The h Bar

12:08 AM

@0celo7 Check that taking the quotient in this way maps open sets to open sets: Any $U_i\times G_i\times V_i$ for the individual sets open in their spaces, they get sent to $\bigcup_{g\in G_i} (U_i\times g^{-1}V_i)$, which is a union of open sets ($g^{-1}U_i$ is open by continuity of the group action) and so is open.

@DanielSank Were all coworkers already proficient in git?

DanielSank

@ACuriousMind Proficient enough.

I did most of the merging.

Anyway, after this all happened, a coworker and I started thinking about all the problems with how academic paper refereeing works.

1) The number of reviewers is laughably small. PRL usually gets two referees. That is not enough people to comment on language/math/figure clarity, judge the correctness of the work and comment on whether the work is properly framed within the field.

2) The time for each revision is ridiculously long.

3) The referees know who the authors are, but the authors don't know who the referees are.

So we discussed ways to fix things up. Our solution was a website which scrapes arXiv entries, making a page for each one. Each page contains comments and issues.

That's it.

Turns out Scirate already exists and does most of what we want, but it doesn't have issues.

0celo7

@ACuriousMind What is $G_i$?

DanielSank

So, we contacted Dave Bacon, and now we're on our way to seeing if issues are a good idea.

0celo7

or, any of the other things?

DanielSank

Probably it's actually better to set up an independent issue tracker which can use Scirate's or arXiv's API appropriately.

Anyway, we'll be discussion all the considerations with the right people soon.

Danu

12:16 AM

@DanielSank It's a feature, not a bug!

DanielSank

^ False

ACuriousMind

@0celo7 With "for the individual sets open in their spaces", I meant that $U_i\subset U,G_i\subset G,V_i\subset V$ are each open.

Danu

@DanielSank Okay, better phrasing: status-bydesign

Potential hostility would be a huge issue

0celo7

@ACuriousMind Yes, but can't we take $U_i=U$ because $U$ is open, and $V_i=V$ because we want the whole fiber, and don't we need all of $G$?

DanielSank

@Danu There are already comments. People can be as hostile as they want in comments.

I do not see a new problem there.

ACuriousMind

12:18 AM

@Danu I would think it would be sensible if the review was mutually blind - why assume the authors might be biased by knowing the referees, but not assume that the referees might be biased by knowing the authors?

Seems like a bug to me sure enough

DanielSank

@Danu If someone files an issue saying "You're an idiot and your results are wrong", then I close the issue as "wontfix".

Done.

Danu

@DanielSank You're not seeing the issue

Say you reject a paper by a heavyweight in your field, and he proceeds to ruin your career.

#rekt.

ACuriousMind

@0celo7 Not sure if I understand what you mean. You wanted to know why taking the quotient in that way is a homeomorphism, not merely set-theoretic. To check that, check that the topologies on the product $U\times V$ and $(U\times G\times V)/{\sim}$ agree under this map, which means checking that all open sets are mapped to open sets.

Danu

So nobody wants that

(and this is a genuine issue---at least part of the "important people" do throw their weight around like this, I know for a fact)

ACuriousMind

I'm not sure if Daniel is suggesting the issue tracker be non-anonymous

Danu

12:21 AM

Ah

His point 3 certainly seems to suggest so.

Anyways, @ACuriousMind's idea of double blind is of course also relatively obvious but it also has some drawbacks---it might be very important to consider context in some cases

ACuriousMind

@Danu If the necessary context is not evident from the paper itself, doesn't that itself indicate a bad paper?

DanielSank

@Danu Fine, whatever.

No bearing on people filing issues.

It's opt-in. If I want community feedback, then I can look at the issues.

If I don't care, then I don't care.

People can already post on their blogs about how stupid I am etc., so it's not like an issue tracker gives people new capabilities for being negative.

Danu

@ACuriousMind No, not in certain (quite specific) cases, e.g. in papers that cover mostly preparatory material for an important theorem that the author wants to publish in a different paper, maybe?

