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8:00 PM
I just read a bit of Hatcher cohomology.... I am in love
 
You can write it as $(A\setminus B)\cup(B\setminus A)$ or as $(A\cup B)\setminus(A\cap B)$
You're a faster typer
 
I don't like the Chapter 3 as much as I like Chapter 2.
 
@Ali As in, you read about cohomology in the book by Hatcher? Or is there a type of cohomology called "Hatcher cohomology"?
 
Probably the first one, not aware of the second
 
Can you see why that would make ${\cal P}(X)$ a group? (What would the identity be? Or the inverses? Is it associative?) @MeowMix
 
8:01 PM
I love the extra structure it gets just from being contravariant
I had never thought about that before
 
Contravariant.
Also you do get that extra structure in homology too. It's just harder to see.
 
Chapter 3 in Hatcher is when I gave up, to be honest
Which is a shame, because chapter 4 looks like witchcraft
 
Its cool to see the links with homotopy as well though
 
The identity is obviously the empty set, right?
 
@AkivaWeinberger I wouldn't say you need to read 3 for 4
 
8:03 PM
@AkivaWeinberger I felt like giving up too. Too much algebra for me.
You do, @Ali
 
Like it would help, but you can do a bit
 
(I have never read Ch 4)
 
Just skim through the results in 4 to motivate the algebra
 
@Akiva It looks like each element is nilpotent, if that makes any sense from a group standpoint
like each element is it's own inverse?
 
Although I am weird for some reason and enjoy the algebra as well
 
8:04 PM
Yeah @MeowMix
 
hatcher talks about a few elementary methods for computing higher homotopy groups
 
Wait, before you continue, I have good news
 
and there is much discussion about the nature of stable homotopy groups
 
that's the standard "hey, cool math" reaction. the actual stuff scares me off, tbh
 
Is it the algebra?
 
8:06 PM
Yeah. The ext and tor and Kenneth-something
Someone correct me on the name of that
It's been a while
 
Kunneth formula. He's talking about homotopy, not cohomology.
 
well both are mentioned
 
@Akiva My math teacher is letting me go on the computer for independent study for the rest of the year.
 
Its just the way its presented that motivates me perhaps
Like it is so open ended and can be applied to so much more
 
8:07 PM
@MeowMix What does that mean?
Like, instead of class?
 
I don't have to do the work
Yeah
 
Whoa, yay!
 
And showing a few tricks (co)homology theories can give you really motivates studying them for different objects
And homotopy is mysterious and irregular which motivates understanding
 
I'm kind of confused as to how that would work, though
 
I have no idea what you mean but ok
 
8:08 PM
Like, what prevents you from just goofing off the whole time
 
What do you mean?
 
For example you can prove the hairy ball theorem pretty easily with homology
 
Oh.
Well yeah, I could
 
There are ten thousand different proofs of the hairy ball theorem.
 
And that group (co)homology is just (co)homology of the Eilenberg-Maclane spaces
 
8:09 PM
Ah, that's something we're gonna do in difftop
 
Homology theory is useful and interesting and non-scary. I have no doubts about that.
Homotopy is scary to me.
 
@Akiva and I might. But I'm glad I'll have the time to work on math I'm interested in.
 
Well, yay!
 
Homotopy is not nearly as well understood as homology
 
I know :P
 
8:10 PM
Anyway, the thing's associative, do you see why?
 
but doesn't its irregularity excite you?
 
Exciting and scary are not mutually exclusive
 
(Also, it's usually written $A\Delta B$ rather than $A\oplus B$, but whatever)
 
@AkivaWeinberger Well, to start, it's also commutative
 
8:13 PM
Have you seen the Freudenthal suspension theorem?
 
Thats a pretty cool result
 
Reminds me of a joke I saw once
 
It says for a specific $n$, $\pi_{n+k}(S^k X)$ are all isomorphic for large enough $k$, @Akiva.
 
Nice one @Akiva
 
8:14 PM
The suspension map $\pi_i(S^n)\to\pi_{i+1}(S^{n+1})$ is an iso for $i<2n-1$ and surj for $i=2n-1$
 
$X$ being a CW complex.
 
