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01:58
In Gallian there is a question where you are supposed to use $S_3$ to solve:
"Give an example of subgroups H and K of a group G such that HK is not a subgroup of G."
But... even if you take $H = \langle (1,2) \rangle$ and $K = \langle (1,3) \rangle$ how do you join them... ah never mind... it's a coset thing
Kind of stupid question as $HK$ is sorta ambiguous in the context of the question as H and K are groups...
nvm
02:12
(Oh it's internal direct product)
02:30
In the equation $(at+1)({q^{1-b}-\frac{Qa(1-b)}{q^b}})=q$, is there a general solution for a when the other variables are known rationals? I am able to find values of a using numerical methods such as Newton-Raphson/Secant/Bisection but I'd like to know if it's possible to get an exact number through other means
Also $b$ isn't guaranteed to be an integer
03:00
@MikeM: Clearly I'm wrong about that leaf space comment. There's a theorem of Haefliger and Reeb that says that the leaf space of a plane foliation is always a $T_1$ space.
03:18
Hi @TedShifrin
@EricStucky Nice new color.
@TedShifrin What we wanted to say was that every closed set that contains one of the non-vertical leaves is either that leaf alone, or the whole space. I guess your claim is probably a consequence of a careful application of Poincare-Bendixon, though I tried to prove it for a little and had difficulty.
But when one starts doing foliations one needs to be extremely careful about what sort of regularity you have, and their result no doubt applies in far more generality than PB.
I recall but do not know a reference to another theorem of Haefliger (he had a couple theorems) that there aren't actually any real analytic foliations of any sphere.
Oh, the statement of PB is differentiable real, not real analytic... oops. I'm not very comfortable with this.
03:40
1
Q: Trouble with algebra in showing a sequence has no convergent subsequences in $C^{1}[0,1]$

Jessy CatThis question is related to one I asked earlier here. Specifically, I am trying to show that the differential operator $\displaystyle Tx(t)=\frac{dx}{dt}$ is not compact when $T: C^{2}[0,1]\mapsto C^{1}[0,1]$. To do that, I am attempting to exhibit a bounded sequence in $C^{2}[0,1]$ whose image ...

03:54
hi
are there some numbers which are integers in base 10 but not integers in other bases?
04:17
@user19405892: Maybe this question would be useful to you
 
1 hour later…
05:32
@TedShifrin OK, here's a better way of saying what we wanted to say: The only closed sets are points and the whole space. Still not what we said. But oh well.
05:44
Still wrong. Blah. I'm going to bed.
G'night, @MikeMiller.
I probably can't sleep. But I'll stop trying to say words that end up wrong.
 
1 hour later…
@JessyCat It was a nice question. No need to delete it.
Well I guess unless you figured out the answer in the meantime.
07:26
Is there a nice elementary proof for the E & U of a logarithm outside of calculus?
E & U?
i.e. $\forall\, a\in\mathbb{R}^+,\;\forall\, b\in(\,\mathbb{R^+}\backslash\{1\}\,),\;\exists !\, x\in\mathbb{R},\;a=b^x$
so existence and uniqueness ^.^
typically this requires IVT but a semi-related question had me wondering what else might work
dunno, I'd just do intermediate value theorem too. blah^x is a continuous function.
The intermediate value theorem is calculus?
@MikeMiller yes. what would you call it?
07:34
"Standard"? "Nearly axiomatic"? If someone pushed me, analysis.
In any case, how do you define $b^x$?
by handwaving
Probably good to start there before asking about how to define its inverse.
i guess it's labeled as a "calculus" topic because of the continuity and is often introduced in undergrad calc courses
'morning
a short notation question
when mentioning differentiability/continuity of curves in space
is it better to use the subscript or superscript? i.e.: C^n or C_n ?
The former.
07:39
perfect, thanks
@MikeMiller seems like that preliminary def requires limits
which makes the "no calculus/analysis" modus a lost cause
@Brody Yes. You are going to have serious trouble defining an exponential without something.
The reason for this is not smart: it's because the definition of $\Bbb R$ is artificial and you have to take it into account into your definition of whatever functions you want to define on it.
Sometimes these functions are trivially built into $\Bbb R$; addition and multiplication, say. But nothing else.
@MikeMiller sure but that's not too relevant for simply proving the statement. why not just assume the engine is running and the belts are turning like usual
I am not sure if "do this thing which naturally requires [something] without that [something]" is a good idea in general. Is there a particular reason you're trying to do it? I mean then maybe we'll be able to help better.
I suspected there was no alternative, but math.se users are good at surprising me. just thought it might bring up something interesting
anyway, I'll just assume all the desired properties and nuances of real exponentiation without issue. otherwise it feels like overkill for me
08:02
@BalarkaSen this question seems awfully familiar...
8
A: Why Is This Set Not a Manifold?

