I've solved my problem!

@Thorgott Its kind of algebraic so maybe you want to hear about it?

or anyone?

sucks

@XanderHenderson you are here? Want to hear about the problem I solved?

is it about an infinite-dimensional cross-polytope?

is the complex matrix exponential periodic o.0

silly: yes, exp(z+w) = exp(z) exp(w) holds for all z and w, and there are nonzero w for which exp(w) = 1 (e.g. w = 2 pi i and its integer multiples)

oh, you said matrix, same answer but with w = 2 pi i * the identity matrix, and noting that exp(z+w) = exp(z) exp(w) holds as long as z and w commute

@leslietownes no, its about the ring $C(X)$

jakobian: for weird X, i hope?

I've shown that if $P$ is a prime ideal of $C(X)$ and $f\geq 0$, then $(f+P)^r = f^r+P$ is a well-defined exponentiation by real numbers in $C(X)/P$

for all $X$

Insert commuting diagram here

And this is cool because I thought I could only take rational exponents

And it behaves like you'd expect from taking powers, for example if $0\leq a < b$ where $a, b\in C(X)/P$, then $a^r < b^r$

I've listed 9 properties that an exponential should obey and it satisfies them. If you replace $P$ by a z-ideal $I$ then $C(X)/I$ is only partial order instead of total order, but exponentiation still satisfies those 9 properties

I had a lot of partial solutions but, although distinct from my other attempts, I finally solved it. And its not that hard

Its a question I posed myself

How to find the set of discontinuities of this function? I don't know how to start.

@ThomasFinley Draw the graph

So the question is, I don't really know why topologist study fiber bundle unless, we know certain manifold can be realized as a total space of some other manifold.

for example like Seifert fibered space although the base is an orbifold

(I'm not talking about vector bundle or principal $G$-bundle or something, just a general fiber bundle)

I know $\nabla f$ has the direction of the fastest increase. I am thinking $\nabla f$ as $\lim_{\Delta \overrightarrow{r} \to 0} \frac{\Delta f}{\Delta \overrightarrow{r}}$ and wondering how can it have the same direction no matter from which direction $\Delta \overrightarrow{r} \to 0$.

i'm not sure i understand where the n is coming from in the denominators of those things. it also isn't entirely clear what "S" and "S'" are, or what their relation is to "x_n"

stepping back from that a moment, that T is an isometry is a stronger condition than that T has norm 1 (i.e., showing that ||T|| = 1 does not show that T is an isometry) so it is not clear why you begin by some stuff that seems to be beginning to evaluate ||T||

to put it another way, you want to show that for ||Tx|| = ||x|| for every x in the space

you might try writing e.g. x = (x_1, x_2, x_3, ...) and using the definition of || || to write both ||Tx|| and ||x|| in terms of the supremum of a set of real numbers involving the x_n

and having done that to argue that those suprema are equal

@giannisl9 how are u dividing by a vector

@nickbros123 like $\frac{\Delta f}{\Delta x}\hat{i} + \frac{\Delta f}{\Delta y}\hat{j}$ since in 2D $\Delta\overrightarrow{r}=\Delta x \hat{i} + \Delta y\hat{j}$

@giannisl9 where can I learn all these stuff?

Oh sorry, I know all these stuff, It is just derivative of vector's x and y components.

@leslietownes they are supposed to be bounded functions from N to some field K (pick R)

@leslietownes they are elements of $ l^\infty (\mathbb{N} $ so they take natural numbers and send them to some image in the field, which i called "x_n"

@leslietownes you are correct, i guess i wanted to show it has norm equal to the norm of S, i got confused a bit

@leslietownes thats what i did. the norm of S would be $ \Vert S \Vert = \sup_{n\neq0} \frac{\vert x_n\vert}{n}$

I got to employ gluing subsets of solutions to diff eq's and I saw goblin gone's question unfortunately no answers

is ted un@able now?

ted shifrin

He is fine, he is taking a long break from the chat

I think I might do the same

but before I leave I want to ask something

nevermind

I'm studying the group $\mathbb R/\mathbb Z$, in particular, this answer.