That can be generalised for $X$ and $SX$ rather than $S^n$ and $S^n+1$
 
@MeowMix Hint: Venn diagrams
 
@Akiva Yeah, that's what I was thinking
I don't know how to put it in words, though
 
Here's one argument
 
8:17 PM
It's kind of like you can define a ternary operation of three sets where the resulting set is an element in only one of the three sets
 
I should write $\Sigma^k X$ but whatever. I don't know how it's proved.
 
And that will be equal to all $A \oplus (B \oplus C), (A \oplus B) \oplus C$
 
@BalarkaSen no I think $S$ is correct
 
The Iversion bracket is $[P]$ where $P$ is some proposition, defined to be $1$ if $P$ is true and $0$ otherwise.
So, like, $[1\in\Bbb R]=1$, but $[3>4]=0$
 
@AliCaglayan $SX$ and $\Sigma X$ are homotopy equivalent if $X$ is a CW complex. You should read Ch 0 of Hatcher, really, before jumping ahead to Ch 4.
$\Sigma X$ is just used in the based category.
 
8:18 PM
You mixed "Iverson" and "Inversion" to get "Iversion"
 
Peter May intensifies
 
Hence, appropriate when writing down their homotopy groups.
 
So, the idea is that: for any $x$, $~x\in A\oplus B$ iff $[x\in A]+[x\in B]\equiv1\pmod 2$.
Does that make sense? @MeowMix
 
I guess
 
If $x$ is in neither $A$ nor $B$, then $x$ isn't in $A\oplus B$
 
8:19 PM
oh yeah
 
and $0+0$ isn't congruent to $1$.
If it's in only one, then it is in $A\oplus B$, and $0+1$ is congruent to $1$.
It it's in both, then it isn't in $A\oplus B$ (by definition), and $1+1$ isn't congruent to $1$.
 
This reminds me of the XOR operator
 
So, in any case, $[x\in A\oplus B]\equiv[x\in A]+[x\in B]\pmod2$.
 
Infact, it definitely is.
 
@MeowMix It should
In any case, using the above property twice on each side, we get:$$[x\in A\oplus(B\oplus C)]\equiv[x\in(A\oplus B)\oplus C]\pmod2$$
 
8:22 PM
Oh, I see
 
and, since the bracket only can have value $0$ or $1$, they must be equal.
 
This is just associativity of XOR, huh?
 
Ted is usually on all day!
 
Not really @Dodsy
 
8:23 PM
Oh.
 
Often yes, but he does have a life
 
Well I'm new, so I have no idea.
 
Since the above two quantities are equal, $x$ is in $A\oplus(B\oplus C)$ iff it's in $(A\oplus B)\oplus C$.
 
yep
 
And since that's true for all $x$, the sets must be equal.
 
8:24 PM
@Daminark have you played bombparty
@Akiva mhm
 
Nope
 
I'm played goat simulator once.
 
Oh.
 
I've***
 
Then who was the other guest?
 
8:25 PM
I'm supposed to be doing chemistry so I can get into Uni next year. Everytime I come on Ted says "Dodsy: CHEMISTRY"
 
Hi @Ted
 
Hi @PVAL.
 
TED!
 
Hi @Zach!
 
Did you receive my messages last night?
 
8:26 PM
@Dodsy: I'm not always on all day :P
 
Ohai
@MeowMix Wut
 
Hi Zach.
Are you talking to yourself again?
 
Rehi @Ted!
 
Rehi Demonark.
 
Hi @Ted
 
8:26 PM
Rehi Balarka. It's almost un-sleep time.
What messages, Dodsy?
Oh, yeah. What you're talking about doesn't make sense.
 
Oh okay. :)
 
How's your sickness @Ted?
 
Thanks for going easy on me, Ted!
;)
 
There's no way to make sense of $n^{\bf v}$ where $\bf v$ is a vector in $\Bbb R^3$. Things that work because of complex analysis are very special to $\Bbb C$.
 
Ah I see.
 
8:28 PM
Still sickly, Zach, thanks.
What work did you do during class?
 
Unfortunate :(
@Ted She told me to just do IXL for today, so I did.
Just did some boring chain rule stuff, because it only goes up to calculus
 
What is IXL?
Chain rule is important though.
 
The problem with IXL is that it is so repetitious.
 
Hi @Alessandro
 
8:29 PM
Sounds Aztec
 
I help kids with IXL stuff at the library.
 
Like Tlaxcala and axotl
 
Once it knows you know how to do a type of problem correctly, say, 3 times, it should move on. Kids who mess it up should get to repeat it another 12 times, but, really ...
 
