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12:03 AM
If we have a sequence of finite disjoint unions of rectangles $\{E_n\}$, I guess we also have a sequence of rectangles $\{A_i\times B_i\}$ and vice versa, and the cover $\bigcup_1^\infty E_n$ with finite disjoint unions of rectangles can also be written as a cover with simply rectangles (it still remains a countable union too, which is important I think). So all's good I guess.
Bob
Bob
12:51 AM
What do people think of the program R?
 
3 hours later…
4:04 AM
If $A,B$ are interger matrices such that all entries of $A-B$ is even, then $\det(A) - \det(B)$ is even?
5:02 AM
@RajeshBhowmick That is a very hard problem. I suspect there is no simple algorithm, just a brute-force search. Just finding all triangular numbers that are multiples of a given triangular number is tricky, involving solutions to generalised Pell equations. See the series of articles by Vladimir Pletser linked at oeis.org/A029549
Joe
Joe
I'm doing a project for my graduate studies on algebraic geometry, specifically something that compares the classical theory (i.e. studying affine algebraic sets over algebraically closed fields), and the more modern scheme-theoretic approach. I was just wondering if there were any notable results than can be stated in the classical language, but whose proofs use schemes in an essential way.
5:17 AM
@onepotatotwopotato any thoughts on this if A-B differs from A in at most one column? i.e. A = B + 2C where C is an integer matrix with nonzero entries in at most one column?
Does anyone have optimism for the direction life moves in? In reference to like progression technologically but also socially
I think the silence says it all
Maybe I ask too much
5:50 AM
hi
i wanted to ask is there an identity that cosAcosBcosC...=1 if A+B+C+D...=90degrees?
hello can you help?
Unable
6:00 AM
I found it thanks for trying
Since cosA is always less then 1 except 0 so it isn't possible
6:23 AM
Fellow mathers I'm now searching for a group of well meaning mathematicians to hand over my ideas to. Does this exist?
I'm not worthy of solving RH
 
1 hour later…
7:50 AM
hi
8:12 AM
@psie I drink coffee whenever I'm stressed
So all the time
@SoumikMukherjee hi
@ZacU. it doesn't exist
8:56 AM
Hi
@SineoftheTime I have a problem, I parameterized the parabola $x^2-y-2 = 0$ , so I wrote $\gamma(t) = (t, t^2-2) , t \in [-1,1]$, so $g(t) = t^2 - t^2 + 2 - 1 \rightarrow g(t) = 1$
9:13 AM
@Pizza so $g'=0$ for all $t$
why is it a problem?
But now I have to calculate the function in -1 and 1?
yes, you have to check the end points manually, but here the function is constant along the parabola under $x=-1$
Since it is constant, I have not understood what I should do
@Pizza what is $f$ evaluated at those ponit?
I got my paperback copy of Hartshorne today
9:20 AM
But then these points $t = -1 , t= 1$, should they be inserted in arcsin(...) or are they ok in g(t)?
first, you see $g(\pm 1)$, then to find the max of the original function you have to substitute in $\arcsin (\dots)$
$g(t)=1$ for all $t\in [-1,1]$ so $f(x,y)=\pi /2$ for all $(x,y)$ such that $x^2-y-2=0$, $x\in[-1,1]$
Ah ok
But then it was "impossible" to replace $t = -1 , t=1$, in $g(t)$ , since it was 1, therefore constant
no it's not impossible, $g(\pm 1)=1$
9:27 AM
But $g(t) = 1$, I should insert $t=1 , t=-1$, but $t$ is not there
I mean, I didn't understand this thing.
@Pizza if $f(x)=2$, what is $f(4)$ ?
ok
same thing with $g(t)=1$ for $t\in[-1,1]$
$g(1)=1=g(-1)$ since it's constant over $[-1,1]$
so we insert 1 in arcsin
9:30 AM
Ok, now I have to do the other parameterization
@Pizza you get $\arcsin(\dots)=\pi/2$ right?
Yes
so what you conclude?
That the function at the points (-1,-1) and (1,-1) is π/2?
what's the nature of these points?