I remember actually reading some cases (maybe on academia?) where it mattered

DanielSank

@Danu @ACuriousMind The point is that I would have loved to get community feedback before going through the insanely slow process of official PRL reviews. Several issues would have been caught, saving me a lot of time. Additionally, more reviewers means better paper.

Danu

@DanielSank Mhm

DanielSank

12:26 AM

Now, I could have just made my github repo public and invited you guys to read it and post issues.

Danu

It's all nice and cool in an ideal world

DanielSank

It's not just ideal. I basically already did this for my last paper and it was awesome.

To prove that, I'm going to make the github repo public.

0celo7

@ACuriousMind So you're saying that the map $U\times G\times V\to U\times V$ is open?

DanielSank

I already got permission from Google to do this.

ACuriousMind

@0celo7 Yes, but you have to pay attention that this map is not the projection, but $(x,g,v)\mapsto (x,1,\rho(g)v)$.

0celo7

12:28 AM

Yes, I know.

Ugh, this is so confusing

ACuriousMind

Good, then I guess that's what I'm saying, yes

0celo7

I want an open map $U\times G\times V/{\sim}\to U\times V$

@ACuriousMind i think this is a different map

different domain

ACuriousMind

@0celo7 Ah! Each surjective continuous map $X\to Y$ induces a homeomorphism $X/{\sim} =Y$ for $x\sim x'\iff f(x) = f(x')$ Now just check that this relation for the map I write down is the same relation you use to define the associated bundle in the first place, and you're done.

0celo7

I feel like this should be way easier.

ACuriousMind

What I just wrote is rather easy it doesn't need most of the stuff I wrote earlier, just the map $U\times G\times V\to U\times V$ itself.

Danu

12:33 AM

HUZZAH, Chapter 2 of Huybrechts is finished!!!

0celo7

I've never heard of what you just wrote and don't know how to prove it

Danu

Exactly 50 pages of notes typed already

ACuriousMind

lol, I was just called "Ms." in a comment

DanielSank

@ACuriousMind Take that.

Danu

Gender freedom is real

DanielSank

12:34 AM

@Danu You just reminded me of something funny.

At Google, everyone has a little profile page.

You get "badges" for stuff you've done. It can be anything and anyone can set up a badge.

ACuriousMind

Well, I guess technically by using a female picture I make that the default assumption, but I'm puzzled why the commenter felt the need to use Mr/Ms at all.

DanielSank

Examples are: being a member of the vim user list, completing various trainings, having successful meme posts, etc.

For various reasons, a coworker and I had to pay the power bill for our lab.

So I made a badge which is a picture of my hand holding a $100 bill in front of an electrical outlet and the mouseover text is "I paid Google's power bill out of my own pocket".

Danu

@ACuriousMind Was from India? They always do this stuff

0celo7

Always do what stuff?

Danu

@DanielSank I wonder how you were reminded of this by "gender freedom" :)

ACuriousMind

12:36 AM

@Danu Can't tell, username is nondescript and no characteristic "doubt"

Danu

lel

DOUBT

that's so accurate

ACuriousMind

@Danu I'm now kinda afraid what they did to the outlets to have to pay the power bill...

Danu

lel

0celo7

@ACuriousMind Ok, so what one shows is that $f:X\to Y$ descends to a cont. bijection $g:X/\sim\to Y$ and that this is a quotient map?

A quotient map is open, so this is a homeomorphism?

ACuriousMind

yep

0celo7

12:39 AM

Ok, we did that in topology.

DanielSank

@Danu Oh, there are badges related to gender identity etc.

0celo7

Ok, so to review: we have a surjection $U\times G\times V\to U\times V$. This induces a homeomorphism $U\times G\times V/\sim\to U\times V$.

Wait, have to check that it's continuous.

That's immediate since the group actions are continuous

Danu

I needda break out of this rhythm of sleeping at 3 AM

0celo7

Ok, so $U\times G\times V/{\sim}\approx U\times V$. I've lost track of why we wanted that.

