I am trying to prove if f has a right inverse then it's surjective

let g be the right inverse of f

then we know f o g = id_B

so let g(b)=a

then $$f(a) = f(g(b)) = (f\circ g)(b) =b$$

but how does this prove that for all b in B, there exists an a in A such that b=f(a)

@LeakyNun yes there is a counterexample, I saw one once but I forgot it

Hi @Ted!

Hi @Mathein

How are you doing?

@Maximus: You have it. You just didn't say it quite clearly. Start with $b\in B$. Now what?

Doing just fine, Mathein, thanks. You?

now you let g(b)=a @TedShifrin

Oh, it's a @Fargle

And then you show $f(a)=b$, so you've done what you wanted, @Maximus.

Heya @Ted

I'm doing well. I'm helping a friend with his bachelor thesis, which is quite fun

Better than doing your own, Mathein? :P

yo yo yo chat

Yo @EricS

@TedShifrin I'll start my own next semester

how goes it

howdy @EricSilva

there's still some stuff I have to do

@TedShifrin yes but I am confused. we let b in B so this means b is a random element in B and we force g(b)=a, how do we know there exists this a such that g(b)=a

Ah, cool, Mathein.

because I took the approach of taking all courses I find interesting instead of taking those that are required :D

Because $g$ is a function from $B$ to $A$, so $g(b)$ is some element of $A$. You just named it $a$.

but how do we know it maps it to an element there

maybe that particular b doesnt get mapped

What is a function?

or we assume it does

@MatheinBoulomenos i did this too! my last year of college is gonna have like no math bc i was a fool lol

Functions always map the entire domain.

oh right

so that means there exists at least an element a in A such that g(b)=a correct?

oh that makes total sense now, since g always maps the whole domain and f takes whatever g mapped then we will get B back, yes?

Not at least, @Maximus. Precisely one. You really need to understand completely the definition of a function.

So, @Fargle, how goes the fun with Hatcher?

yes sorry, that's what I meant

there exists precisely one element

@TedShifrin I'm pretty lost with it at the moment, so I'm trying to give it a break.

You could forget naming it $a$, and just say $f(g(b)) = b$.

Wow, I thought you were doing great, @Fargle.

I also found out that some nonsense I did for fun might actually help in the bachelor thesis of my friend

Oh, that's fun, @Mathein.

@Fargle, you on summer break?

@TedShifrin There's some point somewhere at which something isn't clicking, so I'm kind of just proverbially spinning my tires in mud.

Yep.

Well, I spent my career spinning tires, Fargle, so it's just a sign you're progressing.

yeah that makes sense, Thanks @TedShifrin !

Sure.

I'm sure this is quite well-known, but I didn't saw it before so I tried to work it out on my own: I called it "Tensor products of $G$-sets", like when you have a right $G$-set $X$ and a left $G$-set $Y$ and you use these actions to identify elements in $X \times Y$. I thought this looked like a tensor product, so I tried to work out as much analogous results to those for tensor products of modules as possible

this might help to describe some rep theory stuff that came up in the thesis

Oh, that's sorta like the associated bundle construction, @Mathein.

what do you mean by "spinning tires"

ACtually, it's usually spinning wheels, not tires :P

Hi @Ted

Hi, demonic @Alessandro. All done?

o/

Oh no. It's a @Balarka

Yes, as far as diffgeo is concerned (the exam went very well), but I'll do an algebraic number theory presentation next week which will also be my last exam

I'm proud that you did me proud. :) ... Are you sad that it's almost all over?

Not execissively, I'm looking forward to begin my master in October

I can't believe we've spent so many years in this chat.

If $g$ is a right inverse of $f$, $g$ must be injective.

yes but we never mentioned this in the proof

You can prove that in a similar fashion. You don't need to. It follows from your proof.

yeah I know that g technically has f as a left inverse so it's injective

in what way it follows from the proof?