@Pizza not only at those two points
at all the points $(x,x^2-2)$ for $x\in[-1,1]$
9:35 AM
Oh ok so in that whole interval
along the parabola on that interval
@SineoftheTime shouldn't I finish the exercise to say what point is it?
@Pizza what is the range of arcsin?
[-π/2 , π/2]
Ah so it's a max
yes
this makes sense, since $\arcsin (1)=\pi/2$ and $x^2-y-1=1$ on $x^2-y-2=0$
@Pizza these steps are important, did you fully understood what's going on?
9:48 AM
For now yes
I'm just finishing up, I'll let you know very soon
Ok I found that $g(0) = g(π) = g(3π/2) = 1 \quad g(π/2) = -1 \quad g(7π/6) = g(11π/6) = 5/4$
@Pizza there's something wrong
Now I have to insert them into the arcsin(...) function, right?
how did you parametrize?
... I understood :( , I had written $\theta \in [0,\pi]$, but I went beyond the interval
Anyway I wrote $x = \cos(\theta), y = -1 + \sin(\theta)$
To find $\theta$ I found the points of intersection between the circle and the parabola
$g(0) = g(π) = 1 \quad g(π/2) = -1$
yes
note that $g(0)=g(\pi)$ are the same points you found before
$(\pm1,-1)$
10:00 AM
Yes
so you could have studied for $\theta \in]0,\pi[$
but it's ok also $\theta \in [0,\pi]$
But then on the parabola in the previous interval, are they all maxima, if they always assume the same value?
@Pizza no, that's not the point
since $f(x,y)$ evaluated along the parabola is the max, then all points are points of maximum
the fact that they have the same values does not imply the points are max
they can be min
or you can have that are min and saddle points
But now that I've found all the points, what should I do?
you plug in in $f$ to see the value
since $f=\pi/2$ and $f\le \pi/2$ then the points are max
10:15 AM
@SineoftheTime For $g(0)$ and $g(\pi)$ , arcsin(1) = π/2, for $g(π/2)$ arcsin(-1) = -π/2
yes
but use another letter, you used $g$ previously
I am asked to find the absolute ones
So they would be those at π/2
While the absolute minimums at -π/2?
@SineoftheTime ah ok
@Pizza the absolute min is $-\pi/2$ attained at the point $(0,0)$
the absolute max is $\pi/2$ attained at the points of the form $(x,x^2-2)$ for $x\in [-1,1]$
clear?
But here it only indicates these in the graph
yeah, I don't know why
10:21 AM
That is, they should also be on the "yellow" line? , where the parabola is.
correct
if you draw the 3d graph, all points on $y=x^2-2$ are max
since you have to maximize inside the circle, you select the points s.t. $x\in [-1,1]$
what Is s.t?
but $f(x,x^2-2)=\pi/2$ for all $x\in \Bbb R$
@Pizza such that
tale che
Ah ok
maximize arcsin(Power[x,2]-y-1)
wolfram says the same
$f=\pi /2$ for $y=x^2-2$
10:26 AM
I saw
But then I also have to find the points $(x,y)$ where are the absolute max and min
what?
didn't you already do this?
@SineoftheTime After I wrote $g(0) = 1$ , I immediately wrote arcsin(1) = π/2
ok, then you have to find the corresponding point of $g(0)$
Can I also do it at the end or should it have been done first?
I'd do it immediately but you can do it at the end
10:34 AM
Oh ok, thanks a lot for the help!
thank you for teaching me something new :)
This exercise came out in the June exam
moral of the story: you have to be careful about the domain when maximizing/minimizing a function
Yes, the problem is that I had not parameterized the parabola when I did it
So I was missing some points
yes, it's tricky
10:38 AM
With Lagrange multipliers, do you think it was easier or not?
when the constraint is a curve, parametrizing is pretty straightforward
you can try to do it with Lagrange and see
Yes, maybe I'll try later
@SoumikMukherjee oh nice
$\sum_{n=2}^{\infty} \frac{(x+9)^{n-1}}{(n-1)^2}$ here, to lead to a power series, must we make the substitution y=x+9 and m=n-1?