Danu

Honestly, I think the reason this happened is because it's around 12-3 AM that I can get help in the math chat with my problems :(

ACuriousMind

12:42 AM

@0celo7 Because $\pi_V^{-1}(U) = U\times G\times V/{\sim}$.

0celo7

Hard equals?

ACuriousMind

Yes

0celo7

Ok, let's double check that.

ACuriousMind

@Danu What happened?

Danu

@ACuriousMind Complex Geometry

by Huybrechts

0celo7

12:43 AM

His sleep schedule has turned into yours @ACuriousMind

Danu

that shit is next level for me

yup

ACuriousMind

Oh, I didn't even register that as unnatural :D

0celo7

lel

the vampirism is real

Danu

Also my supervisor gave me some potential topics, I'll tell you about it sometime soon

0celo7

@ACuriousMind Btw this all is trivial if one constructs the associated bundle chartwise...

$\wp$

ACuriousMind

12:46 AM

@Danu Nice, mine still has to get back to me but I guess I'll have to wait for the long weekend to pass now

0celo7

^for reference so I can draw it in my notes

long weekend?

ACuriousMind

btw, warning: You can't shop on Monday

Danu

yes, I realize

god damn germany

0celo7

how does one draw the WP

ACuriousMind

@0celo7 Monday is the day of reunification.

0celo7

12:48 AM

Oh god it's my mom's birthday monday

Apples are pressed

my best shot:

Time to filter a bit

@Danu You should enjoy the tranquility of these holidays :P

0celo7

^sounds like a commie to me

@Slereah Pic?

Danu

12:50 AM

@ACuriousMind Ain't no holidays man

I'm working on this complex geometry stuff all day every day

...till 3 at night!

ACuriousMind

You sound like you need to relax a bit

0celo7

Sounds like me this summer...

@ACuriousMind Ok I think I'm convinced. Many thanks.

@DanielSank Reading your lectures on TeX.

Danu

I'm going to bed now---bye guyz

ACuriousMind

@0celo7 Great :)

@Danu night

0celo7

Oh, maybe instead of messing around with geometry I should look at the QM notes again.

Slereah

1:02 AM

the hooch

0celo7

lol

Slereah

well

applejuice, currently

0celo7

it's dip spit

ACuriousMind

Are you sure a plastic bottle is the right choice?

0celo7

tobacco juice

Slereah

1:03 AM

Very sure it's not

Probably will buy a glass jug for next time

0celo7

get a beaker

Slereah

Well I guess he could help me

He has experience

0celo7

@DanielSank I'm lost already. What is this macro? What is a "directory structure"?

Slereah

Also I don't have a vapor lock

So I'm using a birthday balloon to seal it

0celo7

lol

terrorist

Slereah

1:06 AM

Au contraire

That's to avoid explosions

The birthday balloon is the antiexplosion measure

ACuriousMind

Did you at least sterilize the bottle?

Slereah

Well I cleaned it

0celo7

inb4 Sam dies of botulism

Slereah

Wouldn't go as fancy as "sterilized"

ACuriousMind

@0celo7 I don't think the conditions are sufficently anaerobic for that

Slereah

1:08 AM

Apparently Russia banned alcohol from 1914 to 1923

0celo7

look at Dr. Biology over here

ACuriousMind

But I'll take no bets on what else might grow in there

Slereah

I wonder if that's what triggered the revolution

My ancestors didn't need no fancy sterilized vats to make cider!

0celo7

they also scratched their butts and sniffed thereafter

Slereah

And I keep to that tradition

0celo7

1:13 AM

@ACuriousMind What a good way for formatting a theorem with two parts?

Slereah

Let's put that shit in the basement

So anyway

What is the subset of reals in hyperreals

Is it just $(x_i)$ that converge to a real value

Although that should be like

That value + some infinitesimal

0celo7

...don't you have a book on this?

