrow\infty}x_{2n}=\lim_{n\rightarrow\infty}x_{2n+1}=\infty
\]
show that the sequence $\{x_n\}$ diverges to $\infty$.

My method is to let $M\in\mathbb{R}$, then we know from above that
\begin{align*}
&\exists N_1\in\mathbb{N},\forall n\geq N_1, x_{2n}\geq M \\
&\exists N_2\in\mathbb{N},\forall n\geq N_2, x_{2n+1}\geq M
\end{align*}
Let $N=\max\{N_1,N_2\}$ and consider $n\geq N$. Now, you could use modular arithmetic to show the theorem. However, my professor suggested it was possible to select a max for a $N$ that sh
(If not, I don't think there's really any problems with writing it as I had otherwise planned)
09:26
although with my suggested proof is $n=1$ a problem?
since $\{ x_n\}\neq\{ x_{2n},x_{2n+1}\}$ since $\{ 2n,2n+1\}=\{ 2,3,4,5,\ldots\}\neq \mathbb{N}$
What do I call $\mathfrak{M}$?
What would I say when reading this? Currently I read it as M
Like "If M is a sigma-algebra..."
fraktur?
Hmmm perhaps not, don't know what typeset it is written in
$\mathcal{M}$
saying Fraktur $\mathfrak{M}$ would be as like saying blackboard bold $\mathbb{R}$
@Incurrence did you try that automorphism group problem?
09:34
@BalarkaSen Not yet, I got distracted by functional analysis
ah, i see
\mathcal isn't even standard in LaTeX with its commonly loaded packages - it needs some special package that I can't recall off the top of my head
So I would call it M is what you are saying
yes
although with such a writing it might be common to say "script M" if you desire to differentiate it
What is M though is my main question
$\tau$ is a topology(in that circumstance)
M is a?(collection of subsets of $X$ given three properties blah blah)
It's elements are measureable sets in $X$
Is it the 'collection of measure' or some such thing
@balarka perhaps you can chime in?
09:40
i am not following the discussion : what's the problem?
Hi everyone
I want a name for $M$, We call $\tau$ a topology, what do we call $M$
Where $M$ is a sigma algebra in $X$
what is $M$?
Just wanted to check what is the community's guideline for posting attempted homework questions for comments / reviews
@Incurrence oh, no idea.
09:42
$M=\mathfrak{M}$
Can someone please let me know?
@SanjouOdawali Post the homework with your own working
@Incurrence - fine if I post a snap shot of it?
@SanjouOdawali Definitely not(not trying to be rude)
@SanjouOdawali You will get downvoted to oblivion
@SanjouOdawali Type up relevant work(all of it, if it is answer verification)
Ok no problem, just wanted to check as I am still new here
09:43
@SanjouOdawali The issue wouldn't be the integrity of the content, he is taking issue with presenting it as a photograph instead of taking the effort to rewrite it.
@SanjouOdawali That's fine. Make sure the question is very clear at the start
Thanks all
Even Typing > **Question:** I want the answer verified because of blah blah
At the start
Yes, plus the homework tag as well
@Incurrence Did you see the analysis question I posted ^
(in chat)
if it was $\{n,2n-1\}$ I would be happy with it, but I feel as if there must be a better way now, or at least a justification for it otherwise
09:49
Why does that say $=\infty$? Are you in the extended reals(or complex plane)?
@SanjouOdawali There is no homework tag, and no need to even mention it
$=\infty$ is here just a notation for divergence to infinity
your proof looks alright to me, @GBeau.
not sure why you'd need modular arithmetic. you just need the basic fact that every number is either even or odd.
that's pretty much what I meant by modular arithmetic
That equality with infinity looks really nasty to me(I mean notation wise I would never write that[I hope it is short hand haha])
brb
@Incurrence well, you can just let everything to be in the extended reals. then the notation becomes alright.
09:53
@Incurrence thanks again
@BalarkaSen That's why I asked that first haha
ADG
ADG
hello i want some help
@GBeau Let $N=\max(2N_1,2N_2+1)$, then if $n\ge N$, we have $x_n\ge M$.
