Mathematics - 2020-11-14 (page 1 of 2)

schn

00:46

Suppose $E(X)=\theta$. How come $Var((X-\theta)^2)=Var(X)$?

robjohn

@schn That doesn't seem correct. Do you mean $\mathrm{E}\!\left((X-\theta)^2\right)=\mathrm{Var(X)}$?

That is pretty much the definition

schn

This is on least squares estimators, when the sample is independent but not identically distributed.

@robjohn Maybe the second equality in the red box follows from using the estimator attained for $\theta$?

@robjohn Or it’s a typo somehow.

robjohn

I think the square inside the first Var is not supposed to be there.

But I could be wrong

schn

@robjohn But then, how is $Var(X-\theta)=Var(X)$? : )

Right, since Var(a)=0, for a a constant.

Thorgott

01:03

the variance measures how much the distribution deviates from the expected value, does that deviation change when we just translate the distribution by a given constant?

schn

It doesn’t.

Thorgott

Yup

And that's precisely what $\operatorname{Var}(X-\theta)=\operatorname{Var}(X)$ says

Ted Shifrin

heya @robjohn

schn

@Thorgott But is $Var(X-\theta)$ related to $Var((X-\theta)^2)$, since the author specifically refers to the squared term?

Thorgott

no, I don't know what's going on there

schn

01:07

Crazy.

Ted Shifrin

@schn Be careful. Variance isn't linear.

schn

@TedShifrin Good reminder.

Ted Shifrin

But $\text{Var}(X) = E((X-\mu)^2)$ and $\text{Var}(X-a) = E([(X-a)-(\mu-a)]^2) = \text{Var}(X)$.

You actually don't need $a=\mu$. It's just that mean adjusts when you shift.

Thorgott

precisely, the deviation form the mean doesn't change when you translate the distribution, because you translate the mean by the same amount

which is exactly what that calculation expresses

Ted Shifrin

I didn't even look at the stuff displayed. I just went according to what people said in here :P

schn

01:15

So it is correct?

Ted Shifrin

No, as robjohn said, the first Var should be E.

schn

@TedShifrin Right.

Thorgott

@Ted can you give me an intuitive reason for why a curve fixed by an isometry is geodesic?

Ted Shifrin

@Thor: Because locally geodesics joining two points are unique.

Think of reflection of a sphere across a plane through the origin.

Consider two (not too far) points on that great circle. If there were another geodesic joining those points, it and its reflection would give you two geodesics joining those points.

Thorgott

ah, an isometry necessarily preserves a geodesic joining two points and so if another curve joining those two points is preserved by an isometry, it must be that geodesic by uniqueness

Ted Shifrin

01:21

Well, an isometry certainly can move a geodesic to a different one.

Just translate, rotate lines in the plane.

So be careful with what you said.

I think you're doing a converse or something.

OMG it's @Pedro !!!!

Pedro Tamaroff

Hehe.

Ted Shifrin

Doesn't even say hello. GRR.

Pedro Tamaroff

Hello!

Ted Shifrin

That's better.

Pedro Tamaroff

It's nice to see you're here.

Does anyone know how to show that if $p<q<r$ then for any $f$ we have $\| f\|_q \leqslant \max (\| f\|_p,\| f\|_r)$?

Ted Shifrin

01:23

Oh, isn't there some sort of convexity?

Pedro Tamaroff

This is all in $L^p(\mathbb R)$.

Ted Shifrin

I assume you're on a finite measure space.

You have too many bars, don't you?

Pedro Tamaroff

Yeah, I was asked about this by a friend.

Hm, let me see. My LaTeX is not rendering.

That should be better.

Ted Shifrin

Yes, it's better.

I remember doing this studying for my analysis qualifying. Probably @Thor or @Alessandro knows this off the top of their heads.

Try this.

Yes, you've abandoned all your geometry, analysis, topology ... for formal algebra. You traitor!

Pedro Tamaroff

Hahaha. I never pledged my allegiance to those. But I am teaching people about Fourier transforms and series and other people about L^p spaces. :)

Ted Shifrin

01:28

OK, I won't curse you, then.

Pedro Tamaroff

But my publications are algebra, mostly.

Ted Shifrin

I still remember my first time ever visiting this chatroom was to discuss some multivariable analysis question you had. I'm not sure we ever settled it.

That feels like two lifetimes (and one pandemic) ago.

@Thor: Did you straighten it out (pun intended)?