@Ted One of the funnier ones was "find if a function is continuous given its graph"
 
ITS graph, you mean?
 
8:30 PM
NOOOO
I MADE THE MISTAKE
 
Wait @Meow is your teacher telling you what to study?
 
You are now in jail with Balarka for that.
 
Anyone have any insight on the word invert? It could be obvious, I'm just not a math guy. Pipe invert and obvert: Why is it called invert?
 
@MeowMix What grade are you doing IXL for?
 
its confuzzling
 
8:30 PM
@Wilson: In what context?
 
Inversion.
Inverse.
 
I thought it was more, you'd just be able to go and read Hatcher or May
 
Inside out.
 
:P
 
@TedShifrin Pipes, apparently
 
8:31 PM
What bothers me is when people write "I should of" instead of "I should have"
 
I ofn't seen that very much.
 
@MeowMix "For all intensive purposes" gets me quite agitated.
 
@MeowMix Those are pedantics and not necessarily important in the grand scale of things. Like knowing the name of a bird.
 
My least favorite has to be: "Guess what?"
That's not a question.
 
Oh, pipes. It's some convention that those engineers made up. It has nothing to do with usual uses of the word. In- stands for inside. Ob- usually suggests outside or beyond.
So they're thinking about starting at the bottom of the (inside of the) pipe and measuring from in to ob.
 
8:33 PM
@Fargle What if the purposes are very rigorous and intensive?
 
Engineers muddy the waters.
2
 
@Dodsy 8th grade
 
@Fargle I know right?
 
@Fargle: I thought that was the expression when I was in junior high school.
 
@Wilson I know that the obverse of a coin is the other side
2
 
8:33 PM
Oh, the IXL?
 
the one without the face
 
It's not a grade, it's just calculus
@TedShifrin Anyways, for that one, all you had to do is see if there was a break in the graph...
 
@TedShifrin It's alright, I spent more of my childhood than I care to admit thinking that if a surface is coarse, it's "ruff".
 
Well good for you! I didn't learn calculus until this year, (I turn 23 in 3 days), but I did get a 97% in the class.
 
Another fun word is "awry"
 
8:34 PM
I like "akin"
 
a-wry, not awr-y
 
"Judgment" is a cool word cuz no "E"
 
"a-RYE"
 
@Dodsy Was your course rigorous?
 
@MeowMix Side-question: If you have a graph of a function; $f : [0, 1] \to [0, 1]$ say, and if it has no "breaks", need $f$ be continuous?
This needs a rather general definition of "break" but it's still interesting to wonder about.
 
8:36 PM
If "no 'break'" means connected graph, then…
…well, my favorite class of counterexamples
 
Counterexamples?
Oh noes, it's not always true
 
Depends on what break means, for sure.
 
High school (and many college) teachers seem to mess up the actual definition of continuity. They think the function $f(x)=1/x$ is NOT continuous. It's perfectly continuous.
 
@MeowMix It was alright, I did not learn integrals, but a typical "lesson" looked like a few example questions which you had to infer their meaning from. Here's an example of something I'd have to look at to learn projections: imgur.com/a/nFXsR
 
Yeah, hate that.
 
8:37 PM
Forgetting about these functions is a sin$(1/x)$
 
smacks DogAteMy
 
@TedShifrin I think that's just confusion over the nature of the domain and range.
 
No, @Fargle. Just domain.
 
Well, that's $[-1,1]\to[-1,1]$, and you'd want to add in a value of $f(0)$
 
@TedShifrin and @MeowMix, Yeah, I had wondered if some engineers just butchered some words along the way.
 
8:38 PM
Erp, yeah. I'm tired.
 
Continuous means continuous at every point of the domain.
 
@Ted I mean, over its domain it is
See, I used the correct it[']s!
 
Well, that's what continuous means, Zach.
@Wilson: Etymology is always interesting.
 
but any number from $-1$ to $1$ works as a value of $f(0)$ if you want to keep the graph connected.
 
You can't decide if something is continuous or not at a point it's not defined, is the point.
 
8:39 PM
@MeowMix So it was mere computation and rarely ever proofs. I did poorly in Advanced Functions (93% avg.) because they never explained or even showed a picture of the unit circle. In a sense I am self taught, never having heard a lecture.
 
