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2:12 AM
What's the name of the guy who decided to make differential functional equations?
I just wanna talk to him
3 hours later…
5:13 AM
Good morning everyone, I know that any square matrix $A\in\mathbb{R}^{n\times n}$ can be written as the sum of symmetric and antisymmetric parts
A = \frac{A + A^T}{2} + \frac{A - A^T}{2}
I'm wondering whether we have similar results or similar notion for non-square matrices?
maybe? you'd need some generalization of what 'symmetric' should mean in that setting.
if S is any vector space and f: S to S is any map whatsoever, then for any x in S you have x = (x + f(x))/2 + (x - f(x))/2. so that alone isn't so interesting. if f is a linear map from S to itself that satisfies f(f(t)) = t for all t, then the decomposition just given has the form x = a + b where f(a) = a and f(b) = -b.
and this decomposition is kind of unique. if a+b = a'+b' where f(a)=a and f(a')=a' and f(b)=-b and f(b')=-b' for this linear f, then applying f to both sides of a+b = a'+b' you deduce a-b = a'-b', and adding these last two equations together you deduce 2a = 2a' and hence a = a' and b = b'.
so corresponding to any a linear "involution" [= map satisfying f(f(t)) = t] on a vector space, there is a decomposition of every element of that space into a sum of something fixed by the involution, and something that changes sign under the involution.
if you do this for the involution sending the function g(t) to the function g(-t) on the vector space of functions from R to R, you get the decomposition of such a function into a sum of an 'even' function and an 'odd' function in the usual sense
i can't think of any interesting involutions on the set of non square matrices, but maybe you can
interesting. sometimes innocuous questions lead to herculean mission.
3 hours later…
8:07 AM
Suppose that I have the following set R={(x,y): x,y are in R^n-{0}, y,x are linearly dependent}. I note that if we form a matrix [x y] (n+1 by 2 matrix) with columns x and y, then this matrix has rank at most 1.
So every 2 by 2 minor of the matrix must vanish.
Can I conclude from this information that R is finite?
@Koro Isn't R closed under non-zero scalar multiples?
This automatically implies that it is infinite. (I assume that the second one of the two R's stands for real numbers.)
koro: is the set of real numbers {(1,k): k nonzero} finite?
@MartinSleziak Indeed, the second one is for the reals. Sorry for the confusion.
@leslietownes no, it is not.
if you're maybe talking about the set of ranks of elements of R, or reduced row echelon forms or something, you could maybe squeeze finiteness out of this, but not otherwise.
Actually I am trying to show that R is closed.
8:12 AM
under some operation? or as a subspace of (R^n - {0})^2 with the usual topology?
yes, under the usual topology.
The book uses the fact that: as the zero set of polynomials, R is a closed subset of $(R^n$-{0}$)^2$
huh. the example n = 1 is maybe instructive here.
I suppose that is because all 2 by 2 minors are 0 but I don't understand the details yet.
That's seems like appropriate reasoning for me.
yes, it's a zero set of polynomials, so it is closed, but these are polynomials in multiple variables, so there's no reason for the zero sets to be empty
8:16 AM
Rank of a matrix is the same as the maximum $k$ such that there is a non-zero $k\times k$ minor.
that part is OK, I understand that.
But I don't understand how closedness of R is deduced.
(this is not $\mathbb R$)
What Thorgott wrote.
Preimage of zero under a continuous function is a closed set.
Intersection of closed sets is closed.
(with my f.lux on, without looking at names, sometimes I get confused between Leslie and Thorgott)
@Thorgott Let's fix a $t\ne 0$. Let's look at all (x,y) such that y=tx. We want to find such (x,y). So we solve the equations given by minors n(n-1)/2 of them.
I agree that this set is closed.
But then we have to union over all such t's.
and union(infinite) of closed sets need not be closed.
as you say, there are only finitely many minors
so infinite unions do not appear
If you fix two rows i and j you have a function $f_{ij}(x,y)$ assigning to the given two vectors the corresponding minor.
Every $f_{ij}$ is continuous and you are looking at the intersection $\bigcap f_{ij}^{-1}(0)$.
8:26 AM
but we unioning over all t>0. So that should be taken into consideration. Am I mistaken?
what role does $t$ play?
it seems entirely unnecessary
@MartinSleziak I understand it now. Thank you so much!
@Thorgott Ohh so we just want to find (x,y) such that all 'minors with the specification' are 0. And that doesn't require any t. Thank you so much!
I understand that now.
ok great
I gotta leave for a lecture now, have fun
me too :). Have fun!
1 hour later…
9:41 AM
Does the construction of the universal cover stated in Hatcher have a name? I mean 'the construction is the usual construction of something more general stuff'.