> ...$\mathbb{R}/\mathbb{Z}$ is in $1-1$ correspondence with the set $[0, 1)$. The group structure also carries over to $[0, 1)$, you can check it is addition modulo $1$.

As far as I understand, addition modulo $1$ is simply the positive fractional part of a real number, however, since it is called addition modulo $1$, I wonder if it related to modular arithmetic. The modular arithmetic I'm familiar with is: if $n$ is a positive integer and $x,y$ integers, we say $x\equiv y$ (mod $n$) if $n\mid (x-y)$.

Its related in the sense that the latter is an instance of the quotient $\mathbb{Z}/n\mathbb{Z}$

@psie Same thing here, $x,y$ reals, we say $x\equiv y$ (mod $n$) if $(x-y)$ is an integer multiple of $n$

Here $n=1$, So you get your fractional part definition.

@psie addition modulo $1$ is not simply the fractional part, I think you mean something else

Fractional part of addition of real numbers ?

Note however that unlike in modular arithmetic, $\mathbb{R}/\mathbb{Z}$ is not a ring

ok

Like Soumik says we can define modulo $n$ in $\mathbb{R}$ and this would be an instance of group $\mathbb{R}/n\mathbb{Z}$. But this still has no ring structure, and amounts to the same as $\mathbb{R}/\mathbb{Z}$.

The whole power of modular arithmetic comes from not just the group but also ring structure

(And that its a finite ring)

@psie It is more related to analysis (harmonic analysis, in particular) than to modular arithmetic. E.g. a Fourier series can be thought of as a map from an appropriate function space on $\mathbb{R}/\pi\mathbb{Z}$ to the corresponding function space on $\mathbb{Z}$.

(Where "appropriate function space" might mean continuous functions, or $L^2$ functions, among others.)

interesting

Since we are on the topic, you can look at $\mathbb{R}/\mathbb{Z}$ algebraically, as well as topologically as the circle $S^1$ with multiplication that of complex numbers. The groups $\mathbb{Z}/n\mathbb{Z}$ correspond to $n$th roots of unity then

The union of all $p^n$th roots of unity for prime $p$ forms an important group called Prufer $p$-group

is the space of linear operators over a banach space from itself to itself is just the Endomorphism space? or are we distinguishing the terms used ?

I am noticing that in Functional analysis books, no one of these terms are being used (Endomorphisms, automorphisms, and so on)

So i feel like they are some distinction? or is it just historically related to use different terms

@Mad Yeah, because analysts are all allergic to category theory. :P

@XanderHenderson but they are the same structures?

@Mad I have no idea. I'm an analyst, and deathly allergic to category theory. I don't know (or care, particularly, what an "endomorphism space" is).

Analysts are generally not looking to understand a thing in the most general way possible. Typically, analysts are looking to understand how specific spaces behave, and the abstractification of category theory isn't terribly useful.

@XanderHenderson its a vector space of linear mappings from V to V. it becomes a lie algebra with a commutator, its actually quite neat

So, sure, it might be the same structure. So what?

@XanderHenderson Well i am just wondering why not the same terms are used. thats all.

History is also likely a factor. The categoric point of view is pretty new, while most of the relevant analysis is at least a couple decades older (e.g. Schwarz and Sobolev established a lot of the theory in the early-to-mid 20th century).

There's also the issue that there are various reasonable choices for what the morphisms should be in "the" category of, say, Banach spaces

Not that analysts care about this issue, they know which morphisms they want to work with, they might happen to not be the most categorically convenient choice but so what

(also automorphisms and similar terminology is definitely used in operator algebras)

Or even if they actually do form a category. A friend of mine wrote his phd thesis on a similar topic. It isn't exactly something I understand, but he was trying to understand something about Riemannian and symplectic manifolds in a category-theoretic sense. But these structures don't support push-outs in the way that was required, so it took a ton of work to add enough category-thereoretic machinery to get the right push-outs.

