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6:00 AM
the $\pi_{n}(thing) = H_{n}(other thing)$
 
Oh yeah lol
 
the application you gave sounds like bringing a gun to a knife fight lol
is there any other interesting space which is $SP(something)$
 
They put one of ours in the hospital, we put one of theirs in the MORGUE!
It's the Chicago way.
 
@Daminark I mean after you take the limit it wouldn't matter, but for finite n it changes what SP^n(X) ends up being right?
 
Thanks @anon
 
6:07 AM
It will, and I'm thinking if anything you don't want them to be distinct
Since otherwise you may not get the nice inclusion I mentioned above. If you have a point in $SP^n(X)$ for which one of the coordinates is already the basepoint, how are you gonna fit that in?
@Eric I don't know anything else offhand
But I'm only just starting, originally I was gonna do this paper in the break but Peter was like "Actually how about we put a deadline?"
Wait how do you otherwise prove that $K(\mathbb{Z},2) = \mathbb{CP}^{\infty}$?
 
follows immediately from LES of a fibration
I really would phrase that as saying $\mathbb{C}P^{\infty}$ is a $K(\mathbb{Z},2)$ and not an equality
 
@EricSilva One of the topologists in my math department said using homology to prove invariance of domain was like bringing a nuclear bomb to a thumb war
 
wait why
I don't know how to do it without homology
 
Me too
 
i feel like proving it another way is almost kind of dishonest
because you probably have to disguise algebraic topology as other shit to do it
 
6:18 AM
Kek
What's invariance of domain?
 
Invariance of domain is a theorem in topology about homeomorphic subsets of Euclidean space Rn. It states: If U is an open subset of Rn and f : U → Rn is an injective continuous map, then V = f(U) is open and f is a homeomorphism between U and V. The theorem and its proof are due to L. E. J. Brouwer, published in 1912. The proof uses tools of algebraic topology, notably the Brouwer fixed point theorem. == Notes == The conclusion of the theorem can equivalently be formulated as: "f is an open map". Normally, to check that f is a homeomorphism, one would have to verify that both f and its inverse...
@Daminark You use it to show that a topological manifold $M$ cannot be both of dimension $n$ and $m$ for $n \neq m$ if I recall correctly
 
tbh idr a use other than showing that manifolds have a well defined dimension
 
@EricSilva But's that's a pretty big use of the theorem itself
 
sure i guess
 
Wait why does this give you that? Do you show that all open sets in $\mathbb{R}^n$ are homeomorphic and then given two different dimensions find sets which aren't or something?
 
6:22 AM
"all open sets in $\mathbb{R}^n$ are homeomorphic"- Daminark
5
 
Frick
I just realized that was stupid
 
Nah in the wiki article it says - "An important consequence of the domain invariance theorem is that $\mathbb{R}^n$ cannot be homeomorphic to $\mathbb{R}^m$ if $m \neq n$. Indeed, no non-empty open subset of $\mathbb{R}^n$ can be homeomorphic to any open subset of $\mathbb{R}^m$ in this case."
 
I mean it's better than all compact sets being homeomorphic, which happened (jokingly) last year
I mean it says that but like... how do you prove it from invariance of domain?
 
So like you'd probably reach a contradiction that $M$ isn't homeomorphic to itself when you descend into Euclidean space or something like that
 
do it man
it's an exercise
 
6:27 AM
I guess you could try to go from large to small, then push it back into large
Like
$n > m$ and $f:U\to V$ where $U\subset \mathbb{R}^n$ and $V\subset \mathbb{R}^m$
But then push that back into $\mathbb{R}^n$, at which point invariance of domain gives you a homeomorphism
Which is probably a load of horseshit
Because you need an open ball to live inside a smaller subspace
 
this is essentially the proof
 
Sick
Lol I wonder why none of us ever noticed it and made a fuss about this in Neves' class
Like "w8 w8 w8 what's stopping you from having a bunch of dimensions tho?"
I guess it felt pretty intuitive that this couldn't happen but like, pedants abound, surely at least someone was like :thonk:
 
bc it's a dumb unenlightening technicality
and as neves has said "proofs are overrated"
 
That's a very Neves thing to say
 
@Perturbative I frequently like to refer to it as 'nuking mosquitos'
 
6:38 AM
I've said once "nuke a potato from orbit"
 
My brain autocorrects Neves to Jeeves
I imagine asking Neves a question to be more fun than AskJeeves tho
 
AskJeeves never puts on that strange but somehow nice music Neves listens to
 
he has decent taste
 
See now I wonder what Schlag listens to, if anything
 
classical music a lot
lmao i think im actually pretty familiar with the musical tastes of my potential letter writers
 
6:48 AM
That sounds like an advantage...
 