If $g(b)=g(b')$, then $f(g(b))=f(g(b'))$ because $f$ is a function. Therefore ...

ah right

@TedShifrin Apparently Mather describes the Whitney stratification criterions as an intrinsic definition in terms of blowups, it seems

@Maximus I just meant "unable to gain traction"

Right, @Balarka. See the brief discussion of Nash blow-up and Chern-Mather classes in one of the little papers I sent you. :)

Ahh looking

It's the local Morse theory paper, yes?

No, the Whitney umbrella paper.

Hmm let's see where I put it

Ah found it

Boy, math was so ugly pre-LaTeX.

I kinda like the grainy typesetting fonts

It's classy

This was IBM Selectric typewriter.

And then scanning the reprint.

Ugh.

$\LaTeX$ is one of my top 5 things.

@Mathei That is the right perspective on that construction

pre-Latex is a myth

See the original Spivak 5 volumes, @EricSilva.

Ah of course this captures the idea.

the world has only existed since the original release in the 80s

everything that predates it is just a cover up

And it's about to stop existing.

bye bye world

@TedShifrin i have these on pdf but i dont like looking at old typeface lmao

My preferred typewriter has always been a Clark Nova

No, given the choice, I'd rather read TeX. But I'm not so fond of the default font.

@MikeMiller yeah, I managed to prove Hom-Tensor adjunction and the analog of Eilenberg-Watts (at least the part about tensor products) so far

@Mathein should be proud that I even normalized the Whitney umbrella :P

what's a Whitney umbrella?

A thing with a singularity that never goes away if you blow up any order of times

There's probably some categorical notion that encapsulates all similar constructions but I don't know what it is

Whitney is a really good umbrella manufacturer, they make umbrellas with no holes

truly cutting-edge

pls no cronenberg

@MikeMiller it's apparently "tensor products of functors" which works for monoidal categories

@Mathein: $x^2w-yz^2=0$ in $\Bbb P^3$.

But I haven't checked the details

the arguments for Eilenberg-Watts both in the homological algebra case and in the group action case seem quite specific

@Mathein Ah that makes sense, I remember seeing that come up in bar construction-like things in equivariant homotopy theory

you need coends to describe that and I didn't do coends so far

@EricSilva pls *more cronenberg

Yeah coends are valuable to the categorists

what is the meaning of your existence if you don't like cronenberg fam

@MikeMiller apparently coends also give a really description of geometric realization

i recognize that Cronenberg is interesting but i just get kinda get queasy at his particular brand of body horror

I think there's a coend formula for Kan extensions which you would like

@Mathei Yeah

what kind of body horror do you like

That's encapsulated in what I just said

do you prefer this

Balarka is throwing up at this conversation

The Cronenbwrg is fine though

I have the paper "this is the (co)end, my only (co)friend" bookmarked, I want to read it at some point not too far in the future

@MikeMiller That's some good body horror right there

Uncontrollable vomiting caused by reading nlab

Hellraiser $(\infty, 1)$

@Balarka I'm reading this paper my advisor sent me, you'd love it: pdfs.semanticscholar.org/b44b/…

i dont really have any feelings about the hellraiser movies

@BalarkaSen I looked for a counterexample because Leaky asked me and I just found a concrete and understandable counterexample on nlab

this doesnt bother me tho

I was shocked, to say the least

@Daminark I have seen it

Akiva read it with undivided attention once

@MikeMiller I like that they use integral notation. It's like category theorists are making fun of analysts

Ah, it's so quiet here without DogAteMy.

It really is

Even topologists write an integral sign for evaluation on the fundamental class, Mathein.

I just abuse notation and write the class for its evaluation

LOL

Well, it's well-known that Mike is abusive.

@TedShifrin hmm we didn't do that in our alg top lecture

notation is made to be abused

@Eric Hellraiser 1 is rly good Hellraiser 2 is ok to aight

Well, I smack anyone who writes $\int_a^x f(x)\,dx$.