@Pizza I swear I'm seeing this same exact picture for the fifth time
@Gian'sPizzeria broken latex
you wrote } instead of )
10:43 AM
@Jakobian This is the last time we'll see it
@Gian'sPizzeria to answer your question, you don't need your power series to be centered at zero so subtitution $y = x+9$ actually doesn't matter
@Gian'sPizzeria this is already a power series
@psie We have\begin{align*} \mu\times \nu(D) =\inf\left\{\sum_1^\infty \mu(A_n)\nu(B_n) : \{A_n\times B_n\}_1^\infty \text{ covers }D\right\}.\end{align*} Now, if $\{A_{n} \times B_{n}\}_{1}^{\infty}$ is a sequence of rectangles covering $D$, then $\{A_{n} \cap B_{n}\}_{1}^{\infty}$ covers $X=[0,1]$.
but yeah sure, arguably this is already a power series but in order to make it of the form $\sum a_n (x-c)^n$ you'd like to substitute $m = n-1$
10:45 AM
We don't know if each $A_n\cap B_n$ is $\mu$-measurable. It certainly has a Lebesgue outer measure and since $[0,1]$ is uncountable, this implies that $\mu^\ast\left(A_{n} \cap B_{n}\right)>0$ for some $n \in \mathbb{N}$. Does it follow from this that $\mu(A_n)>0$ also? If yes, then $\nu(B_n)=\infty$ as the Lebesgue outer measure of a finite set is $0$ and we would be done.
Thanks very much
@psie I don't think this is the way I'd try to go about it
Well, somehow one would need to show that any cover of $D$ yields an infinite infimum.
oh actually
yes I think I would go about this how you do this
@psie the claim $\mu^*(A_n\cap B_n) > 0$ follows from subadditivity, sure
and then $\mu(A_n)\geq \mu^*(A_n\cap B_n) > 0$ from monotonicity
yeah it does imply $\nu(B_n) = \infty$ as well
@SineoftheTime did the same thing. drank coffee at night thinking ill study some topology. but I slept almost 20 minutes later lol
10:56 AM
@nickbros123 what do you do in topology right now
@Jakobian ok 👍 I guess $A_n\cap B_n$ is Borel measurable after all, since we equipped $[0,1]$ with the Borel sigma algebra in both cases, once with Lebesgue measure and once with counting measure. All's good.
drank coffee at 10 pm, went to bed at 4 am
@psie I didn't equip it
whats your guys favorite topology fact
@Jakobian i meant metric space topology, but many of the proofs seem to work without making use of distances.. As to where I am, i was to read connectedness and continuity. I think I left off at the theorem: a space is separated if a continuous map exists to discrete set {-1,1}
@nickbros123 separated? You shouldn't say that about spaces
separated means, but very uncommonly, a Hausdorff space
you mean that a space is connected iff there doesn't exist a surjective continuous map onto {0, 1} with discrete topology
11:04 AM
oh right I forgot surjectivity, maybe better word may have been disconnected
but my prof uses separated and disconnected interchangably
@Jakobian well, I was having doubts about you comparing $\mu(A_n)$ with $\mu^\ast(A_n\cap B_n)$. When you say it follows by monotinicity, doesn't one need to speak about the same measure? $\mu^\ast$ may not equal $\mu$ on $A_n\cap B_n$.
fun thing is that on {0, 1} there exist different topologies, for example if you take indiscrete topology then any map into {0, 1} is continuous
and if you take Sierpiński space topology, then the continuous maps correspond to open sets
just those three exist
@psie monotonicty holds for outer measures
ah sure, ok, thanks!
my prof uses (to some extent) this book: Shirali and Vasudeva- Metric spaces (u can google it to find a link, I am unsure on the policy of sharing links, especially copyright material)
another book he uses is Kumaresaan- Topology of metric spaces
and ofcourse, he also uses rudin
cuz the course is actually real analysis :)
Which is bigger: $TREE(3)$ or $f_3(3)$? ($f_a(b)$ is the fast-growing hierarchy)
11:14 AM
@TheEmptyStringPhotographer In response to the question "Which is bigger?", I am very tempted to respond "Your mom."