When I see Neves, I think of Devin Nunez
Or maybe its Devin nunes rather
 
@Daminark Here's a simpler proof. Suppose $M$ is both a $n$ and $m$-dimensional manifold for $n \neq m$. Pick a point $p \in M$, there exists two charts $(U, \phi)$ and $(V \varphi)$ around $p$ for which $\phi$ maps $U$ homeomorphically onto $\widetilde{U} \subseteq \mathbb{R}^n$ and $\varphi$ maps $V$ homeomorphically onto $\hat{V} \subseteq \mathbb{R}^m$., by definition of $M$ both being a $n$ and $m$-dimensional manifold
Put $W = U \cap V$, we have that $W$ is open and $\phi|_{W} : W \to \widetilde{W} \subseteq \mathbb{R}^n$ and $\varphi|_{W} : W \to \hat{W} \subseteq \mathbb{R}^m$. Both of which are homeomorphisms being the restrictions of homeomprhisms. Since $W \cong \widetilde{W}$ and $W \cong \hat{W}$, we have $\widetilde{W} \cong \hat{W}$, contradicting invariance of domain.
$\cong$ just means homeomorphic there
 
@Eric that's no surprise in hindsight
I'd crack up if he was into like, techno or whatever
 
he literally wanders the halls of eckhart playing vivaldi sometimes
 
LOL
(Dunno what Vivaldi is but like LOL)
 
6:58 AM
r u srs
one of the most famous composers of all time
 
The four seasons...!
 
I'm like, really out of touch with music
 
he did a lot of violin stuff, schlag is a violinist, schlag likes vivaldi
 
I thought Vivaldi was like a web browser
 
have you never seen him walking around playing violin?
 
6:59 AM
I quite like Holst the planets.
 
I have never seen him walking around, no, but I've seen the picture
 
lol ;).
 
i like holst too
 
I've almost never seen him at all
 
i guess he was around more for my bootcamp than yours
 
7:00 AM
Also Nyman and phillip glass.
 
so we got to know him more
 
Like, I saw him in class, office hours, that one time when he came to the bootcamp, and like, twice otherwise
 
I don't know if they're classed as classical classical.
 
philip glass lived in my dorm hallway when he was in college
 
Soug and I are best mates but like, I have rarely interacted with Schlag
 
7:01 AM
philip glass is super not classical lol
 
@eric silva: seriously?
 
yeah my house invites him for a celebration every year but he never comes
it's sad
 
Well, he ain't blues, or rock, or rave, or trance - that just leaves classical by elimination.
 
clearly there are only 5 genres
 
@EricSilva: Fames got to his head, like.
 
7:02 AM
i honestly dont like glass that much
it's good studying music i think
 
@EricSilva: there's only five fingers on one hand - I'm a poor mathematician.
 
but i dont really get into it
which i guess is probably why i think it's good studying music
 
Hot take: classical music is kind of bad
 
nah it's ok
 
I probably listened to Glass at the right time in my life - lake late teens.
 
7:03 AM
it's not my favorite but it aint bad
 
Its not bad music, but I really dont like it
 
what do your own tastes run to?
 
i dont really listen to it that often unless im learning to play something
but i will say it's like a lot more fun to play than like pop kinda stuff
 
Sounds like art, I don't like art unless I'm trying to draw or paint something.
I mean I don't like looking at art unless...
 
I'm a jazz man myself
 
7:07 AM
i was literally listening to jazz rn
and I rarely listen to jazz these days
 
I've never been a fan of jazz...I've tried listening to it on occasion to get into it.
Probably not listening to the right stuff.
 