@robjohn I'm not immediately seeing the intuition over my choice
ADG
ADG
anyone?
why do you ignore me?
to all*
10:01
@ADG Speak your question
@ADG People can't chime in to offer help when they have no idea what field it is, or what level
ADG
ADG
i am making an app that draws graphs, i am here for some suggestion, i have many ideas of mine but wish if you all give me some advice, basically it's for android.
@robjohn OOOOOooooohhhhhh. I see it now.
Thanks
@ADG Okay what is the question?
ADG
ADG
@Incurrence nvm, I was asking what minimum should it do and what special things should it do, nvm
Well graphing on an android sounds painful, no offense, but if you want it to be unique(?) that's essentially impossible
Otherwise it should take functional input, and you are good
ADG
ADG
10:16
@Incurrence "painful" why?
none of the apps draw multivariable functions and i'm taking this as my pls point
I personally hate typing on a phone
ADG
ADG
if I make it delicious to type in it?
Who are your market? First year uni, last year highschool?
ADG
ADG
everyone, just cause I cna't find a good one for me indeed.
@ADG It hurts physically to type on a phone, so I think it is unavoidable
@ADG what is a pls point?
ADG
ADG
10:18
which phone do you have.
Samsung galaxy s4 or s5 or something
ADG
ADG
if only my app can draw x^2+y^2=25 which would you choose?
this is plus point
Oh, plus point
I read that as please point lol
ADG
ADG
i disagree with even s4 or s5
LOL :D
What do you disagree with? My physical pains existence?
But not only your app can do that, Wolframalpha has an app that does that
Google can take function input now and any phone can get awesome graphs
Even 3D crap
ADG
ADG
10:20
i would do it even then :D
because i believe in the unbelieveable (pardon the spelling :D)
you $p\to \neg p$
ADG
ADG
??
Just a joke about believing in the unbelievable lol
This account is a year old today
In fact that is as of 2 minutes ago
ADG
ADG
IDC
You don't care?
I just coined the term measurology for $\mathfrak{M}$
ADG
ADG
10:30
IDC*2
@ADG I am talking in general, I don't care if you don't care lmao
ADG
ADG
bye IDC*3
CYa @ADG
Anyone here up for some probability?
I'm looking at this question: math.stackexchange.com/q/1227087/98587
and I think that the answer should just be ((n-1)/n)^n
(Regardless of the OP's requested inclusion-exclusion rule)
Anyone?
10:49
Hi @Incurrence
@JasperLoy Hey JL
@Incurrence I am very worried about next month. I hope I succeed. You must pray for me.
@JasperLoy I shall. I hope you do well
@Incurrence By the way, remember not to make the mistake you made that day, whatever it is.
I try not to remake mistakes in general. But sometimes it is hard
10:57
@Incurrence Yes, I make many mistakes many times.
@JasperLoy I feel like crap right now actually haha. Somewhat related to feeling bad at math again
@Incurrence Sometimes, it is very hard. It is not our fault.
I wish my lecturer told me what I was meant to learn... I am learning measure theory since it is a prerequisite in the text for some thing we were working with, but he never mentioned any of these things
@Incurrence It's OK to be bad at math. You can do something else in life if needed.
(and I am learning for functional analysis, from the Real and complex analysis text by Rudin...)
@JasperLoy It doesn't matter than I am bad for some reason, I find it more enjoyable than anything else
10:58
@Incurrence It is best for you to focus on material covered in the lectures.
@JasperLoy I have covered it all, and the assignment seems unrelated
@Incurrence I will change my email again on 1 May, I will email you again then.
@JasperLoy Okay, what have you chosen?
@Incurrence Bad syllabi and lecturers are everywhere. However, Cambridge has an excellent syllabus.
@Incurrence I am still thinking.
11:01
@Incurrence I think you are a great guy. If I were a girl I might fall in love with you.
@JasperLoy That is good to hear haha :)
@Incurrence You know, my life has been a mess. Until now, I still don't know why I was born.
@Incurrence I used to do weights too, not too much.
@Incurrence I have lost 10 kg over the last 6 months after I put on 10 kg due to overeating.
@Incurrence Which subject do you find the hardest?
@JasperLoy Weights are really fun for me, I find it motivational(it helps that when I started everyone complimented me heaps)
@JasperLoy I don't find any hard I guess, and I am getting great grades, but on here people make me feel bad haha
It doesn't help that I am honest about all of my weaknesses and I think most people aren't
@Incurrence In a way, I have not really started studying math, because in undergrad I was already mad. So when I do get well and start studying, that would be my first real study.