@XanderHenderson Both those values I stated in the question are much bigger than the volume of the Earth and everything that orbits it or is part of it in cubic plank lengths. So I am afraid you are very wrong.
@TheEmptyStringPhotographer I believe that the joke just flew over your head.
@XanderHenderson I am joking to your joke, but if we continue like this, the number of messages in this chat room will be provably larger than $TREE(g(64))$.
@XanderHenderson TRUE
Also, I was watching a video on big numbers and it didn’t mention the fast growing hierarchy!
Anyways, it was an actual question
11 mins ago, by The Empty String Photographer
Which is bigger: $TREE(3)$ or $f_3(3)$? ($f_a(b)$ is the fast-growing hierarchy)
11:30 AM
@TheEmptyStringPhotographer numbers are not the same as physical quantities anyway. A number is much different than something existing in real life so its hard to compare the two
I dislike any comparison of the sort
@Jakobian :) I was waiting for months to buy it, but the cost was too high, finally they reduced the price by 61%+
11:59 AM
you must really like AG
I'm a beginner :), I liked it so far
$\int \frac{1}{\cos(x)}= \frac{1}{\sin(x+\pi/2)}= \frac{1}{2\sin(x/2+\pi/4)\cos(x/2+\pi/4)}=\frac{1}{2\tan(x/2+\pi/4)(cos(x/2+\pi/4))^2}=\ln|\tan(x/2+\pi/4)|$
Is the procedure valid?
not at all
I missed the integrals
Ops
😣
I had made the corrections but the message was duplicated instead of being corrected
$\int \frac{1}{\cos(x)}=\int \frac{1}{\sin(x+\pi/2)}=\int \frac{1}{2\sin(x/2+\pi/4)\cos(x/2+\pi/4)}=\int \frac{1}{2\tan(x/2+\pi/4)(\cos(x/2+\pi/4))^2}=\ln|\tan(x/2+\pi/4)|+c$
Now?
12:15 PM
You can use the fact that $\frac{1}{\cos(x)} = \sec(x)$
$\int \sec(x) dx = \ln|\sec(x) + \tan(x)| + C$
I missed the dx.....
@Pizza 👍
 
1 hour later…
1:33 PM
@Gian'sPizzeria looks good
you can also compute first $\int \frac 1 {\sin x}dx$ and then shift the sine $\sin(x+\pi/2)=\cos x$.
 
2 hours later…
3:08 PM
@Ben Question: If $\mathcal{C}$ is an $\infty$-category, is any functor $\mathcal{C}_{/x}\rightarrow\mathcal{C}_{/y}$ over $\mathcal{C}$ induced by post-composition with a morphism $x\rightarrow y$ (of course, the question only makes sense up to natural isomorphism)
3:22 PM
@Thorgott good question
3:34 PM
Hi
@SineoftheTime But in the demonstration why It Is $\iint_D \frac{\partial{F_2}}{\partial{x}} - \frac{\partial{F_1}}{\partial{y}} dx dy$
Why does it come out - and not +?
If we use $-F_1$
@Thorgott This is not even true 1-categorially: Consider the category $A$ with two objects $A, B$ and two arrows $f, g\colon A \to B$, and consider the functor $S_{/b} \to S_{/b}$ sending $f$ to $g$ and vice-versa.
@Pizza where?
There's no endomorphism of $B$ inducing that functor, even though it is over $S$.
in the divergence?
3:54 PM
@SineoftheTime where it says "Dim." the first formula
this is the divergence of $(F_2,-F_1)$
Ah ok
$\nabla \cdot (F_2,-F_1)=(F_2)_x+(-F_1)_y=(F_2)_x-(F_1)_y$
I have an idea for a number system using primes
5 would be 100, 2 would be 1, and 10 would be 101.
25 would be 200
@SineoftheTime Oh ok, clear thanks
4:00 PM
other questions?
Not for now
@SineoftheTime do you have any
4:21 PM
I keep thinking...celery or fennel? I can't decide which is better. They are so similar in many respects I think, yet different. I think fennel has an aroma that celery doesn't. But I like both of them :)
salary
@BenSteffan hehe, that's good too :)
@psie i cant tell the difference rly
?!