It can be like coffee or beer or scothch. At first it seems real dumb. but if you just tell yourself its amazing for a while, you start to believe it and notice the nuances
 
i kind of unironically liked scotch the first time I had it
 
I'll try that trick next time ;).
My brother told me he had a mystic vision the first time he drank whisky.
 
I mostly stick to single-malt Irish right now
 
7:12 AM
neat?
 
of course
 
I drink it with something like coke or juice - I'm a late developer.
 
I think you should just drink stuff how you enjoy it tbh
 
When Im out I will drink a whisky and coke or a whisky sour
But thats realtively cheap whisky
 
I don't really drink other than on special occasions
 
7:15 AM
I wont mix single-malt stuff because there's just no point, IMO. When you mix it with a greater volume of coke, you cant tell the difference between a $20 bournbon and a $200 bourbon
So why waster you money?
 
so usually when i do drink it's some really nice bourbon i keep on my shelf
 
I drink cheap stuff, like I said I'm a poor mathematician!
 
someone gave me a mimosa once, and I later found out it had Dom Perignon in it and I was flabbergasted
that was a long time ago though
 
I thought mimosa was a flower?
 
isnt it orange juice and champagne
@MoziburUllah the drink is probably named for the flower
 
7:21 AM
ya its a mix of orange juice and champagne
but it is also a genus of flower
 
Yeah - I just checked with google.
 
i like orange juice too much and champagne too little to ever have one
 
@Daminark I too think your application is more underwhelming than theorem.
$\Bbb{CP}^\infty = K(\Bbb Z, 2)$ is actually just a computation
there's a fibration $S^1 \to S^\infty \to \Bbb{CP}^\infty$.
Or, if you want to be r/iamverysmart, $\Bbb{CP}^\infty = BU(1)$ because maps to that classifies complex line bundles/U(1) bundles. U(1) = BZ, so it's a $B^2\Bbb Z$
 
Can I think of points in $S^\infty$ as sequences that square-sum to 1?
 
7:36 AM
@KevinDriscoll sure
 
@KevinDriscoll By being careful about it, yeah. You can think about it as a unit sphere in a Banach space. All of those are homotopy equivalent
 
I just cant even visualize S^3, so for S^\infty i need some point of reference thats familiar
 
But not actually always homeomorphic
 
Just visualize S^n and let n go to infinity
 
Right, standard is to think about it as an infinite union
under the natural inclusions $S^n \subset S^{n+1}$
 
7:39 AM
@MatheinBoulomenos By that logic, all $S^n$ with $n \ge 3$ are equivalent
 
and then give $\bigcup S^k$ the direct limit topology
 
@MatheinBoulomenos it is an inverse limit right
 
@LeakyNun it's a direct limit
 
$S^4 \twoheadrightarrow S^3 \twoheadrightarrow S^2 \twoheadrightarrow S^1$
@MatheinBoulomenos oh you're thinking about the embeddings
$S^1 \hookrightarrow S^2 \hookrightarrow S^3 \hookrightarrow S^4 \hookrightarrow \cdots$
 
What is the natural inclusion of $S^n$ into $S^{n+1}$?
 
7:42 AM
Equator :)
 
@KevinDriscoll attach $0$ at the end
 
@MatheinBoulomenos just stop visualizing
 
best solution
 
I refuse
 
@MatheinBoulomenos what do I get if I do the inverse limit of the epimorphisms?
wait, there is no epimorphism
lol nvm
 
7:46 AM
@Kevin we'll just put the algebraists on ignore
 
I think its funny that Ive somehow cast my lot with teh geometers over here
becaue among physicists im on the more formal side of the spectrum
 
right
 
like I use no physical intuition I just push symbols around
 
That's pretty much physics
 
Failure to have physical intuition is basically what killed me in first year physics
 
7:51 AM
I tried to read Griffth's QM once
that's basically just bad mathematics
there's 0 physical intuition
 
... and this is why kids, don't try to project something as large as a unit ball in banach space into eucledian space:
 