@BalarkaSen "Not continuously extendable" is a thing, I guess
Wait, no
That doesn't work either
 
Good luck drawing it "without lifting your pencil from the paper," DogAteMy.
 
@AkivaWeinberger Can I have a counterexample?
 
@BalarkaSen Right, but it's taught in calculus classes as being merely non-continuous.
 
He just gave one, @Meow.
 
8:40 PM
Sorry I didn't see
 
since they probably wouldn't call $f:\Bbb R\setminus\{1\}\to\Bbb R$, $f(x)=x$ "continuous" either
(though it is)
 
I'd say the sine function is continuous!
 
@AkivaWeinberger $f(x) = \frac{x(x-1)}{x-1}$
 
Relevant terminology to google is "topologist's sine curve"
 
A counterexample so famous, it's the cover of a book of counterexamples.
I really like the TSC.
 
8:41 PM
@Dodsy $\begin{cases}\sin(1/x),&x\ne0\\0,&x=0\end{cases}$ is not continuous at $0$, but the graph is connected.
 
No value is continuous, right?
 
hm.
 
@Fargle AKA the "really wiggly thing"
 
Here's another common misconception that's slightly easier. If I have a graph of a function $f : [-1, 1] \to [-1, 1]$, and I see a "kink" at the graph (like a bend, as in $|x|$) I can immediately say $f$ is not differentiable there. But if the graph has no "kinks", is it necessarily differentiable? (ping @Meow).
 
@MeowMix Yeah, there's no value you could make $f(0)$ to make the function continuous
 
8:43 PM
@MeowMix Have you learned integrals?
 
@Dodsy Yeah
 
@BalarkaSen Ooh, that's sneaky
 
If you add to the topologist's sine curve the entire y-axis from -1 to 1, what you get is a space that's connected but not path connected.
 
@MeowMix I bought a college calc book to learn them, I'm not sure why my HS calc course didn't teach them- I think they should be taught with derivatives.
 
Surely the graph of a continuous function on an interval should be path-connected, not just connected!!
No, @Dodsy. You're confusing antiderivative and integral. BIG conceptual difference.
 
8:45 PM
@TedShifrin …Isn't it?
 
@TedShifrin I closed it, it's no longer a graph.
 
The fundamental theorem of calculus relates those two, however @Dodsy
 
It's also bad pedagogy to dump too much on students at once. Their brains can't handle it.
@Fargle, you did, but DogAteMy threw in just one point.
 
I see.
 
@TedShifrin I was thinking about the derivative of a function which is the instantaneous rate of change, and the integral of a function which is the area under a curve.. I have never even heard of this antiderivative!
 
8:46 PM
@Ted I remember at one point where I didn't know why we put $dx$ at the end of every integral until I realize that it's just an infinite Riemann sum
 
It's still not path connected.
(The function is not continuous)
 
@Dodsy Oh, then you're intuition would be correct
 
Oh, in that case, @Dodsy, way too many concepts at once!!
 
^^ @Ted If there's nothing else I fault the IB with, it's that they literally just define the integral as the primitive, and say a definite integral is evaluating it at the endpoints
 
Is the converse true?
 
8:46 PM
Zach ... Watch those apostrophes, seriously!
 
Does path-connected graph imply continuous? (It should…)
 
@Ted I'm way too tired
 
Demonark: That isn't IB. That's an incompetent teacher.
 
@TedShifrin Perhaps you are correct.
 
Since every point in the bar I added fails to path-connect outside the bar, no function value for it at zero would be continuous.
 
8:46 PM
I've been so tired and appetite-less lately
 
I agree.
 
@Dodsy: I've spent about 45+ years thinking about teaching and pedagogy.
 
@MeowMix You've been smoking too many cigarettes!
 
Well, the curriculum for the IB includes Riemann sums only if you choose additional calc for the optional content
 
Zach, you need to eat and drink fluids even if you're not hungry.
 
8:47 PM
@AkivaWeinberger Yes, it is true.
 
Drink fluids, definitely
Also, sun is nice
 
@BalarkaSen Not in higher dimensions.
 
looks outside With coat
 
Demonark: Really? That's appalling. And they call the antiderivative the integral?
 
The part of calculus that's in the core of the curriculum is just antiderivatives
 
8:48 PM
Ah, thanks, @Fargle.
 