Slereah

Well yes

IN THE MAIL

0celo7

jeez french mail takes forever

Slereah

hurry up postman

Well it is from the US

0celo7

1:27 AM

@ACuriousMind Crap, I think the Hausdorffness thing is proved in Munkres.

One doesn't need my fancy trivialization proof

user218912

does anyone here like the flipped classroom approach to teaching?

0celo7

no

it's terrible

user218912

i think it's dumb.

Newmann

@ACuriousMind It seems that you just are trying to muss up the questions. Here are your comments posted in last 2 hours: I don't understand what you mean by "include density as a dimension in space-time". – ACuriousMind 2 hours ago ; I'm not exactly sure what kind of answer you're looking for for this question - ... – ACuriousMind 2 hours ago

0celo7

whoever approves of it should be shot tbh

user218912

1:32 AM

@0celo7 the math department at my uni is adopting it for some courses.

Newmann

@ACuriousMind ... How can a photon be emitted "radially outwards" precisely at the event horizon? ... Much of the strangeness here comes from the fact that you posit a situation which is ill-defined. – ACuriousMind 2 hours ago ; It's not clear what that means - what kind of "properties"? – ACuriousMind 1 hour ago ; You'll have to be a bit more specific about what you want to know. ... So, what exactly do you mean by "restriction of spacetime geometry"? – ACuriousMind 2 hours ago

0celo7

@Newmann what are you doing?

Newmann

@0celo7 sorry for this, just seems that ACuriousMind tries to muss up all the questions with the same negative comments. Somebody should stop her.

0celo7

Yeah, she does that.

user218912

lol?

user218912

1:42 AM

@Newmann first of all ACuriousMind is a boy, and also I'm sure all of his comments are logical and based off of real physics.

Slereah

Is @ACuriousMind a lady now

Are you hiding something from us @ACuriousMind

0celo7

ACM is genderfluid.

user218912

@0celo7 wow I just realized that was a thing right now.

0celo7

wow

user218912

it's weird.

0celo7

1:45 AM

privileged cis

Slereah

$| ♂ \rangle + | ♀ \rangle$

2

Up to normalization, of course

0celo7

what's the Hamiltonian for that system?

Newmann

@IceLord There are a lot of non-logical and unrelated notes in her/his comments. By the way ACuriousMind is she or he? Just I took a look at her profile.

0celo7

ACM is a she

Look at the profile pic

@ACuriousMind is known as "mamma duck" around here for her motherly qualities

Newmann

yeah

0celo7

1:53 AM

but if you attack her you might regret it

Newmann

I see.

0celo7

the mods like her, so they side with her

Newmann

I see. no, i am not going to lose my time for her.

dmckee

@Slereah In Alabama they'd tell you there is nothing normal about that!

DanielSank

@0celo7 You're trolling, right?

Slereah

1:57 AM

They will collapse that wavefunction

DanielSank

Directory structure means how your files are organized in directories etc. on your computer.

dmckee

I mean, away from Huntsville and Tuscaloosa, anyway.

DanielSank

@Slereah Androgynous state?

Slereah

I thought@ACuriousMind was known as the German warmachine

dmckee

But most of the state doesn't trust those bastions of liberal agenda,

user218912

1:57 AM

@Newmann you do realized you're being trolled right?

0celo7

@DanielSank I don't troll.

dmckee

@Slereah But a motherly warmachine with a fetish for physical precision.

@0celo7 You've got that backwards. You so rarely fail to troll that no one can recall it happening.

0celo7

What?

user218912

yeah I don't get that either.

Newmann

@0celo7 you are fine ;) don't worry

0celo7

2:02 AM

@DanielSank But yeah I don't follow what you're trying to do

user218912

@Newmann lol

0celo7

Ok, let's get the notation down.

$$TM=(B(M),M,\Bbb R^n,\mathrm{GL}(n,\Bbb R),\wp)$$

Slereah

holy shit

5 AM

I'd better go to bed

oh wait only 4

0celo7

wait

yeah

Slereah

still

0celo7

2:09 AM

don't you have a job?