Like I either couldn't draw the picture of a situation at all, even sometimes in basic cases, or I could draw it and I couldn't figure out how to extract any information
 
well when you get to QM you have no chance
 
which is why I do QM. Im not at as much of a disadvantage
 
7:52 AM
I'd stare at some arrows and be like "Um... this registers as nothing to me"
 
I'm prolly going to use and abuse "registers as nothing"
 
I prefer to stare at arrows if they're morphisms in a category
 
@MatheinBoulomenos but they are
 
I am good at pushing symbols around and drawing pictures, but I am bad at things in between
 
yeah i know that ur a sadist @Mathein
 
7:53 AM
this is why I suck at analysis, because too many things are happening at the same time
 
And the topics in that class were classical mechanics, E&M, and then waves. First two quarters were hell because our problems were extremely physical/geometric
 
That I am bad at the things in between is also why I suck at physics
In general, I am bad at any things that has many parameters changing at the same time not in a vectorised manner
e.g. computing limits
By "vectorised manner", I mean there are computations that you can just do by treating the whole problem as an array or matrix problem
In set theory, replacement is like a vectorised computation to me because I am applying functions to all elements in a set at the same time, thus in a sense, there is really only ONE thing that is changing
but dynamics and analysis is different. Unless you have the ability to keep track of uncountably many paths, there is no way to know where you end up
and the worst of them is limits and epsilon delta proofs
(because you have to justify a given epsilon by going backwards)
 
I never much liked epsilons and deltas...
 
I can understand the intuitive meaning of epsilon deltas and also can read the formal meaning, but I often fail to find the required epsilon because there is too many things happening at the same time
However... no maths is worse than driving
 
8:00 AM
@Daminark So, $\text{SP}^2(S^2)$ is $\Bbb{CP}^2$, right?
 
Driving is like doing half of the maths of the human society in the span of 1 second, and if you did any of them wrong, you are dead
 
So hoorah for self-driving cars!
 
which is why during this catch up, one reason I want to do point set topology first and use it to bridge back to analysis is because if I can manage something that abstract and full of infinities, then there should not be trouble for me to deal with something as concrete as epsilon deltas
For example, reading Munkres plus one remark of Steamy and Alessandros does help me understand functions better
 
It's kind of fun to think about $S^2 \times S^2$ as the fiberwise compactification of the tangent bundle over $S^2$
I guess it means the boundary of the tubular neighborhood of the diagonal in $S^2 \times S^2$ is $\Bbb{RP}^3$. Which is pretty interesting
 
8:16 AM
Have you come across nets? They're a useful generalisation of limits, and their dual are filters which are also useful.
I wish I'd known of them when I was studying point set topology.
As well as stuff like initial & final topologies.
 
@Balarka so it seems
 
I briefly read about them, they are very nice to me because you can reason about topologies that are not first countable with them (and I have a tendency to do exercise on not very nice topologies to ensure my intuition being built is general enough), but I have not read very deeply into them yet. I do suspect they will help me to grasp limit computation though due to their general nature
More generally, I like pathological things not just because thy are weird, but because intuition gain on them can often be aplied to nice things easily
This is because pathological things often tell us why some things fail, and hence by contrapositive, why things are nice
I think suffice to say I sometimes find more trouble understanding some nice things compared to the pathological ones
 
That reminds me of what Peter Scholze said about p-adic analysis over ordinary analysis - it just looks more natural to him now, and he has to work to use the ordinary stuff mere mortals use.
he's got p-intuition.
 
The relevant quote is "Now I find real numbers much, much more confusing than p-adic numbers. "
 
Exactly!
 
8:31 AM
Pretty sure that cannot happen for finite vs infinite, because we are mortals, and we always have one more way to comprehend the finite which we don't for infinite: Count them!
 
How does one get into p-adics?
 
By reading a book on p-adic analysis - I bought one once out of curiosity.
 
Any suggested books?
 
I didn't get too far I hasten to add.
 