KeithMadison

Would anyone happen to know how to apply Regula Falsi to multiple dimensions?

I'm looking at this algorithm: stackoverflow.com/questions/36163846/…. Another answer in that Q&A describes the use of what's more or less Regula Falsi, however it isn't obvious to me how it could be used as a stand in for situations like this: stackoverflow.com/a/64719896/14073182 where there's nesting to multiple dimensions (although I know it can be done)

It's likely simple, I'm just rather awful at visualizing this sort of thing

I know it can be done, and I've searched high an low for an in-depth explanation, but all I can find are mentions of it... very frustrating

Thorgott

01:44

I think I got it. Pick $\alpha$ such that $\frac{\alpha}{p}+\frac{1-\alpha}{r}=\frac{1}{q}$. Then $p/\alpha q$ and $r/(1-\alpha)q$ are conjugate and Hölder gives $\int|f|^q=\int|f|^{\alpha q}|f|^{(1-\alpha)q}\le(\int|f|^p)^{\alpha q/p}(\int|f|^r)^{(1-\alpha)q/r}$. Taking the $q$-th root gives $\lVert f\rVert_q\le\lVert f\rVert_p^{\alpha}\lVert f\rVert_r^{1-\alpha}$. Lastly, observe $x^{\alpha}y^{1-\alpha}\le\max(x,y)$ for $0\le\alpha\le1$.

Pedro Tamaroff

@TedShifrin What was the question?

@Thorgott Neat. ;)

trivial math is difficult

math.stackexchange.com/questions/3906621 Can someone look at my thread and especially my comments if they have time?

Ted Shifrin

@Pedro: I don't berember.

Thorgott

@Ted I think I understand what you meant by the reflections now. So if an isometry fixes two points (not too far), then, since it preserves lengths, it maps geodesics to geodesics and, since the endpoints are fixed, it fixes the geodesic between them by local uniqueness. But I'm not seeing how to get the converse.

Ted Shifrin

02:18

If you had a different geodesic joining the points, its image under the isometry would give another. By uniqueness this cannot happen. Thus the fixed point set is totally geodesic.

Thorgott

02:31

Oh, I just realized the converse isn't true at all

but I agree the fixed point set is totally geodesic

ah ok, but if the fixed point is itself a curve, then, being totally geodesic, it is necessarily geodesic

Ted Shifrin

Yup.

trivial math is difficult

02:50

Grr. Being told that my formula is wrong, but the person actually posted (what I think) is a wrong formula instead!

robjohn

03:05

@TedShifrin howdy... just had dinner. How are things with you?

Sal

What's the general behaviour of $\sum_{n=1}^\infty x^n$ for $0 < x < 1$?

I'm pretty sure it converges to some value, but can that general value be expressed as anything other than an infinite sum?

robjohn

@Sal other than $\frac{x}{1-x}$?

Sal

Perhaps.

I'm just not familiar with infinite sums. All I know is that a similar expression $\sum_{n=1}^\infty \dfrac{1}{2^n}$ is just $1$.

robjohn

@Sal yeah, plug in $x=\frac12$

Sal

Is there a proof that I can look up that shows that $\sum_{n=1}^\infty = \dfrac{x}{1-x}$?

Thorgott

03:16

calculate the partial sums

what's $(1-x)\sum_{n=1}^kx^n$?

@Balarka thanks for linking that write-up btw, was a very instructive read

skullpatrol

03:40

4

A: Congratulations: the big thread!

Congratulations Pedro Tamaroff for getting into the 100K club.

CC @robjohn

robjohn

@skullpatrol yes?

skullpatrol

congrats on getting into the club sir

robjohn

@Sal Look up geometric series on Wikipedia. I bet there is a proof there.

@skullpatrol I've been in the 100K club for a while ;-)

skullpatrol

:-D

epic_math

Hullo

skullpatrol

03:51

hi pal

epic_math

I have a weird idea

We usually sum over the natural numbers or any other set of integers

But can we sum over an interval

The function to be summed should have the necessary conditions hold to make the sum converge

Thorgott

the sensible way of summing over an interval is called integration

epic_math

Ya I know that

I just said it for a special reason

Like, to make silly students (who don't have much intuition for integration) think that it's a totally different

Lol

I have another idea

Suppose we a 2×2 matrix whose components are functions

How can we calculate it's determinant

I don't know what has happened to me why I am talking silly things

What is your opinion?

