Mathematics - 2017-12-06 (page 3 of 4)

Krijn

7:00 PM

Any people with number theory/arithmetic geometry background around?

Akiva Weinberger

Hey Ted

Krijn

Otherwise typing out my question would take too long

Ted Shifrin

Right. This isn't bad, since when you repeat the ideal sums of products of elements of the ideal is just the same as ...

Akiva Weinberger

Hey Krijn, Felix

Ted Shifrin

Hi DogAteMy

Hi @Tobias

Tobias Kildetoft

7:01 PM

@TedShifrin Hi

Got the student evaluations today

Ted Shifrin

Oh yeah?

Petar Ivanov

hmm if someone can suggest a number theory book?

Ted Shifrin

That must mean you assigned grades already. How did they do?

Felix.C

@AkivaWeinberger Hey

Ted Shifrin

What kind of number theory, @Petar?

Elementary stuff?

What is your background?

user193319

7:01 PM

@TedShifrin Not sure how to finish the sentence...Would it just be an element in $(r^3)$? I'm somewhat confused.

Tobias Kildetoft

@TedShifrin No, the evaluation is done before the course is fully complete, so that there is time to tell the students about the changes I plan to make based on the evaluations

Ted Shifrin

@user193319: I do believe so. Check it out.

Ohhh ... very different system, @Tobias.

So, what did they say?

I think students appreciate responsiveness to feedback, if their feedback is reasonable ;)

Tobias Kildetoft

@TedShifrin Generally decent scores (I think). The comments were mostly constructive, which was nice

quallenjäger

How can I define a trivial connection on a trivial bundle?

Krijn

They gave me feedback on my socks today: cool

Ted Shifrin

7:03 PM

Just differentiate as usual, using the trivialization, @quallenjäger.

You were wearing socks, Krijn? I'm shocked!

Petar Ivanov

@TedShifrin i am new to that field

Ted Shifrin

Do you know abstract algebra, @Petar?

Krijn

@TedShifrin Not just any socks, cool socks

Tobias Kildetoft

The TAs got amazing scores and great comments, which was good to see, given that the structure of the exercise sessions was not very well liked

Ted Shifrin

Well, they didn't blame the TAs for the department's structure. That's enlightened of them.

Tobias Kildetoft

7:04 PM

@TedShifrin Yeah. A bit surprising, but a good kind of surprise

quallenjäger

@TedShifrin I have read that one can define $\nabla_X e_i=0$ for all $X\inT_pM$ and $e_i$ from smooth section of the vector bundle

@TedShifrin Why can't I do that if the bundle is non trivial?

Ted Shifrin

No, that's wrong, @quallenjäger.

Tobias Kildetoft

Well, not really the department's structure. The structure of the session was planned by myself and the TAs

Ted Shifrin

Oh. Then they should blame you, @Tobias :P

Krijn

What did you do to the exercise class

Ted Shifrin

7:06 PM

@quallenjäger: Consider the trivial bundle $\Bbb R^n\times\Bbb R$ on $\Bbb R^n$. You're saying that you can say that any real-valued function has zero derivative.

Petar Ivanov

@TedShifrin hmm i am not sure since we use other terms here but lets say i dont

Tobias Kildetoft

@Krijn We changed it so that fewer exercises would be done at the board

Ted Shifrin

Do you know about groups and rings, @Petar?

Tobias Kildetoft

In fact, we cut it all the way down to one exercise per session

Ted Shifrin

Hmm, @Tobias. And what else happened?

Krijn

7:07 PM

Ah, yes, I see that they would dislike that

Ted Shifrin

The problem with working homework problems is that the students who haven't made a good effort don't learn much, but they think that by copying solutions they've mastered the course.

Krijn

@TedShifrin This was me, in my undergraduate studies

Ted Shifrin

I used to criticize students a bit who'd come to my office hours without even having read or attempted the exercises.

Bad, bad Krijn.

Tobias Kildetoft

@TedShifrin Well, the way the sessions used to go was that the students would get some amount of time to do the exercises (because they would invariably not have done them beforehand), then the problems would be done at the board, either by one of the students who had done it or more commonly by the TA

Petar Ivanov

@TedShifrin i have some knowledge in it

Tobias Kildetoft

7:09 PM

@TedShifrin Right, that deficiency of the board exercises was why we changed it. So now they would be split into smaller groups depending on how well prepared they were and be told to discuss the exercises in those groups with guidance from the TA

Ted Shifrin

OK, @Petar. A classic book is Hardy and Wright. Beautiful stuff in it, but a bit old-fashioned. A more modern book is by Joe Silverman.