Gouvêa's book is pretty good for a first intro
 
8:33 AM
I am so confident on that claim that I am dared to say the following: Whoever mathematician that claimed "Now I find finite numbers much, much more confusing than infinite numbers." should deserve 5 Fields Medals
 
The book I got was Roberts, A course on p-adic analysis.
 
and the next thing will happen, his or her brain will be dissected
 
8:52 AM
How do I work with something like $(\Bbb Z[X]/(X^2 - 47))/(2)$? Am I right in saying that $\Bbb Z[\sqrt{47}]/(2) \cong (\Bbb Z/(2))^2$?
 
@ÍgjøgnumMeg $(\Bbb Z[X]/(X^2 - 47))/(2) \cong \Bbb F_2[X]/((X+1)^2) \cong F_2[X]/(X^2)$
 
@MatheinBoulomenos So in the first step you've used the third isomorphism theorem to rewrite the quotient, right?
 
Not sure if it's third isomorphism
 
Something like $(R/I)/(I/J)$
 
It is
 
8:59 AM
Yo @Tasty
 
Morning
 
What's up?
 
Hey @Tasty
 
Oi @Balarka
 
Nikolai Ivanovich Lobachevsky is my name
 
9:02 AM
Fucking hell I'm gonna be known for the "all sets are homeomorphic" thing if this keeps up
 
lololol
 
9:22 AM
Aren't they?
 
annulus and ball
 
@Tasty you're thinking what I think I was thinking of
That open balls are homeomorphic to $\mathbb{R}^n$, and thus to each other
 
And any open is the union of open balls
So there's really only open balls anyway
Ohi @Alessandro
 
Yo @Alessandro
 
9:27 AM
Can a continuous function map a bounded set to an unbounded set?
 
@TastyRomeo $\mathbb{R}^n$ with the product topology and the metric topology are the same
 
Think $\frac{1}{x}$ on $(0,1)$
 
hmm..so yes
 
Yeah
 
@TastyRomeo $\prod_{i = 1}^n (0, 1)_i$ the cartesian product of $n$ intervals of the form $(0, 1)$ is open in $\mathbb{R}^n$ and isn't an open ball
 
9:29 AM
@Perturbative he was being ironic
 
@Perturbative every finite product of metrizable spaces is metrizable
(Even countable ones)
 
but $f^{-1}(A)$ is compact for all compact subsets $A $ ?
 
This chat needs an irony filter
 
Since $f$ is continuous
 
Preimages of compact sets need not be compact
 
9:31 AM
only for proper maps
 
Consider a constant map $\Bbb R \to \{0\}$
 
@AlessandroCodenotti So if $(X_i)_{i \in I}$ are a countable family of metrizable topological spaces, the product topology on $X =\prod_{i \in I}X_i$ is the same as the metric topology on $X$?
 
@MatheinBoulomenos Isee
 
@Perturbative what's the metric topology on $X$?
There is some metric inducing on $X$ the same topology as the product one
 
10:03 AM
@MatheinBoulomenos how did you type it
 
Find a nowhere open ball set
More precisely:
Find an open set in $\Bbb{R}^n$ that is not a union of open balls
1
Q: $A$ is open iff it is union of open balls

user124140Suppose $(X,d)$ is metric space. I want to show that $A \subseteq X$ is open iff $A$ is union of open balls. My attempt. suppose $A$ is open, then for every $x \in A$, there exists $r>0$ such that $B(x,r) \subset A$ by definition. We claim that $A = \bigcup_{x\in A} B(x,r) $. To see this, pick $...

=FAIL
7
Q: Are open sets in $R^n$ homeomorphic to $R^n$?

iYOA I am working on exercise 1.1 and I think the way to do this would be to show that open sets are homeomorphic to $R^n$ or open balls in $R^n$. Is this even true? I'm not sure how to go about proving it. btw, the exercise is from Lee's Smooth Manifolds book.

4 hours ago, by Eric Silva
"all open sets in $\mathbb{R}^n$ are homeomorphic"- Daminark
All open connected sets in $\Bbb{R}^n$ are homeomorphic
 
Uh
No
 
actually, how should I express a set with only one connected component?
cause those must be homeomorphic to open balls
 
They don't.
Annulus and open disk in $\mathbb{R}^2$
 
10:18 AM
great, forgot those "holey sets"
 
 
1 hour later…
11:40 AM
Hello
someone here ?
"Let $a,b\in K $ where $K$ is a compact and convex set in a Hilbert space, if $||x-a||=\inf_{k\in K}||x-k||=||x-b||$ why $<a-b, x-\frac{a+b}{2}>=0$?"
 