Mariano Suárez-AlvarezA 1-dimensional connected manifold has the property that when you remove any one of its points you get at most two connected components. Prove this and then use it to show that your set is not a manifold.

The guy's using a different definition of a manifold, and I am not sure if you see the correlation with the definition we used, though. You need to know the implicit function theorem to see it.
is $\mathbb Z_9 \into \mathbb Z_5$ the only group of order $45$?
Apart from $\mathbb Z_45 $ i mean
@BalarkaSen So what it is that you're doing this week?
08:11
Nothing, as of now. A little less pressed, but not quite that much to start learning serious mathematics yet. Sorry.
I'm not pushing you, I'm asking because I'm curious is all.
Well, in that case, I am planning on watching Seventh Seal. That's all I got.
I spent most of today failing to answer an MSE question and watching mega64 videos on youtube.
Fantastic movie.
Hmm, never heard of mega64 videos.
@BalarkaSen are the definitions equivalent spare a few minor consequences?
08:18
Yeah, those are equivalent.
I like yours better :p
@BalarkaSen They're video game injokes, mostly. You probably wouldn't be amused.
Sure, but that definition also has it's uses. It basically says a manifold is something which are locally level sets.
Frequently three guys humiliating themselves in public. They've been terrorizing one particular Subway for a decade.
Ah, like those troll videos?
08:22
Not really.
They're harmless.
Impractical Jokers are well known for public self-embarrassment
ooo mega64 has nice content
They probably have the cops called on them for every one of these videos, even though they don't really do anything to other people.
disturbing the peace is capital crime to some people
lol are they singing the element song from Lehrer?
Probably not, no.
08:37
yeah apparently not.
That channel frequently has adult content. Instead I'll direct you to youtube.com/watch?v=CXerF6crDRs
fun for the whole family
I find this more amusing and entertaining, I'll admit.
Which is what I am watching since 8 in the morning.
@MikeMiller well I can't say it's not relatable
Huy
Huy
@MikeMiller: why aren't you asleep
god knows
why aren't YOU asleep
08:52
who's "asleep"? never heard of them.
we're having a grownup discussion
Huy
Huy
^
it's 11am here
also the sun shines directly in my face in the morning
hence I'm up since 6
aw shit that's gonna do it to me too
asleep is a grownup thing? that makes sense
i should put up some impromptu blinds
08:55
silly diurnals
09:19
Anyone gone skydiving? Will you ever?
We consider the $\mathbbb{Z}$-module $F=\mathbb{Z}$. This is free with basis $\{1\}$.
What is a generating set of that?
09:35
@Huy is $\mathbb Z_9 \into \mathbb Z_5$ the only group of order $45$?
Apart from $\mathbb Z_45 $ i mean?
Huy
Huy
@Paradox101: I'd have to think about that, but I don't think it's true. but I'm otherwise occupied at the moment, so I can't really give an answer.
I have also an other question... Is every set that contains just one element of $\mathbb{Z}$ linearly independent?
Consider for example the set $\{2\}$. Can this set be extended to a basis?
Ok @Huy
10:28
Is $X\vee Y \vee (X\wedge Y) \cong X\times Y$?
Hi @iwriteonbananas
Hi Balarka
I forgot what $X \wedge Y$ is.
It's $X\times Y/(X\vee Y)$
(pointed spaces)
$S^1 \vee S^1 \vee S^2$ is not homotopy equivalent to $S^1 \times S^1$...
10:31
Whoops, of course
Thanks :P
No problem. What's up?
I'm trying to find a simpler proof for the claim here: math.stackexchange.com/questions/301402/…
Ooh, that is a nice fact.
Uh, suspension over a torus is homotopy equivalent to $S^2 \vee S^2 \vee S^3$? That is not obvious to me.
How do I see that?
10:34
God knows
Ugh, wasted an hour trying to find an easier proof but now I give up
Mike referenced that question in his answer here math.stackexchange.com/questions/1785261/…
Before trying to find a proof I'd like to see the torus case.
which is fantastic, btw.
Ah, OK. Probably won't read that right now tho. Thanks.
I didn't say anything for a good while, but it's sad to realize no one noticed the relation of my user number to the golden ration ...
$1.6180339887498948482045868343656381177203091798058 ...$
These simple observations make the difference.
The number that defines me best.
Anyway.
10:41