Slereah

nope

0celo7

wait

is it $\mathrm{GL}$ or $\mathrm{Gl}$

first one.

@IceLord Lee!!!

NeuroFuzzy

Down to four nominations?!

Wow.

user228700

2:47 AM

Hi :-) I've a bit of a homework-tsy question. The problem given is "The speed of a particle travelling in a circle of radius 20cm increases uniformly from 6.0 m/s to 8.0 m/s in 4.0 s. Find the angular acceleration".

user228700

First of all, I note that since the speed is changing uniformly, the tangential acceleration is constant, because of which its value is given by $(V_2-V_1)/(t_2-t_1)$ By doing this, I obtain $a_t$(tangential acceleration)=0.5 m/s², after plugging in the numbers given in the question.

user228700

This, however, is only the *tangential* acceleration. The magnitude of the *radial* acceleration is given as $v\omega$ but what value should I substitute for $v$ here, and which value of $v$ must I use to calculate $\omega$?

0celo7

Homework?? I vote to close

dmckee

The angular velocity is changing just as the tangential velocity is changing. Pick a time and use the tangential velocity at that time. Or average over the motion.

user228700

@0celo7 Yes, homework. You can't vote to close in the chat! Besides, some people here do help with homework as well, if I specify my work on the question and exactly where I think the problem is.

dmckee

2:52 AM

@KaumudiHarikumar He's trying to be clever.

user228700

@dmckee How am I supposed to average over the motion when the initial position is not given..?

0celo7

@dmckee Oh, I think I succeeded.

user228700

OK, I'll wait for @JohnRennie.

user228700

Wait, angular acceleration is defined as the rate of change of speed of the particle?!

dmckee

@KaumudiHarikumar No. The rate of change of the angular speed. But the angular speed is linked to the linear speed by the radius.

0celo7

3:10 AM

Whoops!

dropbox.com/s/8hwh5958bpqpnhv/…

@IceLord I don't know if this will become anything

I need to define manifold.

user228700

@dmckee You mean $d\omega/dt$?

dmckee

Yep.

John Rennie

3:55 AM

@KaumudiHarikumar :: as if by magic John appears ::

Secret

Given magnitude
$$a_c=\omega^2 r$$
$$\frac{da_c}{dt}=2\omega\frac{d\omega}{dt}r$$
$$\frac{da_c}{dt}=2\omega\alpha r$$
Now since
$$a_t=r\alpha$$
Hence
$$\frac{da_c}{dt}=2\omega a_t$$
Therefore the overall radial acceleration is
$$a_c(0)+\int_0^4 2\omega (t) a_t (t) dt$$
where $t=0$ corresponds to $v=6.0$
which I have no idea how to integrate analytically
I must be overthinking this...

user228700

4:11 AM

@JohnRennie :-) Morning sir.

user228700

Oh, you've disapparated back whence you came?

John Rennie

Morning, I'm still here

user228700

OK. So yes, that question... Angular acceleration, defined as $d\omega$/dt$ gives the total(sum of the tangential and radial accelerations) acceleration of the body, correct?

John Rennie

No

user228700

Oh... :/ What is it then?

John Rennie

4:26 AM

Suppose you're accelerating radially away from the centre, then $d\omega/dt=0$

user228700

@JohnRennie I don't understand. What do u mean by "radially away"?

John Rennie

All measurements of angular quantities have a centre about which they are measured e.g. the angular velocity of the Earth's rotation is measured about it's centre of mass.

user228700

@JohnRennie Yes, OK...

John Rennie

If you're moving directly away from that point then your angular velocity is zero, so your angular acceleration is also zero.

So you can have a non-zero radial acceleration and $d\omega/dt$ can still be zero.

user228700

@JohnRennie OK, but that's not exactly circular motion...no?