3 hours later…
12:41 PM
Is it true that every positive integer $N$ has at most one representation $$N=F_m+F_n$$ with positive integers $m,n$ with $1<m<n$ , where $F_n$ denotes the $n$ th Fibonacci-number ? If yes, how can it be proven ?
2 hours later…
3:05 PM
Confusion in 'attaching of spaces via a map': Suppose that X,Y are given spaces. Let A be a closed subspace of X, f: A-->Y be a map. Then define ~ on $X\sqcup Y$ as follows: x~y iff one of the following happens: 1) x=y, 2) x,y both are in A and f(x)=f(y), 3) x is in A, y=f(x).
How do I show that ~ is an equivalence relation?
In particular, why is ~ symmetric?
Suppose that u~v. Suppose we have that u is in A, v=f(u). From here, how do we get u=v or f(u)=f(v) or u=f(v)?
there is another terminology for (1),(2),(3) combined, which is equivalence relation generated by the relations a~f(a) for every a in A.
But I don't understand this terminology yet so I'm working with the alternative (equivalent?) definition.
But the problem is in showing ~ to be an equivalence relation.
Hi @ペガサスSeiya!!
3:21 PM
@Koro What's up
I'm good. what about you
To avoid watching the last of us, I watched a video that tells its entire story in like 30 mins. :-)
3:36 PM
@Koro just play the game
I don't quite like RPGs because the player doesn't have much control over what happens.
@Koro you don't
If I define ~ using partitions of the disjoint set then I get why ~ would be an equivalence relation. But using this equivalent characterisation does not give me the equivalence relation.
well, then it's not equivalent
would it be correct if I introduce 4th condition: v is in A, u=f(v)? I guess, no.
A: Attaching space in topology

Brian M. ScottI’m assuming that $A$ is in fact a subset of $X$, not $Y$. Let’s simplify matters a little by assuming that $X$ and $Y$ are disjoint, so that we don’t have to form $X\sqcup Y$; I’ll return to the general case later. We have disjoint spaces $X$ and $Y$, a subset $A$ of $X$, and a continuous $f:A...

I have posted one comment there requesting clarification.
@Thorgott: I'm trying to understand definition 13.13 in Bredon's AT book.
3:44 PM
what you said earlier is not the same as what is claimed in the answer
you're missing an option
@Koro shooters?
Thorgott: I think you meant this:
11 mins ago, by Koro
would it be correct if I introduce 4th condition: v is in A, u=f(v)? I guess, no.
But still, one won't get v~u from u~v.
sure, one will
Suppose that we have u~v and it is known that u is in A and that v=f(u), $u\ne v$. If v~u were true, then this should give either f(u)=f(v) or 'v is in A and u=f(v)'.
and I don't understand why that will hold.
4:08 PM
no, that's precisely the case u in A and f(u)=v that was missing in your original list
Hello non-math.se user
4:46 PM
I thought that and commented the same thing above. But then I don't know what I thought and deleted my comments also earlier.
yes, that makes sense to me now.
@Thorgott Thank you :-)
5:03 PM
Day 2 without meds
Hotdogs aren't supposed to go into ramen are they? @XanderHenderson
@ペガサスSeiya Why not?
Personally, I wouldn't put hot dogs in ramen, but I don't like hot dogs.
If I'm going to use pork, I'd prefer to add chashu.
Hotdogs are notoriously the most evil things ... full of ... um ... inedible things.
@XanderHenderson Does salad have any place in ramen? Mine seems to have some
@TedShifrin like intestines?
No, like rat droppings.
@ペガサスSeiya "Salad" is not an ingredient. It is a dish.
What do you mean by "salad" in ramen?
5:14 PM
@XanderHenderson well I meant some greens
@TedShifrin Yummy
I've seen seaweed in ramen.
Gourmet even
Bok choi and various asian broccoli can go fine in ramen.
@XanderHenderson nah this is different. This is like an actual leaf
On the other hand, I am an American Jew, raised in the southwest and the midwest. I am not an expert on Japanese food.
5:15 PM
@TedShifrin oh yes! That's what this is!
My understanding is that the defining characteristics of ramen are the broth, the tare, an aromatic oil, and the noodles.
I eats me spinach
@XanderHenderson and beef
As long as you have those things, it's ramen. Add whatever you like.
But, again, I'm not Japanese, so I don't know that I can speak with much authority on that issue.
@ペガサスSeiya No, beef is not a defining part of ramen.
There is lots of ramen without beef.
@XanderHenderson I've even seen peas and carrot in it
And fishcake
5:17 PM
Tonkotsu, ramen, for example, is definitively pork based, if I recall correctly.
I have an unhealthy habit of eating the same thing for weeks on end
Is it true that $(X\sqcup \{y\})/(A\sqcup \{y\})=X/A?$
Why not?