Narcissus jewel

Okay, just read some complex analysis for the first time. It says something, perhaps not-rigorously at this point; 'if $f$ and $g$ are holomorphic functions in $\Omega$ which are equal in an arbitrarily small disc in $\Omega$, then $f=g$ everywhere in $\Omega$.

This sounds something like, take a sheaf $\mathcal{O}_X$ of holomorphic functions. Then if $f,g\in \mathcal{O}_X(U)$ have germ $f_x=g_x\in O_{X,x}$, then $f=g$. Is this correct?

Ted Shifrin

Probably an intelligent change, Tobias, despite their whining.

Balarka Sen

@Narcissusjewel yup

Ted Shifrin

@Narcissusjewel: You need $\Omega$ connected, of course.

Narcissus jewel

7:10 PM

@BalarkaSen Spicy.

Krijn

Hey @Bala

Balarka Sen

This is basically gluing axiom for the presheaf of holomorphic functions

Narcissus jewel

@TedShifrin Good point.

@BalarkaSen <3.

Ted Shifrin

I see Balarka has made it through one more page of Forster. :)

Tobias Kildetoft

@TedShifrin Yeah. The detailed feedback from some of the comments have given me some ideas for tweaks, but the overall idea still feels solid (and whining is always to be expected when a change means the students need to do more work)

Petar Ivanov

7:11 PM

@TedShifrin ok thanks, i will check them out

Ted Shifrin

@Tobias: I usually found that even the students who complained about my being demanding appreciated me a year or two down the road ...

quallenjäger

Well let me try a different way: If the manifold is parallelizable, then I can find a diffeomorphism $\rho:M x \Bbb R^d\rightarrow TM$, $\rho(p,v)=v^iX_i(p)$

Balarka Sen

Well, not gluing axiom, the locality axiom, I guess.

Krijn

@TedShifrin That might have just been survivor bias

Balarka Sen

Hey @Krijn

Tobias Kildetoft

7:12 PM

Other than that there were some general comments about me being inexperienced, but improving during the course. And a general consensus that representation theory is hard

Balarka Sen

Long time

Ted Shifrin

@Krijn: That includes people who got not-so-great grades, but — sure — there were indeed quitters.

Krijn

Yeah, I'm finishing up my thesis and hoped that someone here could help me out with a small proof

Alas, no number theory in this room so I keep on googling

Narcissus jewel

@BalarkaSen For upgrading to a sheaf you mean here.

Balarka Sen

@TedShifrin I have gone through Chapter 1, more or less. I gave Daminark what I consider a quality exposition of analytic continuation a few days ago here

quallenjäger

7:13 PM

Is the mapping $\rho(p,v)\rightarrow \rho(q,v)$ a parallel transport?

Ted Shifrin

@quallenjäger: You're declaring certain sections to be "constant." Having fixed that fixed basis, you define covariant derivative subject to the product rule, yes. So the connection will depend, of course, on the trivialization.

quallenjäger

So at this point I need information of connection?

Ted Shifrin

You should try your colleagues or professors, @Krijn ... Maybe they know more than we do.

quallenjäger

Is the connection set once I have choose the basis $X_i(p)$

?

Krijn

@TedShifrin They left office already

:(

Ted Shifrin

7:15 PM

Yes, once you choose the trivializing sections, the connection is determined.

Good point, @Krijn.

quallenjäger

@TedShifrin Thanks, got It

And I have run through the story with the hyperboloid, did you mean that the hyperboloid at infinity is a cone rather than a cylinder?

Ted Shifrin

What would be the cone point?

Jacksoja

Hi chat

quallenjäger

I mean just "like" a cone at infinity

A cylinder would go straight upwards and that is what bothers me.

Jacksoja

If f is a mapping from S onto T , prove that f'f = identity of S

Krijn

7:19 PM

@Jacksoja What are S and T?

Jacksoja

I don't understand how much is enough to prove this

Tobias Kildetoft

@Jacksoja What does $f'$ mean?

Jacksoja

S and T are sets

f' is f invers

Ted Shifrin

Well, I see your point. The surface is, in fact, asymptotic to the cone formed by the asymptotes. I was sloppy.