My opinion is that these silly writers write anything math is nothing without calculus

RewCie

04:48

Happy Diwali Everyone

2

epic_math

Happy diwali!

love_sodam

06:04

@TedShifrin I mean I understood that at least one of principal curvature is zero. But how can I interpret it geometrically?

robjohn

06:35

@love_sodam That means it is like a rolled up piece of paper, flat.

Ted Shifrin

07:10

@love_sodam See p. 61 of my diff geo text. You'll find exercises about what possible developable surfaces you can have.

love_sodam

Your diff geo text?

C.F.G

@TedShifrin: Fortunately you are online!! I want to know more about your thoughts about math.stackexchange.com/q/3906461/272127

Ted Shifrin

See my profile, @love_sodam.

@C.F.G I've said what I have to say. It's late here now.

C.F.G

@TedShifrin: OK, I'll wait until morning. sweet dream professor!

mathguy

08:01

hello

I am back

epic_math

08:22

in how many ways can we represent a number as a sum of squares?

robjohn

@epic_math It depends on the number and its factorization

epic_math

is there an exact formula for $n\in\Bbb{N}$ given the prime factorization of $n$?

I have a very interesting idea in my mind related to partitions.

robjohn

$1105=4^2+33^2=9^2+32^2=12^2+31^2=23^2+24^2$

Note that $1105=5\cdot13\cdot17$ where each factor is $1\pmod4$

epic_math

we know the formula for representing an integer as the sum of natural numbers, but what about representing as a sum of squares?

Balarka Sen

@Thorgott Glad to be of some use

robjohn

08:37

$(2 + i) (3 + 2 i) (4 + i)=9+32i$

$(2 + i) (3 + 2 i) (4 - i)=23+24i$

$(2 + i) (3 - 2 i) (4 - i)=31-12i$

$(2 + i) (3 - 2 i) (4 + i)=33+4i$

epic_math

can you tell me why you are multiplying complex numbers?

robjohn

Each prime that is $1\pmod4$ is not a Gaussian Prime...

epic_math

okay

robjohn

e.g. $5=(2+i)(2-i)$

so $5=2^2+1^2$

$13=(3+2i)(3-2i)$

$17=(4+i)(4-i)$

combining the various $+$s and $-$s gives different ways to write the product as a sum of squares

That's how I got the 4 ways to write 1105 as a sum of squares

I believe that the number of ways to write a number as a sum of squares is $2^{s-1}$ where $s$ is the sum of the exponents of prime factors that are $1\pmod4$

epic_math

08:53

do you have a rigorous proof?

robjohn

Pretty much using the factorization as Gaussian integers

If you are willing to write $25=5^2+0^2=3^2+4^2$ as two ways

epic_math

0 is not to be included, btw, no need to write it

robjohn

then the count will be different

epic_math

yeah

that formula would be (I think) more complicated

have to do some research

robjohn

You also need each prime factor that is $3\pmod4$ to appear an even number of times

epic_math

08:56

this is why number theory is cool

robjohn

also $2=(1+i)(1-i)$, so we include the exponents of $2$ in $s$

epic_math

hey can you tell me what 'G.F.' means in OEIS?

robjohn

Ah, no we don't count the exponents of $2$

GF? not sure

let me look...

epic_math

hey there is a conjecture on the number of ways to represent a number as sums of squares

let $f(n)$ denote the number of such ways

robjohn

@epic_math Looks like it means generating function

epic_math

09:03

it is conjectured that $$f(n)\equiv cn^{\alpha}e^{\beta \sqrt[3]{n}}$$ where $$c=\frac{\zeta({3/2})^{2/3}}{\sqrt{3}(4\pi)^{7/6}}$$

and $\alpha=\frac{7}{6},\quad\beta=\frac{3}{2}\frac{\pi}{2}^{1/3}\zeta(3/2)^{2/3}$

don't know the symbol for ~ wrote it as $\equiv$

can you prove it?

and btw the generating function of $f(n)$ is $$\prod_{m=1}^{\infty}\frac{1}{1-n^{m^2}}$$

The idea in my mind was a 'generalized partition function' and here I am analyzing the partition as sums of squares