@Koro What could go wrong?
because in / we are identifying and not 'deleting'.
5:19 PM
Should I take my meds by putting them in my ramen bowl?
Well, you haven't specified how you're identifying. I assume $y$ gets identified to $y$.
$[0,1]\cup \{2\}/([1/2, 1/3]\cup \{2\})$
So that's a pointless identification. Pun intended.
Thanks. I read someone on MSE say that when self-studying, you should have an open problem in mind and try applying the techniques to it/learn techniques that would be beneficial for that problem.
There is also the David Hilbert quote "He who seeks for methods without having a definite problem in mind seeks for the most part in vain"
That is meant for people who are already research mathematicians though, right?
@TedShifrin ahh, I am using the following convention: X/A is A identified to a point.
5:21 PM
I mean, unless you are trying to be super f'ing pedantic. I guess, technically, $X \sqcup \{y\}$ is a space which consists of ordered pairs $(x,0)$ (with $x \in X$) and $(y,1)$ (with $y\in \{y\}$). Thus $(X\sqcup \{y\}) \setminus (A \sqcup \{y\})$ is the set $\{ (x,0) : x \in X \setminus A\}$.
My adviser, who was one of the pre-eminent mathematicians of the 20th century, always said that one should be working on a problem and then go learn what's needed for that.
Fine, Koro. So $y$ disappears.
But I don't think that anyone ever really cares about that kind of annoying detail. Since $\{ (x,0) : x \in X\setminus A\}$ is "the same as" $X\setminus A$ for all practical purposes.
@TedShifrin I have given that advice to people.
Where did $\backslash$ come from?
yes, so 'y' actually disappears or disappears 'homeomorphically'?
@TedShifrin Does that apply for a current undergraduate studying ahead though? In that case, you should just try to cover the topics in your courses ahead, not try to learn techniques to solve open problems, right?
5:23 PM
@TedShifrin Oh, shoot!
I am dyslexic.
I misread the problem.
@XanderHenderson 🥲
@ILike What is the point of studying ahead?
Any spy balloons flying above you guys?
@Koro In topology, everything is "up to homeomorphism" :)
Except perhaps for $X\backslash A \cup A = X$.
@TedShifrin Ohh. I was getting confused because the book uses = there.
This is right after definition 13.13 in Bredon's AT book.
5:26 PM
@ペガサスSeiya There are 99 red balloons flying by...
I used to think medical fields had the most confusingly named terms. Then I read my dad's graduate mathematics textbooks
I don't own that book. Indeed, I've never looked at it.
A: Adjunction Space , Collapsing boundary to a point

Alex ProvostYou are correct. By definition of the adjunction space, if $Y$ is a one-point space then $X \sqcup_f Y = X/A$ . In this case, the resulting space is a sphere, $D^2/S^1 = S^2$. Indeed, let $p$ denote a point in $S^2$. By the stereographic projection, there is a homeomorphism $\operatorname{int}...

@TedShifrin Well, you will probably get more out of the lectures since you already know the material. Since you get familiar with the type of problems, you will know how to solve them faster or tackle harder ones.
...and no shortage of captain Kirks.
5:27 PM
this answer also uses it (the = sign).
That's the point of studying ahead I would say
@ILike What's the point if you "already know the material"? I don't get it. I think it makes more sense to go further in depth with things you've already studied than to mislearn and/or be bored in the next course.
As a teacher, I wanted to take my students on a voyage of discovery, and students who already "knew" it and showed off their knowledge were not much fun.
@Koro This is a philosophical issue, I guess. What does it mean to say $X=Y$? Literally identical.
I see. So = is for 'homeomorphic' here, I guess.
No bijective mappings allowed.
Yeah. Does Bredon say "continuous" every time he says "map"?
not everytime
so far it's been clear to me in the book from the context -whether map is continuous or not.
5:34 PM
Most topologists have the understanding that all to use the word "map" the function must be continuous.
Well, then the context should make it clear if two sets are literally identical or merely homeomorphic.
@TedShifrin The category theorists got to them. :)
Before category theory was a thing.
it is common etiquette to write $=$ for a natural isomorphism
@TedShifrin According to every category theorist I've ever talked to, category thing has always been a thing. We just didn't know it yet.
Also, category theory is going to save us all.
Like Jeebus.
I got confused at this point. That's why I asked.
= sign (please see the last example in the above wiki page.)
5:36 PM
Category theory for heathens?
to write $=$ for any isomorphism is more contentious, but topologists do it pretty regularly anyway
anyway, I don't see a 13.13 like that in Bredon
chapter 13 Quotient spaces?
@TedShifrin haha
The point would be grades.
"I think it makes more sense to go further in depth" So you would recommend picking an open problem in that area, like your advisor said, and learn what's needed for it?