John Rennie

4:33 AM

Ah, well, you didn't specify circular motion in your question ...

In circular motion the radial acceleration is by definition zero otherwise the radius would change and the motion wouldn't be circular

user228700

@JohnRennie Oh, no, no! I mean centripetal acceleration, which causes the direction of the body to change w/o changing its speed.

user228700

(I read that the terms "radial acceleration" and "centripetal acceleration" are synonymous.)

John Rennie

If the radius is $r$, then the radial velocity is $dr/dt$ and the radial acceleration is $d^2r/dt^2$. At least that's what radial acceleration normally means.

The centripetal acceleration is $r\omega^2$ so it is not related to $d\omega/dt$

user228700

@JohnRennie But no, the velocity of the body has no component along the radius in the situation that I'm trying to describe...

user228700

user116211

4:41 AM

Did you complete the set theory preliminary of Munkres @secret?

Secret

Still working on exercise 5 and 6 of the function section. I need to figure how to account for the non surjective case when proving the left and right inverses

user116211

ahh! sure.

user228700

^That is the situation that I'm trying to describe.

John Rennie

I think I'm a bit lost ...

Everything on that page looks fine.

user228700

@JohnRennie Yes, it does...

John Rennie

4:45 AM

Acceleration is just $\frac{d}{dt}\mathbf v$ where $v$ is a vector.

Secret

Well it appears I am thinking too much about the issue last night that I end up dreaming about solving some weird set theory problems, including a nonsensical function which maps the empty set into a union of 3 closed intervals

user228700

@JohnRennie Yes, OK, what about angular acceleration?

user116211

Speaking of empty set, did find an exciting post today..

user116211

@KaumudiHarikumar $\dot \omega$

John Rennie

The angular velocity is defined by $$\omega = \mathbf r \times \mathbf v$$

And you get the angular acceleration by differentiating with wrt time.

user116211

4:47 AM

Check this post @Secret:

user116211

4

Q: Is the empty set homeomorphic to itself?

Consider the empty set $\emptyset$ as a topological space. Since the power set of it is just $\wp(\emptyset)=\{\emptyset\}$, this means that the only topology on $\emptyset$ is $\tau=\wp(\emptyset)$. Anyway, we can make $\emptyset$ into a topological space and therefore talk about its homeomorph...

user228700

@JohnRennie OK. I greatly suspect that I'm the one that's lost; this $d\omega/dt$ has nothing to do with the centripetal and tangential acceleration..?

user116211

@JohnRennie After you omit the $\dot r$ term (of course it's zero) ;)

Secret

Well, that map h in the question is precisely the empty function

user116211

Anways, back to reading Langrange's $\lambda$ Method in Lanczos.

user228700

4:50 AM

@JohnRennie Never mind. Is this correct:

John Rennie

In the case of angular motion the radial distance is constant, and after a lot of algebra you end up with $$\frac{d\omega}{dt} = r\frac{dv_t}{dt}$$

Where ${dv_t}{dt}$ is the tangential acceleraion.

user228700

..?

John Rennie

That assumes $d{\mathbf r}{dt} = 0$. Yes?

user228700

@JohnRennie Yes.

John Rennie

4:54 AM

But although $|\mathbf r|$ is constant, the direction of $\mathbf r$ is constantly changing with time. Yes?

user228700

@JohnRennie Oh crap, yes...

user116211

@KaumudiHarikumar No.

user228700

@MAFIA36790 Huh?

user116211

@KaumudiHarikumar Ponder over what John said....

John Rennie

When you grind through all maths you do end up with: $$\frac{d\omega}{dt} = ra_t$$

where $a_t$ is the magnitude of the tangential acceleration.

and obviously $r$ is the (constant) radius.

user228700

5:00 AM

@JohnRennie OK. I will try to arrive at that then. Brb.