@Fargle Well, we can't get everything
 
@TedShifrin Sorry, I didn't mean to give offense. You certainly would know the most about teaching!
 
I'm not offended, @Dodsy. I'm just saying that I don't come by my opinions lightly.
As opposed to yesterday, when I actually was offended by someone.
 
$f(x,y) = \frac{x^2}{x^2 + y^2}$ plus $f(0,0) = 0$ is path-connected but discontinuous at 0.
 
@TedShifrin Let's not dwell on such matters
 
8:49 PM
Right. All kinds of worst things can happen in higher dimension.
 
I did read that, I am sorry about what he said. It was definitely way over the line- and way too infantile for a chatroom like this.
 
Hence, $C^\infty$.
:D
 
I'm just saying I don't get offended by intellectual discussions, DogAteMy :)
 
Yup, the IB has a rather specific syllabus, with allocated (suggested) teaching time for each topic and everything
 
lol you right
 
8:49 PM
@BalarkaSen How do we define a kink?
 
And the material on integration starts with "Indefinite integration as anti-differentiation"
 
@MeowMix A point on the curve (graph) which has multiple tangent lines in $\Bbb R^2$, suppose?
 
Demonark: On the other hand, AP teachers are coached on exactly what to tell their students to write in the free response questions. In particular, on the Mean Value Theorem problem, don't bother wasting your time explaining that the function satisfies the hypotheses of the theorem. Just write the result.
I would take off serious points in college for that.
 
hm.
I lost points on one of my assignments for not showing how I rationalized the numerator....
 
@BalarkaSen Or, rather, no tangent
 
8:51 PM
That's a better phrasing, thanks.
 
@Balarka: That would leave out a cusp.
Oops.
 
@TedShifrin Those can't be graphs
 
Guess again, Sherlock.
 
I am not even sure how to show your work for rationalizing a numerator, I thought you just did it.
 
@Dodsy: I would want to see some steps, probably. Like what form of 1 did you multiply by? I don't want you to write out $(a+b)(a-b) = a^2-b^2$, I guess, but ...
 
8:53 PM
@TedShifrin Cusps are double points?
 
Nope.
 
I forget the words. "X" can't be a graph.
What are those, then?
 
$y=|x|^{1/3}$ or $y=|x|^{1/2}$ would be cusps.
 
You are doing Meow's exercise for him!
 
The two branches have a common tangent line.
 
8:54 PM
I forgot those were called cusps.
 
LOL ... don't yell at me:)
 
Anyway that's exactly what I had in mind when I posed the fake-theorem.
 
An ordinary node is something with two branches with different tangent lines.
 
When Lagrange met Cauchy as a boy, he told his father to not let him see an advanced math text book until he was 16.
 
Cusps are specifically where the derivative isn't bounded in a neighborhood of a point?
 
8:56 PM
Huh? How is there a differentiable function with a cusp?
 
The obverse of a coin is a coin. The inverse of a coin is 1/coin
 
@Fargle: Well, $y^2=x^3$ has a cusp at the origin.
@Dodsy: CHEMISTRY!
 
Oh so that's what a cusp is.
 
@MeowMix Well, $y = |x|^{1/3}$ isn't diff. You can modify it to obtain an example of a function whose graph doesn't have kinks but which is not diff.
 
@Fargle: You presumably agree that both $y=x^{3/2}$ and $y=-x^{3/2}$ are differentiable at 0 with derivative 0.
 
8:58 PM
@BalarkaSen Wait, what was the question?
If it has no cusps is it differentiable?
 
If it has no "kinks".
 
no kinks => differentiable T/F
 
If it has no "kinks" (which I interpret to be points of non-differentiability), then is it differentiable?
 
Kink meaning pointy things like $y = |x|$
 
No, @Balarka. That's not a definition.
 
8:59 PM
I defined it above. Points where there is no tangent.
 
kinks and cusps.
I'm going to open a dirty store called that.
 
Okay, so we want our derivative to be non-continuous but our graph not to have any kinks
 
Are we disallowing vertical tangents, @Balarka?
 
Well, a cusp already lies outside that definition if we consider vertical tangent lines.
Dang it, @Ted beat me.
 
Who said anything about the derivative's being continuous?
 
8:59 PM
@TedShifrin Tangent lines in $\Bbb R^2$. You are allowed everything.
 

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