Hellraiser n > 2 are a clusterfuck

Just read the book tbh

Corporal punishment in math

@MikeMiller it's been years since i watched the first one so ill have to revisit now that im more snobby than i used to be and actually appreciate things

I like hellraiser $\aleph_0$ the most

[garbled radio transmission] REPEAT: DO NOT ENGAGE (with Hellraiser n>2)

@TedShifrin im gonna write this exclusively from now on

Be prepared to be smacked, Eric.

@Ted I write $\int_a^x f$

@TedShifrin that smack is quite well-deserved

and do change of variables anyway

Delearn English just write unparseable integrals

not every abuse of notation is justified

lol

I very much try to stick with $\int_a^x f(x')\,dx'$

@Balarka: That's fine. But what if you don't have an $f$, but a formula? :D

@TedShifrin $\int_x^x x(x)\;\text{x}x$

@MatheinBoulomenos in what sense is it a tensor product?

@LeakyNun why did you quote that?

ah

smacks @Fargle with multiplicity

@MatheinBoulomenos to ask you a question afterwards :)

or as I prefer to write my functions and variables, $$\int_0^1 2(3) d4$$

well, the construction looks the same to me

Well, I did ask for it.

@Semiclassical i dont like primes bc sometimes my x is a function

@MikeMiller i just screeched at this

it's not perfect, no

my usual default in cases like that is $u$

I love how Fargle still wrote \mathrm{x} at the end

i like t and s

What a nerd

i tend to stay away from x tbh

the page refreshed for some reason and I assumed I got banned for that integral

for single variable

lmao

LOL, I've been banned for words less offensive than that.

\mathrm, ugh

i can't get myself to ever bother with that

@MikeMiller honestly u deserve to

@LeakyNun I mean, in the tensor product, you quotient out stuff like $xr - ry$ and in the tensor product for $G$-sets I quotient out by the equivalence relation generated by stuff like $xg \sim gy$

is there a joke here i dont comprehend with my peanut brain @Mike

its bad

@MatheinBoulomenos i see

Also there's a Hom-Tensor adjunction for that construction

real pet peeve: $\frac{\mathrm{d}}{\mathrm{d}d}$

derivative w/r/t distance?

yeah

$\frac{\mathrm{d}}{\mathrm{d}d} d^d$

@LeakyNun if you want, it's on my blog: wlou.blog

hi @loch

@TedShifrin I get the Nash blowup. You're taking the closure of the image of a given stratum (of a smooth stratified space, embedded in a smooth manifold) by the Gauss map to get the limiting tangent planes. What about the secants?

using $d$ as a variable in calculus contexts is a bad time

I see

what about using $\int$ as a variable

Secants aren't there unless you do the secant variety.

Im currently writing on a third blog post with applications in representation theory

and yet you're forced to do it in physics when you talk about capacitors, since people seem to insist on $d$ as the plate separation

$\frac{d}{d\int} = \text{Id}$ is the fundamental theorem of calculus?

@MikeMiller good $\int$hit

use $s$ gdi

there's a nice proof of Mackey's formula with this approach

@TedShifrin It's the crucial part of a Whitney stratified space!

You want limiting secants to be contained in the limit plane

I thought you wanted limiting tangent lines to be contained in the limit plane

@MikeMiller Both

That's condition A and B

I assumed if you had secant lines it would imply tangent linez

Right. So I haven't thought about it, but you could frame it by saying that a map on $X\times X-\Delta$ (for $X$ the smooth stratum) extends to the blow-up.

@TedShifrin Yeah that's exactly what Mather does

@MikeMiller That is correct

Not with bad singularities, Mike. You're doing secants on the smooth stratum.

@Semiclassical $\frac{\mathrm{d}}{\mathrm{d}d} \mathfrak{d}(\partial(d), \mathscr{d}(d))$

flagged

Think about my example to Asaf with the surface $x^3=y^2$. Limiting secants don't get you tangents to the $z$-axis.