Celery and fennel are vastly different
4:28 PM
@BenSteffan Depends on the sense in which you are describing them.
If someone makes a celery salad and fennel salad with a lot of dressing, then it can be hard to tell the difference I think. They have the same consistency kind of.
@XanderHenderson In terms of "as food," methinks
as cathegories
@psie I see what you mean but I disagree
@SineoftheTime obviously celery is a sheaf because it has stalks :)
i cant remember the last time I ate either tbh. theyre both really forgettable
4:33 PM
@BenSteffan That still doesn't really make clear the distinction you are after.
though indians have a practice of cooking with fennel seeds and LEAVING them in the end product. So theyre definitely not forgettable
@XanderHenderson I am not after a clear distinction
I make an infusion with fennel seeds
fennel tea
what's called?
that must be some strong stuff
fennel tea is nice
4:35 PM
@BenSteffan that morphism is not over $S$
@Thorgott ?
that slice category consists of two morphisms with a common target and the projection to $S$ identifies the source of these two morphisms
@BenSteffan fennel cateagory
2
but you can't switch the two morphisms in the slice without switching them in $S$
I don't follow
The slice is a discrete category on two points
4:37 PM
@BenSteffan Both celery and fennel are in the family Apiaceae. They both grow from "bulbs", and have similarly shaped stalks.
ah see, I knew they were related somehow
Celery is milder in flavor, but the two plants are otherwise rather similar in cooking---the stalks provide texture (i.e. crunch).
It has two objects: $f$ and $g$. The functor to $S$ takes both of these to $A$
And fennel stalks are much milder in flavor than the bulbs.
I guess that's not quite true, there's a third object, $\mathrm{id}_B$
ah, okay
I see the issue
4:43 PM
I don't like fennel so I'd say celery
it reminds me of soup
@BenSteffan yeah, that one's crucial
it's actually pretty easy to prove the statement in the $1$-categorical case
im also 100% convinced it's true generally, I just haven't figured out why
I see, $\mathrm{id}_x$ gets send to some morphism $x \to y \in \mathcal{C}_{/y}$ and this is the morphism inducing the functor
but this doesn't translate well to $\infty$-categories because you will have to fill some horns in the process and that doesn't guarantee you get the same morphism (?)
it works up to equivalence, which should be good enough
I suppose we can see that the statement is true under straightening-unstraightening, but there shouldn't be a need for big guns
@Thorgott I was wondering if $\mathcal C_{/x} \to \mathcal C$ being a Cartesian fibration helps
@Thorgott lol I was about to say something like this
lol
@LukasHeger hmm, perhaps it's some fiber transport business
oh wait i think i see it
5:18 PM
an $n$-simplex of $\mathcal{C}_{/x}$ is given by an $(n+1)$-simplex of $\mathcal{C}$ with final vertex $x$, obtain from this an $(n+2)$-simplex of $\mathcal{C}$ by degenerating the edge from $n+1$ to $n+2$, curry this to an $(n+1)$-simplex of $\mathcal{C}_{/x}$ and apply the functor to obtain an $(n+1)$-simplex of $\mathcal{C}_{/y}$, which curries to an $(n+2)$-simplex of $\mathcal{C}$ whose edge from $n+1$ to $n+2$ is given by a morphism $f\colon x\rightarrow y$ (the image of $\mathrm{id}_x$ under the functor), so this curries to an $n$-simplex of $\mathcal{C}_{/f}$.
ah, I can put this more elegantly
the projection $\mathcal{C}_{/\mathrm{id}_x}=(\mathcal{C}_{/x})_{/\mathrm{id}_x}\rightarrow\mathcal{C}_x$ has a canonical section (given precisely by degenerating along the last edge) and then we compose that with the induced functor $(\mathcal{C}_{/x})_{/\mathrm{id}_x}\rightarrow(\mathcal{C}_{/y})_{/f}=\mathcal{C}_{/f}$ to obtain the section of $\mathcal{C}_{/f}\rightarrow\mathcal{C}_{/x}$
5:46 PM
I feel depressed, I can't really focus
I was trying to be happy and not think about anything, but it hit me
I talk to a friend when I feel depressed
@Jakobian have you taken a walk today? Sometimes that helps, getting out. Or take a bike ride, if you have a bike.