@Incurrence The system is such that many professors do not care about teaching but only research, which is bad for students.
@Incurrence The professors get paid high salaries for publishing papers only a few people in the world read and understand, and they don't help the students who paid huge tuition fees.
@JasperLoy That isn't the case here fortunately, my professors seem to love teaching(and researching) - I feel I would love teaching and researching also, and I hope to do this one day
11:13
@Incurrence The past 8 years, my mind has been so confused I don't know who I am anymore. I will see the doc next week to get more meds.
"Once you lose who you are, only then, can you find yourself"
@Incurrence I think many people read the transcript. The whole world knows I am mad by now.
@Incurrence This actually makes sense to me right now.
Essentially noone reads the transcript. How are the meds going?
@JasperLoy My pretending to be wise quotes I am making up?
@Incurrence Makes me less anxious.
@Incurrence That one really does make sense.
That's good haha
11:16
@Incurrence It's true that when you reach rock bottom like me, you then can skyrocket.
@Incurrence They do, that's why there are many flaggers.
@Incurrence Do you use Linux?
"One only truly understands the feeling of rock bottom when they start to ascend"
I don't just Windows 7
@Incurrence This one sounds crap.
@Incurrence I will be installing Debian 8 on Apr 25.
@JasperLoy Oh you are right, there may be multiple minima
Pretending to be wise is a hit and miss business
@Incurrence People without OCD who tell you they know how you feel are talking crap, including the professionals.
@JasperLoy I know, it's the difference between sympathising and empathising
11:20
@Incurrence When you do not follow their methods, they then say you are not getting well because you are not listening to them, when it is actually the case that these methods just do not work.
@Incurrence I do not think the mental health professionals are very smart people. If they were so smart, they would have gone into mathematics and gone mad instead.
Hello everyone.
@Incurrence I am going to eat dinner with my mum now and then I will take a nap, good night.
Hello @Nickolas and aha @evinda.
Goodnight @JasperLoy
Hello @Nickolas and aha @JasperLoy. :p
12:05
Good morning everyoone
Hi @teadawg1337
@evinda Did you see my exciting news yesterday?
No, I didn't.. @teadawg1337 What news?
@teadawg1337 Congrats on research
@evinda I met with my academic advisor last week to find someone to review my work, and she reached out to colleagues and emailed me yesterday. One of the professors interested in reviewing my work is the mathematics department head at Middle Tennessee State University
12:12
@teadawg1337 Congratulations :) What is your work about?
@evinda It's mainly been about special functions thus far, mostly in relation to the Basel problem
I don't think any of it is too groundbreaking, but what do I know?
12:34
Greetings
Greetings
Greetings
I am finally learning what seems like a lot of math at once :)
12:50
what are u studiyng?
Right now, I think topology haha, but it is in my complex analysis book
oh nice, what topology?
@teadawg1337 :)
bananas, did you do 2.1.10?
Topology of the complex plane probably
12:57
Just relatively trivial stuff, like defining a domain, and I was learning a little measure theory intro stuff
And the people emerge lmao
@BalarkaSen no, i tried but havent gotten far
The stuff I am learning right now is trivial because I started this textbook recently
now im learning about relative homology groups
@Incurrence it's not trivial
@iwriteonbananas Domains?
relative homology is good stuff, @iwriteonbananas
@iwriteonbananas ok, let me know if you want a hint.
12:59
@Incurrence it's only trivial after you understand it
everything is trivial once you understand it :P
@iwriteonbananas It is trivial considering what the people in here talk about haha
@BalarkaSen yeah it's really kewl...i dont feel like doing ex. 2.10...
but mb i'll ask u for a hint in a bit lol
I think Balarka couldn't help me for the first time today, so that's something(the measure theory question haha)
@Incurrence people in here talk about their feelings
13:00
try it first. an edge shares only two triangles : how can you make up a local homeomorphism with R^2?
@Incurrence yes, good point : i totally dunno any functional analysis. you're ahead of me on that.