Well i just noticed something.

Endomorphisms are not nessicarily continous if the space is infinite dimensional

So we can not say that $ End(V)=Linear(V,V)$

This would only be true if V has a finite dimension, we have an inclusion: '$Linear(V,V)\subset End(V)$

as Alessandro says, it depends on the category being considered

and that ambiguity is probably part of why functional analysts don't typically adopt the terminology

there's a text on functional analysis with a categorical slant, "Lectures and Exercises on Functional Analysis" by A. Ya. Helemskii

@onepotatotwopotato I don't understand the question.

Fiber bundles occur everywhere in topology. Isn't it natural to understand something that keeps cropping up?

anyone like the Oppenheimer movie?

@RyderRude Haven't seen it yet. There is literally one movie screen within an hour of where I live, and it tends to show more of "family" films (it is rare that anything R-rated, or long and boring, ever shows up here).

oh.. but it's been out for long now. u can stream it.

@XanderHenderson but this movie is great in cinemas, yeah

@RyderRude I will likely stream it once it is on one of the services I already pay for. I don't really feel like paying Amazon or whatever $20 for the privilege, and I certainly am not going to pay for *shudder* Peacock.

@AlessandroCodenotti you want isomorphisms to be either linear isomorphisms or isometries but you want to work with both

So yeah, this pretty much proves the point

About category theory. I find it often very much not insightful to consider mathematical objects as some category. The issue is always with choice of maps. It doesn't seem like an useful data to consider

They are useful for what they are made for

But not to put category structure on everything

tsk tsk, the right context is obviously to work with the category of liquid vector spaces

In particular I don't find useful the view of "the" category of objects

skill issue

I don't have skill to be able to have issues with it

i only have issues

the inter-universal category

Does anyone understand why the part in green is true?

Here $L$ is the Lie functor (for if $G$ is a Lie group, then $L(G)$ is its corresponding Lie algebra, and likewise for morphisms)

@BalarkaSen I prefer to work with plasmas.

One can alternatively use the fact that Polish subgroups of a Polish group are closed, which is more general and is proved with considerably less machinery

@ShaVuklia this is just the definition of the subspace topology together with the fact that $\exp$ is a diffeomorphism on $V$, no?

ah it is indeed!

thanks, I lost my overview of what was going on there for a sec

@BalarkaSen an object in the category of skill issues

I have two cohomologous n-forms, $\omega_0,\omega_1$ on $M$ and a submanifold $i\colon S\hookrightarrow M$ such that

I hate when I do that.

One second as I write my entire statement out...................

I have two cohomologous n-forms, $\omega_0,\omega_1$ on $M$ and a submanifold $i\colon S\hookrightarrow M$ such that $\omega_0|_S = \omega_1|_S = 0$ and are in the same relative cohomology class (wrt $i$). Because they are cohomologous, $d\beta = \omega_0 - \omega_1$ for some $(n-1)$-form $\beta$. Something I am reading claims that I should have $i^*\beta = 0$, but I can't figure out how they get this right from the relative cohomologous statement.

The best I get is that of course $i^*d\beta = 0$.

Using Bott-Tu's definition for relative cohomology, I get that it's something like $i^*\beta = d\theta - i^*\alpha$ for some $\alpha\in\Omega^{n-1}(M)$ and $\theta\in\Omega^{n-2}(S)$, and where $\alpha$ and $\beta$ differ by a closed form.

Err, sorry that should be that $\alpha$ is a closed form, not that they differ by a closed form...I think...