On this notational nauseous note, I leave to eat lunch.

Or should that be nauseating notional note?

@TedShifrin It doesn't satisfy the limit secants condition

good alliteration

Right, @Balarka.

Oh, OK.

what does $ f(x) $ mean

Mike is saying if it satisfies the limiting secants condition, then it satisfies the tangent line condition

OK. Right.

Bye for now.

bye @Ted

Cool example anyhow :)

Bye

See you Ted!

@Semi same, never saw the appeal, I just write $\int stuff dx$

Or sometimes without the $x$, it depends

$\int f$

If it's clear

@MikeMiller yes, and taylor's theorem says that $\exp{d}=\operatorname{Id}$ locally, for functions that are nice enough

@Semiclassical ok in actuality if i wanna differentiate wrt something that's like a distance i use r and write $d/dr$

but hot take: i pronounce it dee durr

yeah

i tend to avoid $r$ for that because one uses $r$ so much in electrodynamics already

I am starting to use $\partial_t$ for $d/dt$ a lot these days

It's really convenient

@MatheinBoulomenos Better to say $e^d = 1$ :D

therefore the derivative operator is identically zero on analytic functions

Please stop @Mike @Mathein

but i also like using $s$ for the radial coordinate in cylindrical coordinates, so it's not entirely consistent

@MikeMiller right. So $\exp$ is not only $2\pi i$-periodic, but also $d$-periodic? thonk

@Balarka i write notation like $\partial_{t}$ for a tangent vector p much exclusively

Taylor's theorem says every section of a smooth fibration satisfies the h-principle locally with respect to any open partial differential relation

That's what Taylor's theorem is

$d/d|x|$ :P

It's just an h-principle

and then i write either $\frac{\partial}{\partial t}$ or $\frac{d}{dt}$ depending

I'm pretty sure I change notation for partial derivatives every 3 days

i pronounce the $\partial$ symbol as th-

hehehe mathein got rekt in algebra

so $\partial/\partial t$ is thatheet

@BalarkaSen it's analysis

it's h principle

$f_x$, $D_xf$, $\partial_x f$, $\frac{\partial f}{\partial x}$, $\int^{-1} f dx$, and so on

@Mathein I bumped into the algebraist that gave me the inverse limit problem today at lunch

@BalarkaSen ah

"I bumped into the algebraist" did you survive?

He liked my ideas/approach, told me not to spend too much time on it, and told me an obscene solution by Waterhouse

@Daminark No I didn't

F

He started teaching me what a Deligne Mumford stack is

It was like 3 hours long conversation

zombalarka

what's so obscene about that solution?

Ended with the conclusion that geometers don't understand orbifolds

These are obscene people man

whatever

lmao don't take it personally. I'm just memeing

I thought it was a clever solution

@Daminark i love how literally all of these are fine till you get to the last one

and then u take a sharp turn into crazytown

But algebra people have always ever been obscene. I have been reading a book on intersection homology. Beautiful stuff developed by topologists, Goresky and MacPherson. But here comes butting in the algebraists with their leader in charge Deligne and does perverse things with it

yes, Deligne did some stuff with perverse sheaves

@Balarka if anything topology is the obscene thing, like check out this picture of cobordism

That's just lewd

If a Mickey Mouse pair of pants is obscene to you I recommend getting help

Mickey Mouse?

No no no he's studying to do surgery on people, not on manifolds

This is just anime pants

"the category with compact oriented 2-dimensional cobordisms as morphisms, is the free symmetric monoidal category on a commutative Frobenius monoid object"

TQFTs are pretty cool though

That's how I learnt basic category theory and how to relate it to geometry

relate it to geometry sometimes people just miss the point...

Yo anyone know the assumptions needed on this theorem in order for it to be true? I would like to prove it. $e^f$ is measurable if and only if $f$ is measurable

Perhaps something to do with Borel measure?