@iwriteonbananas Thoughts. I suspend the torus.
Look at a meridian. That suspends to an $S^2$. Same for longitude.
Yes, true
So I get a subspace which looks like $S^2 \vee S^2$ inside $\Sigma T^2$.
@robjohn this year was EPIC to me in terms of mathematical discoveries (by far the best one). I wonder if I can do more than what I did this year (I also wonder if there is a point where a possible regression may happen - that scares me a bit).
The attaching map of the 2-cell in $T^2$ is $S^1 \to S^1 \vee S^1$ given by $[a, b]$. This is not nullhomotopic, that is why $T^2$ is not htpy eq to $S^1 \vee S^1 \vee S^2$. So it seems it's worth understanding what resolves when we suspend. What does this map become in the suspension?
Is it suspension of $S^1 \to S^1 \vee S^1$? Ish.
Something will become nullhomotopic after suspension. I want to understand what.
10:47
@BalarkaSen Yeah, we attach a 3-cell to $S^2\vee S^2$ with attaching map the suspension of $S^1\to S^1\vee S^1$
Eh, suspension of $S^1 \vee S^1$ is $S^2 \vee S^2$? Wedge commutes with suspension? Am I being dumb?
Wedge does commute with suspension, right?
Just a moment.
Suspension is a left-adjoint functor and forming a wedge is a pushout
What Does It Take to Become a Good Mathematician?
10:49
...
@iwriteonbananas I think that is true upto homotopy equivalence.
Yeah, agreed
"We uncover some surprising facts, such as the absence of age related decline in productivity and the relative symmetry of international movements, rejecting the presumption of a massive ”brain drain” towards the U.S."
"Contrarily to a widely held belief (among both scientists and lay people) the rate
and quality of mathematical production does not decline rapidly with age. For
mathematicians who remain scientifically active, productivity typically increases
over the first 10 years, then remains almost constant until the end of their ca-
reer. However there is a substantial attrition rate (i.e. mathematicians who stop
publishing) at all ages."
@iwriteonbananas Are you sure about this? I am feeling a bit queasy.
Yeah I'm pretty sure that $\Sigma(X\vee Y)\simeq \Sigma X\vee \Sigma Y$
How to detect a promising mathematician?
"There are some indicators of future success for a young mathematician, among which,
obviously, the quality of his publications."
10:55
I trust you on this one. In any case it is obvious that suspension of $S^1 \vee S^1$ is htpy eq to $S^2 \vee S^2$, and that's all I need.
So, we just want to prove that suspension of $S^1 \to S^1 \vee S^1$ given by $[a, b]$ is nullhomotopic?
Ok, so I guess now you should try to prove that suspending that attaching map gives something nulhtpic
Ok, this is definitely doable. Hmm.
Higher homotopy groups are abelian, if that helps
That is good point.
Very nice point, actually. But let me try to explicitly see the map - it doesn't seem too hard.
11:02
Yeah, it is exactly $S^2 \to S^2 \vee S^2$ given by $[a, b]$ ($a$ and $b$ are the element of $\pi_2(S^2)$ represented by the identity map on each of the two $S^2$s)! Which is 0 because what you said!
No more loss of time. Back to my work.
Explain what you mean by "given by $[a,b]$" please
Ok, you edited it in
Man it's raining like crazy here
see for suspending $S^1 \vee S^1$ you first look at $(S^1 \vee S^1) \times I$ which is two copies of $(S^1 \times I)$ glued along a parallel $\{x\} \times I$.
Right
It seems if we collapse $\{x\}\times I$ in the suspension we get exactly $S^2\vee S^2$
when you do that, the map $S^1 \to S^1 \vee S^1$ becomes $(S^1 \times I) \to (S^1 \vee S^1) \times I$ which wraps around the two cylinders like a commutator, but now the basepoint is whole of the $I$ the two cylinders are glued along instead of a point ('cause it's a map from $S^1 \times I$, duh)
Then after we pinch the tops and bottom and collapse that $x \times I$, it becomes $[a, b]$, and basepoint is the point we pinched it to.
I suppose that's a bit vague but that should be the picture. I think you already see it.
11:08
Yeah, makes sense
Ok, I'm happy.
But note that this nontrivially uses a fact about higher homotopy groups. So I wouldn't be surprised if the general case is hard.
I was hoping to do it with some abstract nonsense but my argument didn't pan out :/
awww yeah abstract nonsense sucks sorry to hear that.
@robjohn I almost forgot to ask you: did you manage to calculate my limit?