Let me ask it on MO I will link it here please see it

robjohn

@epic_math That generating function will include $5^2+0^2$ as well as $3^2+4^2$

epic_math

okay didn't notice

robjohn

Let me think on that

epic_math

so what I want is a formula not depending on the prime factorization

robjohn

09:20

Oh... you are asking about any sum of squares... I was only thinking as the sum of two squares. Sorry

epic_math

sums of two squares is very easy

robjohn

That is $3=1^2+1^2+1^2$

epic_math

I derived it myself

robjohn

Yeah, when I saw the generating function, I realized that

epic_math

your help was still good no worry

@robjohn in my question, I will ask about the distinct partitions, unlike what I mentioned here

epic_math

09:37

would ask later don't I have much time today

maybe I will ask about non distinct partitions which are more important for me

I will do some research and post it in the analytic number theory room (owned by me) which is in MO.

epic_math

10:39

Hello

How can we prove this:

Where:

epic_math

11:12

Analytic number theory proofs are sometimes hard

epic_math

11:28

Can nobody solve it?

Leaky Nun

Jacobi triple product

In mathematics, the Jacobi triple product is the mathematical identity: ∏ m = 1 ∞ ( 1 − x 2 m ) ( 1 + x 2 m − 1...

@epic_math ^

nonsensation

11:47

Good morning all
I have a small question where I am stuck on:
given this exp equation: $I = I_0 * e^(a*x)$ I want to solve for $a$:
I can either divide first by $I_0$ and take the $ln$ : $ln(I/I_0) = ax$
OR
take the $ln$ first and divide then: $ln(I)/ln(I_0)=ax$
The results differ then
I cannot find where I am wrong, they contradict

Merosity

second way is wrong, you're breaking log rules

ln(I) = ln(I_0) + ax

subtract

nonsensation

Thank you very much.. I overlooked that one

Merosity

you're welcome

epic_math

12:13

@LeakyNun thanks!

love_sodam

If sigma(t,\theta) = (x(t)cos\theta,x(t)sin\theta,z(t)) is a surface patch (surface of revolution) where t->(x(t),0,z(t)) is a regular unit speed curve with x(t)>0,z'(t)>0, then in my computation, the Gaussian curvature is -1 if and only if x(t) = x''(t)

And I saw that the persudosphere (gamma(t) = (sin(t),0,cos(t)+ln(tan(t/2))) is also has Gaussian curvature -1

So I parametrized the persudosphere by arc length. the arc length was cot(\theta).

But x(t) = sin(cot^{-1}(t)), x(t)\neq x''(t)

What is the problem?

I will show my computation just a second

Now (x'z''-x''z')z'/x = -1 <-> x'z''z'-x''(z')^2 = -x and as x'x''+z'z'' = 0, (x')^2x''+(z')^2x'' = x so that x'' = x

e,f,g over comes from the second fundamental form

Oh it's pseudosphere and by the way in t\in (\pi/2,\pi) in that pseudosphere

Secret

12:47

@Merosity yes

Merosity

wot

Secret

It is from a density functional theory paper that goes a bit more mathematical on its foundations

Merosity

ah right

that scary index thing lol

user2103480

13:51

are universal covers important for singular (co-)homology?

Secret

"Weak people take revenge, strong people forgive, and intelligent people ignore."-not albert Einstein
"and gods make you unable to ignore without gone rage"

epic_math

Hullo

Secret

hi

epic_math

@Secret are you new to this room?

Secret

nope

epic_math

13:55

Oh

I don't see you much in this room do I guessed that

Secret

I have been in some other communities

I used to be active before 2018

epic_math

Okay

Lol

user2103480

@epic_math it would be more helpful if you included all the conditions in the picture

epic_math

@user2103480 the photo is not mine btw

It is from reddit

user2103480

@epic_math yuck

@BalarkaSen at some point you gotta explain to me how to obtain any finitely presented group as the fundamental group of some space. The easy part is of course just the wedge product of each generator, but glueing it in the right way to obtain the relations didn't work out for me. I tried it by induction but failed so meh

It was an exercise of last semester's topology 1 course and that sounds a bit ambitious without substantial hints lol

Leaky Nun

14:16

@user2103480 no

epic_math

@user2103480 yuck reddit is really yuck

user2103480

@Thorgott in topological spaces, working with maps $I \times I \rightarrow X$ is way more chill than with $I \rightarrow X^I$. This has got something to do with exponentials not working well in that category, right? Like, in comparison to set theory, where I could curry the hell out of each function whenever I felt like it

@LeakyNun ok nice

Then bundles are probably also not so important

Different constructions of spaces are probably more important in the regard, to actually compute the stuff