Jacksoja

I can see how if we take s in S f(s) = t

and f' (f(s)) = s

Tobias Kildetoft

7:21 PM

@Jacksoja the inverse is only defined for invertible, i.e. bijective functions, and it is defined such that the identity you need to prove holds

quallenjäger

@TedShifrin Thanks!

Jacksoja

@TobiasKildetoft so what is it exactly I should to show?

Krijn

What if $S = \{ 1,2,3 \}$ and $T = \{ 1,2 \}$ with $f: 1 \mapsto 1, 2 \mapsto 2, 3\mapsto 2$? @Jack?

Tobias Kildetoft

@Jacksoja The statement you have written is not true

Krijn

You need more information on $S$ and $T$ here

Jacksoja

7:22 PM

f is a 1-1 mapping sorry did not mention that earliar

f is a 1-1 mapping from S onto T , prove that f'f = identity of S

quallenjäger

I have the feeling that one need to be extremely cautions in differential geometry.

Its very easy to make mistakes

Krijn

I am capable of making very easy mistakes in most of mathematics.

Ted Shifrin

Me too :P

quallenjäger

All these minus sign, you wish sometimes they disappears, but they simply don't

best way is to erase them and hope no one will see it

Tobias Kildetoft

@quallenjäger You should try doing Lie algebra cohomology. So many signs

I had a lecturer write up some of it, then say "there is a formula with signs. I refuse to put signs"

quallenjäger

7:26 PM

@TobiasKildetoft :D

Jacksoja

@TobiasKildetoft f is a 1-1 mapping from S onto T , prove that f'f = identity of S

@TobiasKildetoft what I said before makes sense now?

f(s) = t

Ted Shifrin

In complex geometry you must keep track both of signs and of powers of $\sqrt{-1}$.

Lots of folks get both wrong at times.

Tobias Kildetoft

@Jacksoja Except that once the inverse is defined, the identity is the definition of the inverse

Ted Shifrin

@Tobias: Your use of "the identity" is very confusing there.

quallenjäger

Oh dear thats why I never touched complex stuff.

Tobias Kildetoft

7:27 PM

@TedShifrin ahh, right

quallenjäger

Get real

Jacksoja

Yes exactly

there is also identity on T

Ted Shifrin

People are so used to $1/2\pi i$ in complex analysis they tend to mess up in complex geometry ... when it should be $i/2\pi$ to get signs right.

Tobias Kildetoft

@Jacksoja By "the identity" I meant the equality (i.e. the one you are trying to show)

Jacksoja

But the proof of it , how should one write that ?

Ted Shifrin

7:28 PM

I won't name some very famous people that have gotten it wrong. ;)

Jacksoja

I see why the composition sends any eleemnt of s to itself

Tobias Kildetoft

@Jacksoja how have you defined the inverse?

Ted Shifrin

@Jacksoja: What is the definition of the inverse of a function?

in chorus

Jacksoja

well yeah I know what it is

every 1-1 and onto function has an inverse

quallenjäger

I even forgot what $1/i$ is

Ted Shifrin

7:29 PM

And what is the definition?

That's a horrible thing, @quallenjäger.

Surely you jest.

Jacksoja

f composed with f inverse = identity

Ted Shifrin

In both directions, @Jacksoja.

So the equation you wanted is part of the definition.

Jacksoja

so ff' =i_T , and f'f = i_S

in keeping track on what identity map

Ted Shifrin

Backwards maybe?

Jacksoja

we just need to see what is the first product act on right?

f :S-->T

Ted Shifrin

7:32 PM

Better :)

Jacksoja

@TedShifrin Right corrected it now

Ted Shifrin

Right. And you asked how to prove one of those earlier. But it's half of the definition. That's what Tobias was saying (several times).

Jacksoja

all right thanks, these questions on set theory part are easy to see but hard to give proper proof

like if g and f are onto functions

show that the composition is also onto

Ted Shifrin

That's a good question.

Jacksoja

I did this part for injective

but surjective was hard to write

Ted Shifrin

7:34 PM

No. It shouldn't be.

It should be easier than injectivity.

Eric Silva

sup chat

orbit-stabilizer

Let $E\subseteq \mathbb{R}$ be a bounded set and suppose $f:E \rightarrow \mathbb{R}$ is uniformaly continuous Prove that $f(E)$ is bounded.