It's instructive to draw it out
You're essentially saying that if points $a$ and $b$ are equidistant from $x$, then the line between $a$ and $b$ is perpendicular to the line between $x$ and $\frac{a+b}2$, their midpoint
That said, I think you could do it algebraically by expanding it with the distributive property
 
i have $<a-b,x-\frac{a+b}{2}>=\frac12<a,a>+<a,x>-\frac12<a,b>-<b,x>-\frac12<b,a>+\frac1‌​2<b,b>$
 
Hm, idea: to simplify the calculations, define $a'=a-x$ and $b'=b-x$
Then $a-b=a'-b'$ and $x-\frac{a+b}2=-\frac{a'+b'}2$
and the hypothesis that $\|x-a\|=\|x-b\|$ is just that $\|a'\|=\|b'\|$, or $\langle a',a'\rangle=\langle b',b'\rangle$
Thus, you eliminate $x$ from the calculation.
You just need to prove $\langle a'-b',-\frac{a'+b'}2\rangle=0$.
That expands into $-\frac12\langle a',a'\rangle+\frac12\langle b',b'\rangle$
or $-\frac12\|a'\|+\frac12\|b'\|$ or $-\frac12\|x-a\|+\frac12\|x-b\|$
which equals $0$
QED
@Vrouvrou
 
12:01 PM
$<a-b, x-\frac{a+b}{2}>=<a'-b',-\frac{a'+b'}{2}>=<a',-\frac{a'+b'}{2}>-<b',-\frac{a'+b'‌​}{2}>=-\frac12<a',a'>-\frac12<a',b'>+\frac12<b',a'>+\frac12<b',b'>$
 
And now those middle two terms ($-\frac12\langle a',b'\rangle$ and $\frac12\langle b',a'\rangle$) cancel
 
$=-\frac12||a'||^2+\frac12||b'||^2=\frac12(-||x-a||^2+||x-b||^2)=0$
 
thank you very much @AkivaWeinberger
 
You're welcome!
 
12:11 PM
If $\widetilde{M} \to M$ is a universal covering of a smooth manifold, is it clear that the natural map $\pi_1(M) \backslash T\widetilde{M} \to TM$ is a diffeomorphism?
 
I go to look up to look up what a homotopy fiber is and I get this: ncatlab.org/nlab/show/fiber+sequence
Someone's having a giggle
 
@AkivaWeinberger we can deduce from this that $-||x-\frac{a+b}{2}||^2+||x-a||^2=||\frac{a-b}{2}||^2$ or we must calculate ?
 
If I have a function $h$ which is a shifted version of another $g$ and I have a Fourier series approximation for $h$, what can I say about a Fourier series approximation for $g$?
 
@abenthy It's a one-to-one local diffeomorphism which is also surjective, hence a diffeomorphism.
 
I think I have to at least shift in an equivalent way (to the way $h$ is shifted from $g$) the Fourier series for $h$ in order to make it approximate $g$
 
12:23 PM
And now all of a sudden Concise makes sense, the fuck?
 
It does?
 
Just for the definition of a homotopy fiber
 
Ah
I half-remember the definition of a homotopy fiber. If $f : X \to Y$ is a map, I make it into a fibration using the mapping cone or something, right?
 
For me to understand the nlab article will require clicking through links for so long that once I know what the definition is, the paper will be due
Concise defines $Ff = \{(x,\chi)|f(x) = \chi(1)\} \subset X\times PY$
There is a way I think to make this what you said
 
hmhmhm
So I have a map $Ff \to Y$
By sending $g(x, \chi) = f(x)$
I think this dude is a fibration
And the homotopy fiber is then a generic fiber of this dude
It's making the domain of $f$ larger without changing the homotopy type so that $f$ becomes a fibration, in some way
 
12:30 PM
Well, I could see $\pi:Ff\to X$ as being a fibration
But your $g$ is just $f\circ \pi$ and that this is a fibration feels a bit more suspicious
 
Your $\pi$ should be a homotopy equivalence
 
What?
 