Leaky Nun

14:32

Chern class

In mathematics, in particular in algebraic topology, differential geometry and algebraic geometry, the Chern classes are characteristic classes associated with complex vector bundles. They have since found applications in physics, Calabi–Yau manifolds, string theory, Chern–Simons theory, knot theory, Gromov–Witten invariants, topological quantum field theory, the Chern theorem etc. Chern classes were introduced by Shiing-Shen Chern (1946). == Geometric approach == === Basic idea and motivation === Chern classes are characteristic classes. They are topological invariants associated with ...

complex line bundles = elements of H^2

Mike Miller

That's an application of cohomology to line bundles tho and I think usernumbers was asking about the other way around

user2103480

14:56

Yeh but no worries.

Lifting a homotopy between paths is the same proof as the one for lifting a path, just that we patch together squares, right?

Mike Miller

Pretty much

Since you know the "left-side" of the square already, and you know lifts of paths are unique once you specify the start-point, the lift of the square is actually already determined by what you've already done

You're just checking continuity

And that can be done locally

user2103480

Ok cool thanks

Mike Miller

The general statement is that if p: E -> B is a covering space, and X x [0,1] -> B is a homotopy for which you have specified the lift X x {0} -> E, there is a unique lift of F to a homotopy on E with the specified starting-data

for X a compact metric space

Gotta use Lebesgue number somewhere here I believe

Actually I'm skeptical now, I think you can do this pretty generally

user2103480

I only considered the case with X = [0,1] so it works in general there

Leaky Nun

15:30

a homotopy between paths is just a path in the path space

user2103480

@LeakyNun that's the reason why I asked about exponentials in topological spaces; this is a useful analogy but isn't used in proofs of the more simple facts

Mike Miller

You can curry when what you curry is locally compact

AKA Y l.c. => C(X x Y, Z) = C(X, Z^Y)

user2103480

15:50

Ah cool, so can that be used to simplify proofs about homotopies and if not, what are the obstructions?

Thorgott

Right, $\mathbf{Top}$ isn't cartesian closed. The idea of currying is that we want to have bijections $\operatorname{Hom}(X\times Y,Z)\cong\operatorname{Hom}(X,Z^Y)$ (and the category theorist asks this to be natural in all variables). In more fancy language, this is requiring that taking products with $Y$ and exponentiating by $Y$ are adjoint functors and there are abstract nonsense reasons why this can't happen in $\mathbf{Top}$.
I believe algebraic topologists restrict themselves to the slightly adopted category of compactly generated weak Hausdorff spaces for this reason and slightly mo…

compactly generated weak Hausdorff spaces are rather obtuse to define, but I believe they include every nice space you ever stumble across

mathguy

Hello

Couldn't talk for a while

Balarka Sen

@user2103480 this seems like the right idea where did you get stuck

user2103480

16:13

@BalarkaSen I would say I didn't even think it through enough to really get stuck. I know that by the construction that you showed me, the application of SvK, I can get A^nB^m=e for any two generators and n,m in Z. My next step would to just try an induction step to get three generators, but there I'm not so sure if I should glue three spaces at once in a way, or glue two spaces and then glue the resulting two that are left

Mike Miller

16:25

@user2103480 No why would it be useful

I've never met a map that was easier to show is continuous as a map to a goddamn function-space

@Thorgott Nerd

user2103480

@MikeMiller Instead of doing a slightly different glueing argument every time one might be able to do some shit like lifting paths in path spaces

Mike Miller

It's exactly the same argument in both cases

I mean that's what currying says right

If you can curry then proving one means proving the other, you can immediately translate proofs

user2103480

For translating proofs I need to check that I get a covering p:Y^I -> X^I and such

Thorgott

isn't that what you use the whole loop space suspension thing for

feynhat

What is Z^Y?

C(Y, Z)?

user2103480

16:30

So it's not obvious to me that the proofs are translated immediately

Thorgott

ye

it's the iNtERnAl hOm

feynhat

why aren't you writing it as that?

I mean as C(-, -).

user2103480

to make it more ambiguous

jk

C(-,-) is clumsy

Thorgott

cause it's supposed to be an eXpoNenTiaL oBJecT

Mike Miller

@user2103480 Meh fine

I don't like this approach

user2103480

16:33

@MikeMiller I quote Mike Miller: "I don't care"

Mike Miller

Good man

Balarka Sen

@user2103480 Add a cell for every relator in the group

Use SvKT at each step, yeah

feynhat

Why would you write C(-,-) and the exponential notation both in the same expression when they're supposed to mean the same thing?