Ted Shifrin

heya Eric, orbit.

orbit-stabilizer

hello

How are you?

Ted Shifrin

Doing just fine, thanks. You haven't had your exam yet, I take it.

Rehi DogAteMy.

orbit-stabilizer

7:35 PM

It's today.

Akiva Weinberger

Rehi

Jacksoja

well g :S-->T ,f : T--->U , i need to show that for every u in U, there is s in S, such that gf(s) = u @TedShifrin

orbit-stabilizer

In a few hours :/

Ted Shifrin

Aha.

Right, @Jacksoja.

So?

Jacksoja

f(s) = t for some t in T

Ted Shifrin

7:36 PM

This makes no sense at all.

We started with $u$. That's all.

Jacksoja

we know all of T is used up because it was given that f is surjective

yes but what i meant to say

Ted Shifrin

What does $gf(s)$ mean?

Jacksoja

g(f(s) =g(t) = u for some u in U

Ted Shifrin

No. $u$ is fixed to start with.

Jacksoja

and we also know that g is onto

Ted Shifrin

7:37 PM

You have to get your language right.

Eric Silva

i should really learn some complex algebraic geo

Jacksoja

all right

Ted Shifrin

Yes, $g$ is onto is the right thing to use first.

Add it to your list, Eric.

Jacksoja

let u be in U, u =f(t)

t=g(s)

Ted Shifrin

NO.

Oh, your function composition doesn't even make sense, does it?

Jacksoja

7:39 PM

that is not how we start this?

Ted Shifrin

Hang on.

Either your function composition doesn't make sense or you got $f$ and $g$ reversed.

Jacksoja

well the composition is S-->T-->U

Ted Shifrin

And which function is first, and which function is second?

Jacksoja

g:S-->T , f :T--->U

Ted Shifrin

And what does $gf$ mean?

Jacksoja

7:41 PM

we do f first

Ted Shifrin

Then this makes no sense.

Details, man.

Jacksoja

right we do fg

u=g(t)

Ted Shifrin

OK. So let's start over. You said that for every $u\in U$, we want to find $s\in S$ so that $fg(s) = u$.

Jacksoja

right

Ted Shifrin

NO ... how can $u=g(t)$? Look.

Jacksoja

7:42 PM

got it backwards again sorry

Eric Silva

yeah im wondering how to compute homology for some smooth algebraic curve of degree $d$ in $\mathbb{C}P^{2}$ using pure hatcher stuff so i can show they're orientable

Jacksoja

u=f(t)

Ted Shifrin

Pictures help!

Jacksoja

hang on ill draw one now

Ted Shifrin

Right. Since $f$ is onto, there is $t\in T$ so that $f(t)=u$.

And you can even put a $t\in T$ with an arrow going to $u\in U$ :)

Jacksoja

7:44 PM

right and t = g(s)

Ted Shifrin

Oh, Eric, that's a good question.

Jacksoja

f(g(s) = u

Ted Shifrin

@Jacksoja, the words "for some $s\in S$" are crucial.

Do you see the picture with the mappings?

Jacksoja

and the only argument here is that we are guanteed to have such an element mapped to u because of ontoness

Balarka Sen

@EricSilva Degree-genus theorem tells you what genus that dude has, probably?

Jacksoja

7:45 PM

I drew the picture like it was bijection and that is bad

Ted Shifrin

Right, ontoness of both.

Balarka Sen

I don't know algebraic geometry

But this is definitely nontrivial

Ted Shifrin

It's just Euler characteristic for a covering map with branch points, Balarka, Eric.

But is that "in" Hatcher?

Balarka Sen

Ahh OK

Nope, Riemann-Hurwitz is not there

Or whatever that's called

But it's a good exercise to prove it

Ted Shifrin

How 'bout implicit function theorem to get that it's a complex $1$-manifold, therefore orientable?

@EricSilva: Where did this question come from?

Balarka Sen

7:47 PM

ya that doesn't take much argument

Eric Silva

my prof posed it

Ted Shifrin

Or give a nowhere-vanishing $2$-form on it :P

Eric Silva

im trying to use stuff from just this quarter of alg top is the thing

Ted Shifrin

Good luck with that.