@psie yes. And I was twice in the shop
Bml
Bml
Hi everyone, sorry to bother, could you kindly explain why this edit was rejected? The post has typos in grammar, and MathJax formatting is not correct. Why?
6:02 PM
@Jakobian have you tried talking to someone? not just anyone, someone who can listen to you without judging
@Bml Are you familiar with the concept of "polishing a turd"?
@XanderHenderson it spreads on your hand?
Bml
Bml
@XanderHenderson No, sorry, I don't understand what you are talking about. Could you explain in more detail?
Verb: polish a turd (third-person singular simple present polishes a turd, present participle polishing a turd, simple past and past participle polished a turd)
  1. (idiomatic, vulgar) To work on a time-consuming and ultimately pointless or impossible task; to try to perfect what is inherently bad.
@SoumikMukherjee its not a problem that needs solving you know... you can't just approach it like that and expect everything to be okay
6:04 PM
I think the edit is fine, but yeah
I will (hopefully) stabilize sooner or later
"Polishing a turd" is the act of trying to improve something which is fundamentally not worth improving. In your edit suggestion, sure, you made improvements to the question, but the question still isn't appropriate for the site. So it was kind of a waste of your time to write the edit, and kind of a waste of the reviewers' time to look over it.
I see, I think I can understand. I hope you feel better soon.
So the rejection might mean "you tried to polish a turd, please don't do this".
@Jakobian I know it sounds stupid, but when I used to play piano when I was younger and the piece that I was playing was hard, then I would just play really really slowly, and after some time, I'd be able to play faster again. I apply this strategy to life sometimes. Take it just really slowly. Tomorrow might be different and you might be able to focus better.
Bml
Bml
6:08 PM
@XanderHenderson OK, but it should be one of the objectives of the site to improve the posts of new users, no? I understand that it is time-consuming, but I believe it is always worth it. Personal opinion.
@Bml Even after the edits, the question doesn't meet site standards.
Bml
Bml
@XanderHenderson OK, let's see this other edit. It is a wiki tag edit and it corrects a typo in grammar. Why was it rejected?
In depressing times, I talk to a friend not to have my problems solved, but so that I won't feel alone. I think that being alone adds to one's depression.
@Bml First off, I don't actually know why either edit was rejected. I am only guessing.
With regard to the second edit, it honestly isn't clear what, exactly, you have changed in a lot of places. It is not clear how your edit actually improves the tag wiki. I would guess that the reviewers had the same problem.
6:16 PM
Has anyone chatted with robjohn lately?
@XanderHenderson Sorry but what's unclear about it?
@PM2Ring Thanks for the link & taking interest in the problem.
It fixes a number of grammar errors and misplaced hyphens
Honestly, it looks like you removed a space---it isn't clear what the other edits did.
Bml
Bml
@XanderHenderson Yes, sorry, I didn't mean that you knew why the edit was rejected. I meant that it is frustrating to have edits that correct grammatical errors rejected with ‘The edit does not improve the quality of the post. Changes to the content are unnecessary or make the post more confusing.’
@XanderHenderson No, I fixed many misplaced hyphens, corrected a typo in grammar ("might call" instead of "might called") and removed various spaces.
6:20 PM
@Bml Look at what the reviewers are shown. It doesn't look like you have made any real changes. Reviewers generally reject such minor edits.
Fixing a grammar error is a real change, and it improves the entry, even if it's minor
And, frankly, there is no point in arguing with me or trying to convince me that your edit was, in fact, worth keeping. I am not one of the reviwers.
You asked why it was rejected, I have tried to guess.
Why are you arguing with me about it?
Bml
Bml
@BenSteffan Yes, that's the point. I have submitted the edit again. If you can approve it, I thank you wholeheartedly :-)
@XanderHenderson No, I am not arguing with you about that, I expressed myself wrongly. Although I responded to your comment, it was addressed to everyone in this chat. Sorry for the misunderstanding.