11:11
Can I think of $X \times Y$ in general as attaching stuff to $X \vee Y$? Don't know, just a wild idea ($X$, $Y$ are CW complex).
Not sure if the argument generalizes in any way.
I would be surprised
Me too, I guess.
I mean the quotient (X smash Y) doesn't look nice.
You could probably think of it as attaching a bunch of product of cells of $X$ and $Y$ to bits of $X \vee Y$. One could try to carry out the torus argument for each of those pieces.
Even if it works, it would be a tedious combinatorial argument - figuring out what can be all the possible scenarios.
Man I love this song.
Smash products are pretty ugly
But it has similar categorical properties as, say, the tensor product of vector spaces
@BalarkaSen TIL that the world population was ~3billion in 1967.
11:28
I don't think we have discovered a bomb yet which wipes out all the civilization.
That being said it is unlikely that he meant world population when he said that.
It was 3.4-3.5 billion at the time
ok, surprised. i thought that number should have been a lot bigger.
11:45
What're you working on today, Balarka?
Watching cartoon. :)
How dare you enjoy your childhood
Who says one needs to be a kid to watch cartoon.
Hey guys.
I think I'm just being a bit thick about something
But, like, just to be sure: is it, in fact, true that if a sequence converges under the uniform norm, then it also converges under the l_1 norm?
@BalarkaSen Hey, easy homology question for you
I'm asked to show that $H_n(X)\cong H_{n+1}(SX)$ where $SX$ is the suspension ($n\geq 0$)
11:56
Mhm.
I have one big problem with this:
Shouldn't I be doing this using the long exact sequence on homology?
You should use the Mayer-Vietoris sequence, yes.
Since this is, I think, a "good pair", I would do this with relative homology
But I suppose you can also use the long exact sequence.
But then doesn't this imply that $H_n(X\times I)=0$ for all $n>1$ at least
(which is false)
11:58
@Danu What is a good pair?
Closed, nonempty and defo retr. of a neighborhood
(the subset must be)
No, I mean, what is your pair in the first place? What are you long exact sequencing?
I think you've got the wrong pair in mind.
Oh, I wanted to use $X\times I$, $X\times \{0\}\cup X\times \{1\}$
Yes, not the pair I would use. But sure, you can use it.
It doesn't imply $H_n(X \times I) = 0$.
How are you getting that?
Yeah, it doesn't, I think
Well, what I wanted to do is this
12:00
I don't think it's a good idea, the quotient space is not the suspension of $X$ (sorry to interject so rudely)
@iwriteonbananas It's not?
Okay
Oh, yeah
You're right.
No, in the suspension you succesively collapse each of those subspaces to a point, not both of them to one point
That too.
I just realized
Maybe try using $(CX, X\times \{1\})$
12:02
Why don't you use $(CX, X)$ instead?
high five
So why can I just ignore one of the copies?
You collapse the top of $X \times I$ to get $CX$ and then collapse the bottom.
Doing the identification step by step realized $SX$ as a quotient of $CX$.
Oh, sorry, I don't know what $CX$ is ;)
But I guess it's the cone
The cone on $X$. $X\times I/X \times 0$.
12:04
What does $\mathbb{C}[x]$ mean?
So then the cone should have trivial homology, right?
Ring of polynomials in single variable $x$ with complex coefficients, @JesterTran.
@Danu Yes, in fact it's contractible.
@BalarkaSen thx
@BalarkaSen Obviously, sigh. Thanks.
It feels so bad when you miss the nice tricks.
You have to ponder on it a bit first.
12:05
I did, but I was fixated on using $X\times I$
...which is hopeless, since $X\times I$ doesn't have trivial homology
"obvious" is dangerous word. something is obvious once you understand it, not before.
so I was weirded out.
@BalarkaSen Obvious that it's contractible, I mean.
(that is quite obvious)
@Danu What made you think that it is contractible?
@Danu X times I has the same homology as X in fact.
Not relevant, just worth noting.
ooh I do have a new color :D
I have no idea why that happens
12:15
@Mambo What?
@BalarkaSen Yeah, of course (since it defo retracts)
12:26
@user1618033 No, I'm sorry. I've been busy. I will be out of town today as well.
Hello @robjohn
If $u \in W^{3,p}(\mathbb{R}^{+})$, I want to construct the catoptric extension $Eu$ of $u$in $\mathbb{R}$ such that $Eu \in W^{3,p}(\mathbb{R})$. Could we pick $Eu(x)=u(|x|)$ ?
 
