Reading mathematical claims in an eng book after reading math books makes you want to cry

@EE18 Who's this "you", white man?

me myself and i

xander: oh, and the kids are going to get that reference. OK.

@leslietownes Okay, so we're both old.

can anyone please explain where the contradiction is?

god, what a convoluted proof

anyway, the contradiction comes from writing down the induced maps on cohomology rings

In past, I have done it using fibrations.

think about what $f$ does on the generator

But this time fibration sequences are not allowed [I didn't impose this condition].

anyways, I was thinking: f o h= g so passing onto cohomology: h* o f*= g* and on S^{2n+1}, g* being nullhomotopic so 0 so h*o f*=0 on S^{2n+1}.

this gives j* o pi* o f* =0 whence j* o (f o pi)*=0

But since f o j= id, we have j* o f*= id

this contradicts the earlier equality.

But there must be something wrong in it. :(

the equality $f\circ h=g$ doesn't imply anything meaningful at the level of homotopy invariants, because $D^{2n+2}$ is contractible

you can derive a contradiction from $f\circ j=\mathrm{id}$ and that alone

if pi: E-->M is a fiber bundle, can it be said that M is a deformation retract of E? If yes, then why?

hi @koro!

Hi @copper.hat!!

hi @copper.hat!

and @koro!

hi @leslietownes!!

I think you're all hi!

hi @leslietownes @Koro, lo @robjohn

@koro About 2-dimensional problem of this, $\{x\in\mathbb{R}^2:\text{exactly one of }x_1,x_2\text{ is rational}\}$, then this is easily seen to be totally disconnected since its complement contains $(r_1, r_2)+(\pm x, \pm x)$ where $r_i$ are any rationals and $x\in\mathbb{R}$

But this argument doesn't quite work for $\{x\in\mathbb{R}^3 : \text{exactly one of }x_1, x_2, x_3\text{ is rational}\}$

Because such lines don't enclose a 2-dimensional shape

It was Derivative who asked the question

@SoumikMukherjee I know that

But Koro asked about the 2d version

I don't have dementia

Is this solution correct? I've looked up the solutions to this problem and every solution uses more complex approaches so I'm unsure if there is some basic flaw in this solution that I'm missing out on.

you write b <= a, c <= a, d <= a in succession (where a = cos(x) and b = a^4 and c = a^2 and d = a), then you write just below that, "implies" b - c + d <= a. what is going on with the "implies"?

@Swan its not correct

note that if c <= a then -c >= -a, but this inequality "goes the wrong way" for it to be so easily added to the others

that's what jumped out to me. i haven't looked at the rest of it

But if you argue based on cos^4n <= cos^2n then it should be fine

Oh I missed that koro asked for 2d, my bad

@leslietownes Yeah, I got it. Sometimes I do glance over these very fundamental errors and need assistance to spot the mistake. Thanks :)

@Jakobian Thanks :)

S^1----> S^2n+1 ----> CP^n is a fibration so writing its long exact sequence gives: pi_k S^2n+1= pi_k CP^n for all k>2

this isomorphism is induced by pi.

Hopf theorem: pi_k S^2n+1= Z when k= 2n+1.

But if pi were homotopic, then pi* would be the 0 map which can't result in isomorphism.

that's one way to look at it.

note that f*=0

so id =0 is the contradiction.

that's another way to look at the question I asked.

@Jakobian thanks. I'll look into it in a while.

@Koro in general, this question is not quite well-posed because $M$ is not a subset of $E$. the LES in homotopy implies that a necessary condition for $\pi$ to be a homotopy-equivalence is that the fiber $F$ is weakly contractible (hence contractible under mild technical assumptions). if the fiber $F$ is contractible, it turns out (under mild technical assumptions) that $\pi$ has a section which turns $M$ into a deformation retract of $E$.

(in particular, this is true for vector bundles, where the proof is of course much easier)

@Koro yeah, that's a nice argument

another argument that uses cohomology instead of homotopy, but is more convenient than the one you've screenshotted, is that if $\pi$ were null-homotopic, you would get homotopy-equivalent attaching spaces $\mathbb{CP}^{n+1}\simeq\mathbb{CP}^n\lor S^{2n+2}$, but these have non-isomorphic cohomology rings