@BalarkaSen i dunno but it seems obvious
@BalarkaSen I don't even know if it is functional analysis, it was in my real and complex textbook by Rudin
ok : i dunno any measure theory either ;)
i know measure theory if u got a question
13:02
Oh I coined the term today, a measurology, I am pretty famous now
@iwriteonbananas
2
Q: What do we call $\mathfrak{M}$?

IncurrenceI am starting to learn some measure theory, and I was wondering if there is a name for $\mathfrak{M}$. I have the definition: A collection $\mathfrak M$ of subsets of a set $X$ is said to be a $\sigma$-algebra in $X$ if $\mathfrak M$ has the following properties: blahblah If $\mathfra...

@iwriteonbananas surely, but you have to set up a homeomorphism. it's not hard, i agree that it's obvious. but this is only the first part of hatcher's claim in the paragraph on pg 108-109
@BalarkaSen how do you want me to write down an explicit homeo?
well, that interior points of the triangles are manifold points is obvious. but what about the points on the edges of the triangles? what about the vertices where several vertices of the triangles gets pasted?
@Incurrence what do you mean? $\mathfrak M$ is a sigma algebra
@BalarkaSen it seems obv that the points on the edges have locally euclidean nbhds
ok? prove it?
what are the nbhds you're going to choose?
13:05
ok if a point is on an edge it has a positive distance to the nearest vertex. let's choose the ball with radius half that distance
it's gonna look like the plane if the edge is only shared by two triangles
how is that homeomorphic to R^2?
[ok, it works. but you still have to prove it]
what do you wanna do?
@iwriteonbananas I mean that i)$X\in \mathfrak{M}$, ii) If $A\in \mathfrak{M}$ then $A^c\in \mathfrak{M}$, iii) All unions are in $\mathfrak{M}$
How do you write the cool 'M' letter
\mathfrak{M}
13:08
@Incurrence yeah i know, but u asked what we call $\mathfrak M$....we call it a sigma algebra
It is called a measurology
i dont get the question :D
@iwriteonbananas i'd choose a small disk around the point so that i can set up a homeomorphism to R^2 piecewise...
$\mathfrak{@Incurrence}$
@iwriteonbananas Is it the only sigma algebra?
13:08
@Incurrence not necessarily...in fact in all interesting cases it isnt
@iwriteonbananas Then it is a measurology :)
@BalarkaSen what do you mean by piecewise? isnt that what i did?
you didn't set up a homeomorphism at all :P
Are there sigma algebras not based on measure theory @Iwrite
13:09
piecewise on the interiors of the triangles, i mean
@Chris'ssis: out of curiosity, did that integral i posed prove at all interesting to you?
you choose a disk, split it up into two bits, one lying in one of the triangles, the other in the other triangle. now you map these two individually to R^2 and ensure continuity by gluing lemma (or whatever it is called)
@BalarkaSen ok, but do we really need to write that out? the interesting case seems to be when we consider a vertex where several vertices of a triangle get pasted
@Semiclassical Yeah, sure, but I didn't have a chance to work much on it. Trying hard to finish my book with the problems already prepared.
@BalarkaSen yeah
13:11
makes sense
@Incurrence people use them to avoid running into trouble with the banach tarski paradox
@iwriteonbananas yes, writing out stuff is necessary. not everything is as obvious as it looks : consider the jordan curve theorem.
@Semiclassical Did you manage to get some results? (excepting for $t=0$)
anyway, use my argument to prove that vertices are manifold points too
you can do that now
eh, not analytically
numerically there are some obvious features
13:12
@iwriteonbananas Sigma algebras circumvent this?
for example, the large $t$ behavior appears to be exponential (though perhaps with different exponents)
but pinning down what those exponents are isn't obvious
@BalarkaSen yeah we just do the same thing u did, except we split the small disk into several bits, one for each triangle that's glued together at that vertex
then appeal to gluiing lemma
@Incurrence indeedio
precisely. now get back to hatcher paragraph on pg 108-109
@Semiclassical Well, I'd be very interested to see the way the closed form for the generalized $t$ looks like. I don't feel there I might find such a thing, although I never like to start doing things thinking like that.
@BalarkaSen what am i looking for now? what's left?
13:15
"Thus $K_\xi$ is a manifold, locally homeomorphic to $\Bbb R^n$, except at a subcomplex of dimension at most $n - 2$." Does this make sense now?