Feeling like Magnus I don't believe in fortresses Carlsen

is there any benefit in reading a proprietary metric space book, rather than going through "basic-topology" chapter in Baby Rudin?

i mean if life gave me the time id do both ofc

@nickbros123 What do you mean by "proprietary"?

hi guys

Also, I have read several books on metric spaces, but, like, my research area is about metric spaces which carry measures (sometimes quite pathological measures), so I am interested in the specifics which might be contained in a book on metric spaces, rather than a general text on analysis.

@Sahaj I'm not your guy, pal.

@XanderHenderson i meant exclusive or specialised or something

@XanderHenderson oh man. :(

for eg: Metric spaces by michael O searcoid

this one was recommended to me by an old but wise man

Yeah, I have a copy of that text running around somewhere. It is very... undergraduate-y.

Which might be what you want?

Like, it is a good text for an undergraduate.

i suppose yes, im an undergraduate

It doesn't really assume much in the way of background knowledge---you don't need to have taken analysis or topology to get the gist of it.

Xander: I remember you were trying to get a copy of a calculus book online for your students some weeks ago. I think the one by Thomas. Were you able to obtain it or was it blocked off?

@Sahaj ?

@nickbros123 Yeah, baby rudin is kinda bad for learning, so chances are a random book on metric spaces will have better pedagogical value

@anak Oh, yeah. That, too.

@XanderHenderson do you think there would be a tangible benefit for me if I spend time on that book rather than just going through Baby rudin's topology chapter? I do understand an exclusive book ought to contain a lot more than a 30 page condensation, so I dont doubt that I will be learning more, but I am strapped for time

I was taught undergraduate analysis out of two books: Folland wrote an undergraduate text (which is exactly as accessible as his graduate text), and there was another text by two authors... Dangelo and something, maybe?

@nickbros123 what's the rush? Why do you have a time limit?

abebooks.com/9780395959336/…

Seyfried. That's the other author.

I am not sure that I would recommend either, but they are better than Rudin.

For recommendations, Royden is fine, and Tao's book is pretty reasonable.

Carothers is another nice metric space one.

Though Tao does seem to get weirdly pedantic in a few places (I seem to remember a couple of somewhat weird idiosyncrasies, though I couldn't identify them off the top of mind right now).

@anak I asked for a reading project to a prof in a university, and I told him I finished 4 chapters of rudin already .....

@XanderHenderson nevermind. I just searched it up in your old messages.

kinda stupid i know

What is the reading project on, @nickbros123?

Complex analysis.....

Topology and analysis in R^n is probably enough for that. One doesn't need the entire generality of metric spaces.

maybe i screwed up

you won't understand topology if you don't understand metric spaces

or at least something similar like $\mathbb{R}^n$

I don't think that's necessarily true. It's great scaffolding for topology, but by no means actually necessary. All nick seems to need is topology/analysis of R^n though, not general topology.

what is going on after the 5th line?

is that hatcher's filthy font

it's from Hatcher's

I don't like Hatcher 😅

I learned topology out of Armstrong (link.springer.com/book/10.1007/978-1-4757-1793-8), in a department with a lot of knot theorists. The connection to metric spaces was... tenuous.

@anak I did tell the prof I know next to nothing in complex analysis, and he agreed to guide me through complex analysis in the summer. Seems as though I need the first 7 chapters of rudin, or equivalent

what is deformation of f near x_i?

@XanderHenderson Armstrong's is an excellent book.

@Koro I'm fond of it.

I like how quickly it develops the concepts.

sure, you can understand any math without good motivation for it

contrary to Munkres which spends too much time on things like order topology etc..

@Koro I like Hatcher, but I can certainly see why someone else wouldn't. He is a fairly informal, and leaves out a lot details (or hides those details in extensive exposition).

I like his style, but I can definitely see how another could bounce right off of it.