@Bml I would advise you, in general, not to resubmit suggested edits. To a moderator, this looks like an attempt to circumvent moderation (i.e. "I know that I'm right and the reviewers are wrong, and I am going to game the system until I find two people who agree with me!").
If it is that important to you, answer questions and earn enough XP to be able to edit posts without needing review.
Bml
Bml
@XanderHenderson OK, sorry. By now I had submitted that edit and can no longer withdraw it. I will no longer submit an edit a second time. Thanks for the advice.
6:24 PM
@Bml I believe there's also a guideline for people without sufficient reputation to make edits that don't have to be reviewed to only make "major" edits; this feels very minor.
Bml
Bml
@BenSteffan I don't understand what you are trying to tell me, sorry. I haven't been feeling very well lately.
@Bml I'm trying to tell you that you probably shouldn't make edits like this (the community wiki one) that improve the entry only in marginal ways (minor grammar errors, formatting & punctuation), since these will still have to go through review.
You are free to make edits of course, but they should be impactful enough, e.g. correcting factual errors.
@Bml Don't make minor edits.
That is what @BenSteffan is telling you.
...that's the best way to put it
6:30 PM
@Thorgott nice
Once you have sufficient XP, you can make edits without requiring other people to review those edits. Once you have that much XP, it is generally okay to make minor edits (as long as you don't do it too often). Until then, don't make minor edits.
Bml
Bml
@XanderHenderson So I cannot even submit them, even if they are then rejected?
@SineoftheTime In short, very bad. I have many physical problems that led me to temporarily drop out of university, along with chronic depression and panic attacks. It's a bad period for me.
@Bml Can you submit minor edits? Sure. Should you? Well, you seem to be unhappy with rejection, so probably not.
@Bml sorry to hear that. I've been in this situation
maybe still I am in it
Take a break from the internet.
Bml
Bml
6:37 PM
@XanderHenderson No, I have nothing against rejections, I only asked because I felt my edit did not deserve it, that's all. It was just a request for clarification. I'm having serious health problems and depression at the moment, so I'm trying to do something helping the community with various edits (even minor ones), and I'll also try to formulate questions and answers in the coming days. That's all.
@SineoftheTime I am very sorry. I hope that moments like these pass quickly.
Don't edit too many old posts because they'll bump in the front page
and I saw many times users compaining about this
Bml
Bml
@SineoftheTime Yes, I know, I have already been through this with Physics SE. That's why, when there are no new question or answers, I go for the wiki tags...
Physics.SE is a lot more strict with their policies.
6:42 PM
@BenSteffan belated observation, but straightening+unstraightening + Yoneda lemma in fact gives you a homotopy equivalence $\mathrm{Fun}_{\mathcal{C}}(\mathcal{C}_{/x},\mathcal{C}_{/y})^{\simeq}\simeq\mathrm{map}_{\mathcal{C}}(x,y)$
I don't understand what completeness has to do with part (b) of the above exercise. Here's an attempt:
the hands-on fact is just that this is a bijection on $\pi_0$
Attempt: Let $E \subset X \times Y$ be a null set such that $f(x, y)=0$ for all $x \in X$ and $y \in Y$ such that $(x, y) \notin E$. If $x \in X$, then $f_{x}(y)=0$ for all $y \in Y$ such that $y \notin E_{x}$. Hence $f_{x}=0$ almost everywhere, so $f_{x}$ is integrable with $\int f_{x} \,d \nu=0$, for almost all $x \in X$. Similarly $f^{y}$ is integrable and $\int f^{y}\, d \mu=0$ for almost every $y \in Y$.
@Thorgott nice
I wish I had the time to learn about un/straightening :')
I've not read a proof either, they all look kinda un-fun
6:50 PM
I think Land (?) has a reasonably ok proof?
At least state-of-the-art
(maybe outdated by now)
sorry, no, it's Hebestreit-Heuts-Ruit
arxiv.org/pdf/2111.00069 only 41 pages, you can basically read it in an hour :^)
probably shouldn't have said "ok"
yeah, I know of that paper
but it's just like half a dozen model structures and I can't tell where the actual math is happening
@psie ignore my attempt above. I think part (b) follows from part (a) and indeed, completeness.
"fun with model categories"
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