You have just proved it, right?
i'm not certain there is either, frankly. (i have some reasons to think it might be related to the Painleve equations, but not in an obvious way)
@BalarkaSen didnt we prove our space is a manifold at every point?
I see.
i say that b/c of the context in which this problem arises, and prior results in that area. but even for those the link isn't terribly obvious, and here it'd necessarily be even less so.
well, our space was about pasting triangles. $K_\xi$ is obtained from pasting $\Delta^n$s. If $n = 2$, this is a manifold as we just proved. And in general this means $K_\xi$ is a manifold away from it's $n-2$ dimensional points, right?
13:18
so, kind of annouying
@BalarkaSen yeah
btw maybe a dumb question
but
why cant 3 triangle share the same edge after gluing?
@iwriteonbananas in $K_\xi$?
@BalarkaSen yeah, or the space from the exercise
well, in the space from the exercises, we were given that we are pasting edges of two triangle in pairs.
oh yeah of course
what about $K_\xi$?
13:21
for $K_\xi$, have a careful look at the definition in the paragraph on pg 108-109. keyword : cancelling pairs.
nvm, i probably misinterpreted your "ok".
$$-\frac{\pi t^{a+b} \csc (\pi (a+b)) \left(-2^{a+b+1} \Gamma (a+1) \, _2\tilde{F}_1(1,a+1;a+b+2;2)+e^{i \pi b} \Gamma (-b)\right)}{2 \Gamma (-b)},\Re(a+b)<0\land \Re(t)>0\land \Re(a)>-1$$
that's the result? (!)
@Semiclassical above the evaluation done by Mathematica without that $e^x$ in denominator. :-)
13:23
oh poo
what does he mean when he says "maximal collection of such canceling pairs" ?
yeah, unfortunately that e^x is such a pain
very bottom of page 108
well, it combined with the other factors anyhow
@iwriteonbananas he's taking all the collection of such cancelling pairs. at least that's how i am interpreting it now, but i was sure i stumbled upon something uneasy proceeding with this idea, i just can't remember what that is. proceed with this idea, see if you're getting stuck anywhere.
13:28
@BalarkaSen i dont understand...what do you want me to do next? btw. i find that whole paragraph pretty confusing
well, did you understand what $K_\xi$ is? [i agree that the paragraph is confusing, but there is a hidden gem in there]
let me re-read it
Answer question, obtain rep:
0
Q: How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$

Incurrence How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$? We have: $p$ is a boundary point of $S$ means that $$\forall r\gt 0, \exists a \in S, \exists b\in S^c| a,b\in N_r(p)$$ Now the definition of a limit point is: $$\color{green}{\...

the general idea is this : if you have an $n$-chain $\xi$, you can write it up as a sum of the simplices and looking at the pairs that cancel in the sum [<-- nonrigorous] when you consider $\partial \xi$, you identify the sides of your bunch of $\Delta^n$s likewise.
but when $\xi$ is a cycle, $\partial \xi = 0$ by definition, so no sides of the $\Delta^n$s are left unidentified
since $\partial \xi = 0$, your simplices must have cancelled somewhere in the sum, and it must have done so in pairs ($x + (-x) = 0$).
hence, you identify the sides of $\Delta^n$s by pairs.
and the space you end up with is $K_\xi$.
@BalarkaSen ok it's slowly starting to make sense but what exactly do you mean when you say "identify the sides of your bunch of $\Delta^n$s likewise?
13:36
let's take a chain. $a + b$, where $a$ and $b$ are both singular $1$-simplices with boundaries $x_0 - x_1$ and $x_1 - x_2$ respectively.
$\partial(a + b) = x_0 - x_1 + x_1 - x_2$
so your $x_1$s cancel in pairs.
now, corresponding to $a, b$, you pick two $\Delta^1$s.
and seeing the cancelling pair, you identify two ends of your $\Delta^1$s
this is your $K_\xi$
13:38
so you see, for a cycle, everything must cancel, so you identify all the sides of $\Delta^n$s by pairs.
yeah yeah
ok, good.
We have proved that $K_\xi$ is a manifold away from it's $n-2$ dimensional points.
yeah
then hatcher says "a closer inspection shows that/...."