The idea should be to slurp the paths like noodles back to their basepoint
 
@BalarkaSen Ah I see, and it's one-to-tone because $\widetilde{M} \to M$ is a local diffeomorphism hence its differential at any point is an isomorphism, right?
 
Wait think about it
 
12:33 PM
Oh wait, I don't really understand (yet) why its one-to-one.
 
So you have a map from $PY$ to $Y$ just by taking $p:I\to X$ and spitting out $p(1)$
And that's an ultra fibration
Then you have $f:X\to Y$
$Ff$ is the pullback, and $\pi$ is gonna be the pullback of $p_1$, so that's a fibration (this from Concise)
 
@Vrouvrou You can use the same trick to calculate that… but remember that if $\langle X,Y\rangle=0$ (that is, if $X$ and $Y$ are perpendicular) then $\|X\|^2+\|Y\|^2=\|X+Y\|^2$. This is the Pythagorean theorem. (You can prove this formulation of it by expanding $\|X+Y\|^2=\langle X+Y,X+Y\rangle$.)
 
@Daminark Maybe I am confused. Let me think for a while
 
Well, you specifically have it as $-\|X\|^2+\|X+Y\|^2=\|Y\|^2$, but same thing.
 
But yeah this is in the context of trying to understand a quasifibration
 
12:37 PM
Ah, ok. Here's a suggestion. Maybe my fibration is $g : Ff \to Y$, $g(x, \chi) = \chi(0)$?
 
There are two definitions floating around and I'm trying to understand why they're equivalent (also the definitions themselves)
Yeah that's a fibration
 
@Daminark I still think $\pi$ is a homotopy equivalence. I can agree it's a fibration, but what if the fibers are contractible?
Consider the homotopy inverse to be $X \to Ff$, $x \mapsto (x, 0_x)$ where $0_x$ is the constant path at $x$
one composition is just identity on $X$. the other composition is $Ff \to Ff$, $(x, \chi) \mapsto (x, 0_x)$
which can be homotoped to the identity
just by slurping along the path
 
You like your slurps
 
mmm
smacks lips
The point is the $PY$ component doesn't contribute to the homotopy type much. It's a contractible object
 
@BalarkaSen I'm trying to see injectivity. If I am given vectors $v,w \in T_p\widetilde{M}$ with $d \pi_p(v) = d \pi_p(w)$, how do I see that $v$ and $w$ are in the same orbit of the action of $\pi_1M$ on $T_p\widetilde{M}$?
 
12:52 PM
@abenthy $d\pi_p : T_p \tilde{M} \to T_pM$ is an isomorphism, so $d\pi_p(v) = d\pi_p(w)$ in fact means $v = w$.
The $\pi_1M$-action only works on the base manifold, not on the fibers of the tangent bundle
 
Oh, I see
 
1:04 PM
Hi!
Again, I have to ask about Fourier series...
I am really new to Fourier series and I get the idea of approximating a periodic function with basic periodic functions such as sines and cosines.
I have the following exercise to solve.
And I already wrote something for the proof, but I am stuck...
Here's what I wrote.
I have $h$ which is a "shifted" version of $g$.
And now I also have a Fourier series approximation for $h$.
But I have somehow to get to the Fourier series for $g$.
I am not sure if I have to use any property of Fourier series, if that's even a good idea.
 
1:23 PM
Wait...
 
1:39 PM
Can someone tell me that in mathematical induction how we can assume that $P(m)$ is true?
 
If there's no contradiction regarding a statement, then you can assume it is true whenever you want.
Once you find a contradiction in a logical argument that follows from that statement, you know it isn't true anymore, thus you can't assume it is true in your successive arguments.
As simple as it sounds.
 
Can some1 review the following question? https://math.stackexchange.com/questions/2549516/question-on-proof-regarding-unique-homomorphism-in-finitely-generated-vector-spa
I am still wondering if I have to find a contradiction for i)one example ii)all vector spaces iii) every family of vectors in every vector space
 

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