John Rennie

But proving this properly is more effort than you might think. It's most easily done by switching to polar coordinates.

user228700

@JohnRennie Oh great :/ I'll try for around 15 minutes anyway.

user116211

Well, if you like @KaumudiHarikumar, this can be done in Cartesian coordinates too. I may help if you say so.

user228700

@MAFIA36790 Thanks :-) I will ask. Let me try...

user228700

Yeah, no, I keep getting

user228700

5:15 AM

user228700

(The other term becomes zero 'cause the cross product becomes zero...)

user228700

..?

John Rennie

@KaumudiHarikumar I can't remember the proper expression for the angular acceleration, so that might well be correct. If you calculate the magnitude you should end up with: $$a = \frac{a_t}{r}$$ where all the quantites in the equation are the magnitudes.

$a_t$ is the linear tangential acceleration

user228700

@JohnRennie But I've got $a_{net)$ on the L.H.S :/

user228700

@MAFIA36790: Can u please dispense that help u were taking about..?

John Rennie

5:27 AM

What do you mean by $a_net$? The angular acceleration is a vector in the same direction as the angular velocity.

Are you getting mixed up with the centripetal acceleration again?

user228700

@JohnRennie Not in non-uniform circular motion...

user228700

@JohnRennie I think so...

John Rennie

Centripetal acceleraion is a completely different and unrelated quantity. If you're trying to compute centripetal acceleration this isn't the way to do it.

user228700

@JohnRennie No, I'm trying to relate tangential acceleration and angular acceleration; properly, using vector notation and all...

John Rennie

Back to basics: the angular velocity is a vector with a direction normal to the plane of rotation. Yes?

0celo7

5:31 AM

proof?

user228700

@JohnRennie Yep.

John Rennie

And the direction of $\omega$ is always the same regardless of its magnitude i.e. always normal to the plane of rotation. Yes?

user228700

@JohnRennie Yes.

user228700

Oh, crap. I see where u're going with this.

John Rennie

So when we talk about $d\omega/dt$ we mean the change in magnitude of this vector.

Because the direction never changes.

user228700

5:34 AM

Yeah, yeah!

user228700

Damn. It's right here:

John Rennie

So the angular acceleration is a vector normal to the plane of the (circular) motiion.

user228700

John Rennie

$\mathbf a_t = \alpha \times \mathbf r$ is correct yes.

user228700

@JohnRennie Okay. Okay! Problem solved. I'm an idiot. Thanks so much sir! :-)

John Rennie

5:37 AM

>I'm an idiot
The story of my (physics) life :-)

user228700

@JohnRennie Haha, yes, that may just be true for every student out there :-)

Manashi

Even in this case, checking dimensionality in formulas like $$\frac{d\omega}{dt} = ra_t$$ is still strongly advised.

John Rennie

@KaumudiHarikumar Understanding and accepting this is the secret to a happy life. A scientific career is a long series of screw ups with occasional flashes of inspiration in between :-)

user116211

@KaumudiHarikumar It's not $\rm net\,.$

John Rennie

@Manashi particularly as I missed out the factor of $1/r^2$ in that equation :-)

user116211

5:44 AM

@KaumudiHarikumar So, do you want to derive the general results?

user116211

Tell me when you want to.

Secret

Got 5a figured out

I have one way to argue about the conclusion, and one way to prove it using previsouly derived lemmas

Proof:
Let $f^{\leftarrow}$ be the preimage of f
We knew from a lemma that we previosuly proved that
$$ff^{\leftarrow}(B) \subset B$$
$$f^{\leftarrow}f(A) \supset A$$
with equality when f is sujective and injective respectively

Now the conditions of the inverses mean
Left inverse: $ff^{\leftarrow}=\text{id}_B$
Right inverse: $f^{\leftarrow}f=\text{id}_A$

Therefore subbing these in we get
$\text{id}_B(B)\subset B$
$\text{id}_A(A)\supset A$
i.e.
$B \subset B$
$A \supset A$
Thus
$B=B$,$A=A$
Therefore, respectively the fs have to be surjective for existence of right inverses and injective for…

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