@XanderHenderson I don't feel confident with that book because my understanding of a statement could be completely different from the intended message. That's the problem with Hatcher being informal. :(

For example in my above question: deformation could be interpreted in more than 1 way.

this confuses me a lot. 😅

getting into topology for beginners is sort of easy. But you have to sit a bit to understand the concepts to be able to think about those things effortlessly

@Koro I think they just mean that for a small enough neighbourhood around x_i, you can easily deform the k-stretched segment corresponding to the image of this neighbourhood back to the original neighbourhood/segment. The deformation is the homeomorphism between the segment and the k-segment. Since there is no overlap (the neighbourhood only has one x_i in it), the (local) degree has to be 1.

@anak I see. Thanks a lot.

Does that make sense to you, Koro? I don't think it's super clear what he's saying, because there is a question about the sign of the local degree, but I guess since everything is being deformed in the same "way", you will get a degree of \pm k.

I swept this under the rug in my original explanation.

imgur.com/1YYTZ8t

I guess the positive degree is used because of this idea of a rotation.

In this image, I calculated local degree at (1,0).

Then the question is to prove rigorously the existence of similar charts at other points too.

@anak Hatcher seems to being doing the same thing then. "You can easily deform the k-stretched...". The easily part is what I was trying to prove rigorously.

So it makes sense now.

@anak rotation is homotopic to identity so positive degree. :)

@anak In the image, I have shown that the sign is as same as that of n at the point (1,0).

Hatcher assumes n to be positive. But in my case n can be assumed to be any non zero integer.

@XanderHenderson I disagree with the notion that Hatcher's book is particularly informal. It is, however, often presented to the wrong audiences.

@Koro the argument is: if $\varphi$ rotates $x_i$ to $y$, then $f$ and $\varphi$ are homotopic rel $x_i$ in a neighborhood of $x_i$, hence they have the same local degree at $x_i$

how to find the degree of a given map from torus to torus?

if you pick charts appropriately (which you seem to have done already), this is easy to see, because $f$ looks like multiplication by $n$ and $\varphi$ looks like the identity

Torus is T= S^1 \times S^1

So the question of finding the degree of such map makes sense.

@Koro the same way

hmm

$f: T\to T: (x,y)\mapsto (ax+by, cx+dy); a,b,c,d\in \mathbb Z$

@Thorgott I think most of the books that students learn out of are presented to the wrong audience. Rudin for undergraduate analysis, for example, is unforgivable.

very much so

@XanderHenderson my professor teached me Lebesgue integrals out of Rudin

(i.e. told me to read it, and then verified my knowledge)

@Jakobian You are far from being the typical student.

@XanderHenderson I was a first year undegraduate

just because I have some knowledge now it doesn't mean this was always the case

@Jakobian Irrelevant. Indeed, the fact that you were studying Lebesgue theory as a first year undergraduate rather reinforces my point: you are far from being a typical student.

yeah, Lebesgue theory is reserved for the 3rd semester

@XanderHenderson I self studied it and loved it. But since I'm at college now, I feel what you are saying.

A typical American student doesn't see Lebesgue theory until maybe their last year as an undergraduate. Some of them might see it in their second-to-last year, maybe. Many won't see it until grad school. I can't imagine that non-American systems are so wildly different from this that Lebesgue theory is considered appropriate for the typical first-year.

@Koro You, also, are likely not a typical student.

what I just said is standard syllabus here

Honestly, I would imagine that anyone who is bothering to spend time in chat on Math SE is quite atypical.

certainly not first-year, but the next best thing

I spend time on mathse because contribution to my learning from classes here at my college is 0 so self-study+online sources like mathse, discord etc. is the combination that works for me.

@XanderHenderson they taught us measure theory and probability on second year because the whole department was probability and analysis oriented

I still remember how my professor gave me a paper about Feynman-Kac operator to understand with my little probability knowledge. I am scared of probability to this day

@XanderHenderson I suppose one (if they are a student) would be here because they are trying to understand things deeper and not just for mugging up for the exams.

college exams are easy: just memorise the solutions to assigments and done.

that may be why I seem like I'm good at analysis to some people, because I had analysis oriented background

I'm not good at anything in particular, actually

well. If anyone ever needs something explained about rings of continuous functions, then I can help with that

But I doubt

Dumb question maybe, but what is the Lebesgue measure of the unit circle $\mathbb T=\mathbb R/\mathbb Z$? My quick browsing has resulted in some sources say it is $0$, others say it is $1$. How does one compute it? I'm just asking for a sketch, unless there's something even simpler.