13:40
now let's skip hatcher's next statement about it being orientable.
yes, hatcher then says that $K_\xi$ is also a manifold at it's $n-2$ dimensional points!
this is nontrivial and nonobvious!
can you prove it?
cant we use the same argument from the exercise?
we can't ;) try it.
i dont understand why not
well, this time we have to prove that if you paste a bunch of $\Delta^3$s by pairs of faces, we get a $3$-manifold.
how do you propose we use your argument?
oh yeah i see what u mean
13:43
mhm :)
it's a bit hard, but you can adjust our argument a bit to fit the need.
oh hey wait.
wait
it does work right?
:P sorry, i have misread the statement
no, no, hatcher's first statement was that $K_\xi$ is a manifold away from it's $n-2$ dimensional points, but that is obvious as it amounts to saying that pasting 1-simplices along pairs of it's faces (endpoints) is a manifold. this time he says that it's a manifold on it's $n-2$ dimensional points, so that means that he asks to prove that pasting 2-simplices by pair of it's faces (edges) is a manifold, so it's just exercise 2.1.10 (a) :P sorry
ok ok
whats next?
@iwriteonbananas so what you should think about is that if it works for $n-3$ dimensional points too.
i claim that there is a $K_\xi$ which is not a manifold at it's codimension 3 points.
anyway, back to the point :
yea hatcher says that
13:49
@iwriteonbananas and you should try to find a counterexample :)
ok : for every cycle $\xi$, we have a compact, orientable space which is manifold away from it's codimension 3 points.
@Chris'ssis: the thing which makes it really difficult, i suppose, is that 1) the branch point structure of the integrand near zero is messy, 2) there's an infinitude of branch points along the imaginary axis due to the E^x.
plus, if we map each of the $\Delta^n$s with which $K_\xi$ is made of to our space $X$ by the maps coming from singular n-simplices, we can get a continuous map $K_\xi \to X$.
agree?
yeah sure
oh now i get that statement
"induces a map...."
corollary : $1$-cycles in $X$ are images of circles in $X$ (a corollary of which is the Hurewicz theorem) and $2$-cycles of $X$ are images of compact orientable $2$-manifolds in $X$.
13:53
@Semiclassical Indeed. Did you check the table of integrals to see if anything like that (some similar stuff) you find there?
oh yeah that's cool
@iwriteonbananas isn't it? that's a good geometric way to think of homology.
yeah, i looked through gradshteyn & rhyzik.
i might've missed it, of course
now the thing is, this great geometric realization doesn't work for $n > 3$ (a counterexample you have to find by yourself)
@Semiclassical Do you have the last version? I don't have it ...
13:55
reminds me of $\pi_1$ and $\pi_2$
not sure if i've got the most recent one
so how about taking something simple : redefine your cycles to be maps from compact orientable 3-manifolds to $X$.
the thing i had initially thought was that, if i couldn't get it explicitly, i might at least be able to expand it in some infinite series
which could be integrated term-by-term
you can define a pretty odd homology theory from these cycles. it's called bordism. google it if you want.
yeah ive read that word before
its in hatcher too
13:56
technically, you won't get a homology theory because the dimension axiom will fail miserably, but it's nice.
@Semiclassical You mean you have a way to go there?
@iwriteonbananas except that $\pi_2$ is about images from spheres, whereas $H_2$ is about images from arbitrary compact orientable $2$-manifolds :)
indeed but compact orientable 2 manifolds are homeomorphic to the sphere or sphere with n handles attached
well, i had in mind the usual way one shows that an integral like $\int_0^\infty \frac{x}{e^x+1}\,dx$ is equivalent to an infinite sum
13:58
so it's somewhat similar :P
yes, but the handle attached thingys are not very homotopically nice. however, you do indeed get a map $\pi_2(X) \to H_2(X)$. the generalization of this map is $\pi_n(X) \to H_n(X)$, called Hurewicz map for arbitrary dimensions.
@Incurrence what's your def of a boundary point?
i.e. pull out $e^x$ from the bottom, and use the fact that $0<e^{-x} <1$ for $0<x<\infty$ to expand in powers of $e^{-x}$, and integrate term by term
@BalarkaSen yeah i've seen that
for the proof that $H_1$ is the abelianization of $\pi_1$

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