1 hour later…
r9m
r9m
13:29
@robjohn It was from quite a while ago :) I figured how to get it .. (in fact managed to get a problem related to it in I&S here ) :-) Thanks!!
@user1618033 It has a nice closed form?! :D I'd be surprised!!
whoo boy
@BalarkaSen In case you're around, I've got another small point of confusion:
Collapsing a subset to a point does not induce an exact sequence of chains, does it?
Since the chains in the subspace will just map to constant chains, but not zero necessarily
@r9m Are you back? :-)
Without that, I don't see how to get to the long exact homology sequence
13:44
@Danu Eh. Collapsing $A \subset X$ to a point gives $X/A$.
@robjohn OK. It's better to take that when you are in an inspired day, it's not that easy.
For good pairs $(X, A)$, that gives you an LES $H_*(A) \to H_*(X) \to H_*(X/A)$.
'Cause you have a short exact sequence $0 \to C_*(A) \to C_*(X) \to C_*(X, A) \to 0$ at the level of chains.
Okay, that's fine.
So not sure what you're asking.
But in fact, on my problem set, this exercise I was talking about earlier
it comes before talking about "good pairs"
So let's say I cannot work with relative homology yet
Then do I somehow get the LES anyways?
13:47
I can't help you if you don't say what tools you have.
@BalarkaSen Okay, so let's say I'd like to avoid $C_*(X,A)$
@Danu What version of long exact sequence do you know of?
The standard one includes relative homology terms.
@BalarkaSen Eh.. The one I get from a short exact sequence on chains.
@BalarkaSen Sure, I know one for relative homology, too
But we have not discussed the relation between $H_*(X,A)$ and $H_*(X/A)$
@Danu tell me about $A$. Is it contractible?
Er, OK. Do you know Mayer-Vietoris?
13:51
@BalarkaSen Yes
Oh, good!
Then try that out for $SX$.
@MikeMiller No; I'm proving that $H_n(X)\cong H_{n+1}(SX)$; Balarka already showed me how
gotcha
Now I was just having some small doubts about how to go to the LES
I would actually have thought to use MV long before relative homology
13:52
Because the "topological quotient" doesn't induces a SES on chains
I am just overly fond of relative homology long exact sequence :)
...it was my suggestion, too.
0
Q: How to solve Fredholm Integral Equation of the Second Kind in $C[0,1]$