Eric Silva

cause i know how to do it if i go beyond that stuff

ya im puzzlin

Balarka Sen

7:48 PM

Ted's $\chi$ hint is pretty good

Ted Shifrin

Lefschetz hyperplane theorem :P

Balarka Sen

Once you know genus you are all set

Eric Silva

ill think about it

just wanted to throw it out there as a cool question :P

Ted Shifrin

You should be able to define a branched covering map to $\Bbb CP^1\subset\Bbb CP^2$.

Draw pictures :)

Balarka Sen

@TedShifrin Yeah the projection map to a hyperplane should be the branched covering in this case

Ted Shifrin

7:49 PM

But draw the "pseudo-real" pictures that complex geometers draw.

Yup, Balarka.

Eric Silva

lol

is rick miranda's book any good

if you're familiar with it

Ted Shifrin

I don't know the book, but I've known Rick for decades. So it's no doubt good. I also like Griffiths's little book on algebraic curves (lectures from China).

quallenjäger

@TedShifrin I still cannot get my head around about trivial connection. Basically my frame $X_i(p)$ would still move from point to point. How can I see that the covariant derivative of the frame would vanish?

Eric Silva

oh i own a copy of that one

miranda's book is big

Ted Shifrin

Miranda does more than curves, right?

@quallenjäger: In the trivialization, each $X_i$ appears to be constant (like the standard basis in $\Bbb R^k$).

Eric Silva

7:53 PM

unsure

Ted Shifrin

LOL, I figure it's time for you to read one of my most popular answers on MSE, @quallenjäger.

Eric Silva

the answer to Zev?

Ted Shifrin

Yup. This @quallenjäger.

quallenjäger

@TedShifrin So I need to define $\nabla_Y X_i$ is equal to zero? Or this can be shown?

Ted Shifrin

That's defined. You're setting up what your "constant sections" are ...

quallenjäger

7:55 PM

@TedShifrin Lol I thought this can be shown, thats why I am struggling.

you mean this one?

Balarka Sen

@TedShifrin Can I steal Figure 2.5 from Chapter 7 of your book with acknowledgements for an answer I am writing on MSE?

Ted Shifrin

WTH is that, Balarka?

quallenjäger

109

A: How do you respond to "I was always bad at math"?

To put a slightly different spin on this (before @Zev votes to close :) ). When folks ask me what I teach, I ask them to name their least favorite subject. Occasionally I do get English or history or science. But 90% of the time I get math ... to which I reply "bingo." :) I then engage them in a ...

Balarka Sen

Fubini's theorem on a triangle :)

Ted Shifrin

Oh good grief.

Balarka Sen

7:57 PM

The \int \int sin(x)/x dx dy thing

Ted Shifrin

No, @quallenjäger. The one I just linked you to above.

@Balarka: That's been answered dozens of times on MSE already.

Balarka Sen

It's a different question

But I need the picture :P

Ted Shifrin

Sure. You may steal.

quallenjäger

@TedShifrin Thanks! This is very very helpful to get my head around

Balarka Sen

Thanks. I am a thief with conscience however, so I shall acknowledge whom I stole

Daminark

7:59 PM

Hey everyone!

Balarka Sen

Hi @Daminark

Ted Shifrin

hi Demonark

You found the right link, @quallenjäger?

quallenjäger

Yes, thurston 37 th way to think about derivative

?

Ted Shifrin

Right.

quallenjäger

I am lack of intuition, thats why I still struggling at different point. I have been looking for many sources for the intuition of the derivative. Thanks for the link

Ted Shifrin

8:01 PM

That post actually took me more than a few minutes to write :P

Back later ...

Daminark

See you Ted!

@Balarka okay so it seems I don't completely understand the path lifting property so, here's something

orbit-stabilizer

$f$ cts, $X$ not compact, but $f(X)$ compact. What's an example of this?

Daminark

Let $E$ be the union of the two axes in the plane, and let $B$ be the x-axis, let $p$ be the projection. I'm trying to show that this map is not a Serre fibration.

@orbit constant map

orbit-stabilizer

No... a singleton is closed

Balarka Sen

@Daminark Okay

orbit-stabilizer

8:09 PM

sorry, that's what you wanted

thanks!

Daminark

No problem!

So we need to find a map $f:D^n \to E$ and $g:D^n\times [0,1] \to B$ such that $g(x,0) = p(f(x))$ but for which you can't lift $g$ to $E$

Balarka Sen

Yup

Vrouvrou

@TedShifrin bonsoir, vous allez bien ?