Balarka Sen

Lmao C(X, Y^Z)

Thorgott

ok good point

Mike is officially silly

feynhat

16:36

Its just confusing... or maybe there is some categorical justification for that which I don't understand.

Balarka Sen

Prove that C(-, X) is a sheaf in an appropriate setting

Mike Miller

(Y^Z)^X is ugly

C(X, C(Y,Z)) is ugly

feynhat

no

Mike Miller

Use each symbol once per expression challenge

Alessandro Codenotti

Looks like I joined chat at the wrong moment lol

Balarka Sen

16:38

yeah im leaving

Thorgott

well, I guess you could make the point it's supposed to be external hom, sometimes people write stuff like $\mathcal{C}(C,D)$ for the morphisms from $C$ to $D$ in a category $\mathcal{C}$

but in Top, when it works, you even have the adjunction internally, I believe

Mike Miller

now I'm gone

Mike

Hello, does anyone know how to find spectral resolution of this operator?

feynhat

I've never seen that Y lc => C(X x Y, Z) = C(X, C(Y, Z)). I know if the evaluation map C(Y, Z) x Y -> Z is continuous then, is continuous, then we have that homeomorphism.

that first equality is a homeo btw.

@user2103480 I quote Mike Miller: "Lets rob a bank".

Why does this look like hom-tensor adjunction that we have for modules (abelian categories even?)?

Thorgott

both are adjunctions with the internal hom

the internal hom in modules is adjoint to the tensor, the internal hom in nice enough top spaces is adjoint to the product

tensor and product are symmetric monoidal structures on those categories respectively

user2103480

16:44

@feynhat one is external, one is internal hom

Also, category theory is one grand hidden project to just flex on the uninitiated

@MikeMiller no

Thorgott

also, I feel like you need hypothesis on $X$ and $Z$ as well, not just on $Y$, but I don't wanna think about it

feynhat

@MikeMiller ^

@feynhat yikes. I type like a moron. No clue why I wrote 'is continuous' twice.

Mike Miller

@Thorgott You don't dude as Feynhat says it relies on evaluation being continuous

Thorgott

ah, it seems we don't need any on $Z$

Mike Miller

So long as locally compact means "for any x in an open U, there is a compact K in U with x in int(K)", it is a straightforward exercise to show the evaluation map is continuous

I put something like this on a midterm. For most people the hard part was showing that a compact Hausdorff space is locally compact in this sense.

Thorgott

16:50

nevermind, just checked, it does appear to in fact be true

Mike Miller

To be clear C(X,Y) has the compact-open topology

Man just prove it it's not hard

feynhat

Yeah. Its easy.

Thorgott

no i dont wanna do topology

feynhat

I have to read Sard's theorem.

Thorgott

just lemme quote the nlab and enjoy the company of my own pretension

Mike Miller

17:01

@feynhat Nobody can make you do that.

user2103480

17:14

@Thorgott so you also failed no nlab november

Thorgott

was one of the first to fail

user2103480

17:28

We've had the following theorem without proof in our lecture: Let $p \colon X \longrightarrow B$ be a fiber bundle, $f \colon (I^k, \partial I^k) \longrightarrow (B,b_0)$. Then there is an $\Tilde{f} \colon (I^k, \partial I^k \setminus I^{k-1} \times \{ 0 \}) \longrightarrow (X,x_0)$ with $p \circ \Tilde{f} = f$.

Thorgott

tilde without caps

Mike Miller

widetilde

user2103480

@Thorgott works for me, weird

Edward Evans

oversquiggly

user2103480

@EdwardEvans nice

Thorgott

17:36

huh

Edward Evans

;)

user2103480

Is the proof worth looking at, or is it just "intuitive"? And how do I imagine this? I'd think of $p^{-1}(\{ x_0 \})$ as the "spine" of the fiber bundle, and the mapping's "slices" going continuously along that spine such the projection gives us the original mapping

Mike Miller

The proof is exactly the same as the proof for covering spaces

You just lose uniqueness

Covering spaces are like twisty versions of $p: X \times D \to X$, where $D$ is discrete

Fiber bundles (what most kinds of fibrations are) are like twisty versions of $p: X \times Y \to X$, where $Y$ need not be discrete

This result is obvious for both of those

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$Mathematics$ Mathematics