Jessy CatI need to solve, in $C[0,1]$, the equation $\displaystyle x(t) - \lambda \int_{0}^{\pi}(\sin t \cos s)x(s) ds = \sin t$. Adding the integral part to both sides, I obtain $x(t) = \sin t + \lambda \int_{0}^{\pi}(\sin t \cos s)x(s) ds$, which is, I believe, a Fredholm Integral Equation of the secon...

I'm going to try to just integrate the thing but I'm not sure that's how you're supposed to do it.
@MikeMiller @BalarkaSen In my mind, MV is somehow stored as "consequence of LES + excision" but I guess that's wrong
No, that is right.
13:57
Oh, so how do I not need to first get an LES?
It's excision with steroids. In it's working form.
So, in my mind, I need SES -> LES -> MV
I don't understand the question. You presumably got a long exact sequence from a friend in the mail already. Now you just use it.
@Danu I am confused. What are you trying to do?
So if I don't have a SES I don't know how to get to MV
@MikeMiller Sadly, I didn't :P
I don't see how collapsing the subset to a point induces a SES on chains
(since constant chains are not zero, right?)
13:59
I mean if you know M-V (as in you have covered it in your course) then you're through. Just use it.
Mayer-Vietoris has nothing to do with the relation between $H_*(X/A)$ and $H_*(X,A)$. The latter doesn't come from an exact sequence, it's something you sort of just prove with your teeth.
Right---the SES from MV is not the same SES that I was thinking of using
r9m
r9m
@user1618033 yes ... now I am! Sorry .. I missed your second message ..
There is no direct chain level SES coming from $A \to X \to X/A$.
Exactly
That's what I was worrying about
14:01
But I thought you were not trying to use that?
But I don't need that for MV, as I see now
Yeah, I was confused why you'd bring that in MV.
Thanks for bearing with me, both of you.
@MikeMiller We've finally arrived at principal bundles now, in math. gauge theory :D
No problem, @Danu.
Always glad to be of (no) help.
14:05
So what's happened in the past n hours
There is a quick proof of $H_*(X/A) \cong H_*(X, A)$ I can tell you if you want after you settle that $SX$ thing with MV, @Danu.
Or maybe you'd want to figure it out yourself at some point.
@BalarkaSen Well, that's my second exercise.
Ah, great, then.
r9m
r9m
@user1618033 come to yahoo chat why don't you ..
@MikeMiller bananas told me that $\Sigma(X \times Y)$ is homotopy equivalent to $\Sigma X \vee \Sigma Y \vee \Sigma(X \wedge Y)$. I was pretty surprised initially as I couldn't even see it for $T^2 = S^1 \times S^1$ but a closer look told that it holds because the attaching map of the 2-cell in $T^2$ suspends to a nullmap as the commutator in $\pi_2$ (which is abelian!) is trivial.
Pretty charming, to me. I don't know how to prove the general thing though.
(he linked a MSE answer too but it looked complicated)
14:10
I used that in an answer I posted yesterday. There's a very short proof in Hatcher, let's see if I can remember it.
OK, take the join X * Y. Glue cones of X and Y onto the ends of the join. Collapsing both cones (which gets you something homotopy equivalent, since they're contractible) gives you $\Sigma(X \times Y)$. But if $x_0, y_0$ are the basepoints, then collapsing the cones $x_0 * Y$ and $X * y_0$ also gives you something homotopy equivalent, and looking at it carefully you see it's precisely $\Sigma(X) \vee \Sigma(Y) \vee \Sigma(X \wedge Y)$.
the cones on the ends are the first two terms; the join becomes the second term
The join is the tetrahedral thing right? $X \times Y \times I$ with one end $X \times Y \times 0$ collapsed to $X \times y \times 0$ and the other end $X \times Y \times 1$ collapsed to $x \times Y \times 0$? I am rusty on this.
yeah, this is one of like two places I have seen the join as a worthwhile notion.
The homotopy theorists tell me it's useful, I guess.
@MikeMiller This makes sense. Really cool idea, no idea how I could have come up with it.
Well, we're not Whitehead or whoever found it.
It seems easy for me to believe that $\Sigma(X \times Y)$ should simplify somehow, that's why I googled it in the first place... and then some honest hard working homotopist figured out precisely how
There's an insane amount of stuff in the last appendix.
You mean section 4L?
14:22
No, I mean in section 4.*
It certainly seems to contain a lot of things I do not know. At some point of time not anytime soon I should sit down and read section 4.
I know what's in 4 but there's a lot of very clever stuff in the appendix I do not know.
That's intimidating.
But I have a feeling I should keep watching cartoon for now. Reaching a maximum cartoon watching streak seems important somehow.
I'm going back to bed.
I guess my wisecracks aren't panning out.
14:38
Follow-up: From the exercise it seems that $H_0(SX)=0$. Is that true?
what does $H_0(A)$ mean, for any $A$?
Reduced homology, yes.
@MikeMiller The constant simplices that are not boundaries?
What does being a boundary mean for formal combination of 0-simplices? Geometrically?
Connected by a path, I guess
14:41
Bingo.
Ah, and $SX$ is path connected
Yes.
Note that this means nonreduced homology $H_0$ is $\Bbb Z$. But by definition reducing it gives $0$.
So $\tilde{H}_0(SX) = 0$.
Yes, thank you :)
@r9m I try to finish my stuff ... (and better stay away from such a chat)
14:55
Almost forgot ...
If $P$ degree$>2$ and $P'$ has all real roots, then

$$\frac{\displaystyle \int_a^b \frac{1}{P'(x)}\ dx}{\displaystyle \int_a^b \frac{1}{P''(x)}\ dx}>\frac{P'(b)-P'(a)}{P(b)-P(a)}$$
Be talkative please ...
I'm out now.
@user1618033 If you'd like to appeal this, use the contact-us link at the bottom of every page.
Blatant abuse toward moderators isn't going to be tolerated.
@Undo Where is the abuse? Tell me a single word of the abuse! Or you don't care and think that if you're a moderator then you're automatically right?
"my opinion is that you are pretty incompetent as a moderator. "
@Undo May I have an opinion? I have this right?
... but I'm not going to argue with you here. Use this page to appeal this.
14:58
@Undo Please look for what reason I was kicked/banned.
r9m
r9m
@user1618033 'kay :)
00:00 - 15:0015:00 - 00:00

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