I'm writing $\mathbb T=\mathbb R/\mathbb Z$ because that is how it's written in the notes I'm reading.

you need to more context. there is something called lebesgue measure in the plane, and as a subset of the plane, the circle has measure zero. there is also something called lebesgue measure on R, and when you realize the circle as R/Z you get another thing that you might call lebesgue measure.

and the measures you get, for those very different measures, are 0 and 1 respectively.

there's also an issue of 'normalization' that sometimes pops up with circles (maybe not how you have realized it, but in general), i.e., when the full circle has nonzero measure, whether that measure should correspond to 1 (and thus be 'normalized') or to the length of the circle (which might suggest e.g. 2pi, if you thought of the circle as the unit circle in the plane and not as R/Z)

too late for me to edit a verb into my first statement, 'you need to have more context.'

if someone is doing harmonic analysis on the circle (as one example, inferred from past chats) they are probably not thinking in terms of planar lebesgue measure.

the normalization issue is also present in the harmonic analysis context, and leads to factors of 2pi or sqrt(2pi) appearing (or not) in different ways in different formulas scattered across the internet.

if you write $\mathbb{R}/\mathbb{Z}$, you don't mean to think of the circle as a subset of the plane

you would at the very least be writing $S^1$ if you were

anyway, that's the explanation for why "the" circle might have different "lebesgue measures" depending on where you look.

@leslietownes ok, makes more sense now. So what subset of $\mathbb R$ could they mean by "the" circle? After all, $\mathbb R/\mathbb Z$ is not a subset of $\mathbb R$, right?

they are thinking of a quotient of R by an equivalence relation where two points are equivalent if they differ by an integer. and you're right, formally speaking, this quotient is not a subset of R

not just formally speaking, this is simply not a subset of R

but, for example, [0,1) (or any interval of length 1 with one endpoint missing) is a complete set of representatives for this relation, so in a lot of contexts you could think of R/Z as just this interval (or more likely [0,1], if that last point doesn't change about anything you are doing)

i guess this is an isomorphism of measure spaces

so someone might say 'lebesgue measure on the circle' and in the next moment suddenly it's just as if the measure space is the interval [0,1] with the usual measure on that interval, and functions on 'the circle' are just [complex?] valued functions on [0,1], integrated the same way you would do riemann integrals in calculus 1

psie: the analogy you might think of here (which is only a loose analogy because it fails to preserve a lot of interesting things) is that one can think of "the integers mod n" as Z/nZ, i.e. a quotient of Z by a subgroup, or as {0, 1, ..., n-1} [a set of representatives] with funny operations (i.e., interpreting everything "mod n" as if it is required to turn a result that is one of those representatives)

something similar with "the circle"

@leslietownes I would challenge that person to a duel. There is the 2d Lebesgue measure of a circle in the plane (which is zero), and there is the 1d Hausdorff measure of the circle (which is the circumference, depending on normalization), and there is the natural Haar measure on the circle (with complex multiplication). But Lebesgue measure?!

PISTOLS!

AT DAWN!

well, i live in a state that has outlawed dueling as a method of private dispute resolution

@leslietownes I am not entirely certain that the same can be said for my state.

This seems like the kind of place where duels are still technically not illegal.

maybe just not in town

or in the governor's house, or something

ok, then they are probably thinking of "the circle" as $[0,1)$ if they say it has measure $1$ :) thanks!