Daminark

And given that $E$ is a bit of a tight space, chances are we can do it for $n = 1$

Wait hold on I'm no longer convinced that this isn't true

Balarka Sen

Try $n = 0$

Leaky Nun

8:16 PM

@orbit-stabilizer $f=\sin$

$X=(0,100)$

$f[X]=[-1,1]$

Daminark

Because you could just define $h(x,t) = g(x,t)$ and that'll be in $E$

Oh hmm

orbit-stabilizer

@LeakyNun yeah, that works too. Sighhhh

why sigh?

good evening, all

failure is imminent

8:17 PM

@orbit-stabilizer hence compact

SoumyoB

indeed for me @orbit-stabilizer

I have my real analysis exam today and I haven't been able to sleep for the last two nights because of pseudo insomnia

orbit-stabilizer

I have mine today as well!

Daminark

@Balarka wait $D^0$? What?

orbit-stabilizer

And it's not looking good

SoumyoB

neither is mine

I'm so sleepy right now I can barely keep my eyes open

yet can't sleep

anyway best of luck @orbit-stabilizer

Balarka Sen

8:18 PM

@Daminark Yes? Consider $f(\{pt\}) = (0, 1)$ mapping to a point on the $y$-axis in $E$ with nonzero height.

orbit-stabilizer

What does yours cover?

you too

SoumyoB

several topics, the last ones being extreme value theorem, generalized mean value theorem etc

orbit-stabilizer

Rudin?

Daminark

Oh... I've never heard of people phrasing it as $D^0$ before, like that just didn't register as a thing which existed

Balarka Sen

Ah

Yeah D^0 is a point

Daminark

8:22 PM

Okay so you can map $g(*,t) = t$ and $f(x) = (0,7)$

Sorry that was a mess

Balarka Sen

Yup

SoumyoB

@orbit-stabilizer kinda
I'll catch you guys later, the exam starts in 12 minutes!

orbit-stabilizer

wow, Good luck!

Daminark

Oh so the reason we can't lift $g$ is because it won't make the $f$ triangle commute?

Balarka Sen

@Daminark Do you have to parse it in vile language like this?

Daminark

8:24 PM

Yeah because if $h$ lifts $g$

Balarka Sen

Like, is that absolutely necessary?

:P

Daminark

Kek

That's the first thing that came to mind

Balarka Sen

What an obscene person

But yeah, go ahead

Daminark

But yeah so if $h$ lifts $g$, you need $h( * ,0) = f( * ) = (0,7)$ to make one side work

Vrouvrou

@AkivaWeinberger hello, please do you know something about Differential equation of Clairaut ?

Daminark

8:28 PM

But then $h( * , \epsilon) = (\epsilon, 0)$

So that fails continuity

Okay sick

Balarka Sen

yup

Daminark

So that's a quasifibration which isn't a Serre fibration

Balarka Sen

Yup

Wait

That's a qifib?

Oh

Of course

Leaky Nun

I've encountered an interesting metric on $\Bbb R$ (on math.SE): $d(x,y) = \begin{cases} |x|+|y| & x \ne y \\ 0 & x = y \end{cases}$

cc @Secret

$0$ is the closest point to any point other than itself

I wonder what sort of topology this induces

$\{a\}$ is open for each $a \ne 0$

and so is $(-x,x)$ for all $x > 0$

in particular $\{0\}$ is not open

any set containing $0$ is closed

Balarka Sen

@Daminark did you prove serre fibrations are qifibrations

Daminark

8:45 PM

Yeah

Leaky Nun

@Daminark any idea?

Daminark

I proved that $p:E\to B$ is a quasifibration iff the inclusion of the fibers into the homotopy fibers is a weak equivalence

Balarka Sen

yeah

Daminark

And you know that if $p$ is a Serre vibration, that inclusion is a homotopy equivalence

You could just do the argument directly though

Balarka Sen

by lifting, yeah?

Daminark

8:48 PM

What I had in mind was that you take the homotopy sequence of the pair $(E,p^{-1}(b))$ and compare (compair it kek) to the homotopy sequence of a Serre fibration

Balarka Sen

@Daminark hmm, I see you have other accounts to post silly questions from in MSE

Daminark

Pah, I'm a cat person anyway

Balarka Sen

i posted two show off comments lmao

Daminark

Fucking hell $\mathbb{R}^n$ is absolute garbage

15

Balarka Sen

lool

-Amin, 2017