Maybe we could use airsoft pistols?

did you guys mention how translation of structure works

seems like it wasn't mentioned out loud

basically, if you have a bijection, in this case $f:\mathbb{R}/\mathbb{Z}\to [0, 1)$ given by $f(x+\mathbb{Z}) = \{x\}$

and you have, say, Lebesgue measure on $[0, 1)$, call it $\lambda$

then given a subset $A\subseteq \mathbb{R}/\mathbb{Z}$, you can define $\mu(A) = \lambda(f(A))$ as long as $f(A)$ is measurable

and this makes $\mathbb{R}/\mathbb{Z}$ into a measure space, with sigma-algebra also given by this bijection $f$

similarly you could translate group structure and so on

if we have a bijection we can assign structure to a set without any structure of this kind

the function $\mu$ could be termed the "Lebesgue measure" on $\mathbb{R}/\mathbb{Z}$

Or the pushforward of the Lebesgue measure...

that's neat

well, I was thinking for example of group structure, or of topology on a set

not necessarily a measure

If I want to find the degree of f:R^2--->R^2 defined as f(x,y)= (x^2-y^2, 2xy), then clearly (0,0) is the only point which is not regular value so I can assume f to be from R^2-0---> R^2-0. R^2-0= S^1 so we may assume f: S^1---> S^1.

det(f)>0 so Df is orientation preserving.

Also I believe that $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ is often called a torus in analysis, and not circle

deg(f) =2 by local degrees.

@Thorgott is it correct?

@Jakobian That is correct. In general, $[0,1)^n$ (viewed as $\mathbb{R}^n/\sim$ for the right notion of $\sim$ that I am too lazy to write out) is the $n$-torus.

uh, in the n = 1 case, i do think that someone would look at you funny if you called R/Z a torus, even if you were calling it T in your notes

at least, i would look at you funny

@leslietownes $T^1 = \mathbb{R}/\mathbb{Z}$ is the 1-torus. :P

I like getting funny looks from you.

note that i'm not saying that it isn't the 1-torus

and I would look at you funny with double intensity

i'm saying, were say yitzhak katnzelson to refer to a result for T as being about "the circle", you probably wouldn't raise your hand and ask "uh, don't you mean the torus"

and we'd have a contest on who can last longer

@leslietownes I would. Just to be an ass.

you'd be in pretty good company

I mean, there is measure-theoretical ambiguity in calling $\mathbb{T}$ a circle

I suppose there's also an ambiguity in whetver we mean $[0, 1)$, $[0, 2\pi)$ or whatever else

i reject the several copies school

so the author would presumably clarify what $\mathbb{T}$ is in the beginning, so calling it a circle would be justifiable

@BalarkaSen there is only one copy of everything

@Thorgott that's why I tried to make sense of degree in this case. For me, degree is defined for compact oriented manifolds given that codomain manifold is connected. :)

@Thorgott b) for me local degree is defined for preimages of a regular value. 0 is not a regular value in this case so b) is not valid to me.

local degree is just the integer by which the induced map on $H_2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})=\mathbb{Z}$ is given

this is really the definition one should be using, the stuff about regular values and what-not is a method of computation that is established post hoc

@Thorgott does post hoc mean the same thing as ad hoc?

@Jakobian No.

Post hoc means "after the fact".

what Xander said

Ad hoc means something more like "on the fly" or "without a plan".

I believe it is literally about need (hoc is something about necessity, I think).

so post hoc means the same thing as a posteriori?

@Jakobian Not quite, no. Post hoc tends to have a connotation of filling in the gaps after the fact, or producing reasoning to match data which has already been collected. It is sort-of kind-of a logical fallacy.

A posteriori is more about logical deduction.

Oh. So post hoc is kind of like ad hoc but with past knowledge?

No, not really.

Post hoc is "after the fact", ad hoc is "as necessary".

A posteriori refers to a process of reasoning.

I've been trying to read about this in google but I don't get the difference between a posteriori and post hoc

I'm guessing the difference is only in the context in which they are usually used?

I've asked in the English language chatroom