He was talking about graduate students and more advanced. But my opinion is that even the students who have gotten A's from me in undergraduate courses (including challenging courses) had plenty left to learn to master the course, and there was plenty of room left for exploration. Go back and work challenge problems from the course; really understand things you only got 75% during the course.
ah, I don't see that claim there, but I do see the chapter
I was looking too far ahead, the fact he resets the numbering is one of the few inconveniences of that book
5:42 PM
I should have said chapter 1 section 13.
Then just before propostion 13.14: 'Note that if Y is a one point space, then $Y\cup_f X= X/A$.'
My view after 40 years of teaching is that students generally did less well in courses than I would have liked because they really didn't know the prerequisite courses. In differential geometry, they really didn't know multivariable calculus and linear algebra sufficiently well. In algebra, they really didn't know basic proof skills and linear algebra well enough.
@Koro Where $f$ maps $A$ to the one point, I presume?
yes, indeed. $f: A\to Y, A\subset X$ is closed subspace.
I might think of that as an equality, even though it is not a literal equality, of course. The sets are not identical, but they sure are homeomorphic.
I wonder if Bredon has a comment early in the book or in the preface about his use of the term.
If they had covered the course beforehand though, studying ahead, worked through the problems beforehand, they should've seen what they were missing though and could've learned that before taking the course.
Doing this takes a lot of time though, yes
well, if $Y$ is a one-point space, there's only one way to map $A$ to $Y$
5:47 PM
Thorgott: did you find the section I am referring to?
Honestly, most of the courses I taught students needed me (or someone) to teach them. A student who could learn them on his own belonged in graduate school.
@ILike Have you gotten past standard calculus-type courses to real mathematics?
wiki page uses =, the answer uses =, Bredon uses =. Ted suggests $\simeq$ and I believe that to be true. It's 3 -2.
the score is uneven :-).
yes I see it
@Koro I say, use both.
I'm saying that most of us use $=$ to be interpreted in the "category" in which one is working. I write $\Bbb R^k\times\Bbb R^\ell = \Bbb R^{k+\ell}$ without hesitation. But it is not a literal equality.
5:49 PM
I'm dealing with a quartic polynomial. But its a depressed quartic polynomial. How do I make it happy?
literal equality as a concept is the opposite of helpful
I agree.
@ペガサスSeiya multiply it by 0.
On the other hand, I won't let students say that $\Bbb R^2\subset\Bbb R^3$ unless they specify the embedding :P
I won't let students write $\subset$.
5:50 PM
It is not automatic to set the last coordinate equal to $0$, although once you say that, I have no problem with it.
@Koro now its really depressed
'Cause I'm an asshole.
I insist on $\subseteq$ or $\subsetneq$.
well, the linear ones are all as good as another
but you better specify if you want a non-linear one
I don't, @Xander, unless I'm teaching into to higher math, in which case I follow the textbook. I use $\subset$ for $\subseteq$ and write $\subsetneq$ if I want to mean that.
(That's not really true. But I don't use $\subset$ in my own writing.)
5:51 PM
I think that is far less ambiguous.
I'm currently in high school, we are doing basic calculus and will be doing some linear algebra (if you could call it that, it's only vectors, some operations with them, ...) and probability soon.
For now, I'm mostly self-studying ahead for next year and undergraduate.
So no.
though I admit that earlier this semester I annoyed my professor by asking whether $\lim\mathrm{GL}(n)$ is formed by filling in $1$s top left or bottom right
@TedShifrin I think that $\subset$ is ambiguous, as there are authors who use it for strict containment, and others who do not.
@ILike Your notions come from a high-school student perspective, not from a serious college math student perspective.
Whereas $\subseteq$ and $\subsetneq$ are completely unambiguous.
5:52 PM
so what's the answer to my question?
@XanderHenderson I am the same
I believe 'homeomorphic' as opposed to actual =.
Is my understanding correct ?
Yes. Move on.
@TedShifrin The self-studying ahead notion?
It's just not worth this.
5:53 PM
thanks a lot :-).
Yes, @ILike. And certainly in high school calculus and supposed "linear algebra" there is plenty to master beyond the level of the course. So many problems and sections not covered.
pedantic me would offer the third option of "natural homeomorphism under $X$"
US high schools move way too slow when it comes to mathematics
in fact, rather than pedantic, it's important to internalize, but only when it's actually needed
Well, there are some people who've watched my videos who complain that the course moves too slowly. That said, it moves at double or triple the speed of most US courses and with far greater depth of material. Even my best students, who have gone on on to do PhDs, were plenty challenged.
I would say that for students like @Thor in his youth, my course was probably too slow/easy.
5:58 PM
let me explain the under $X$ part perhaps: there's a canonical map $X\rightarrow X/A$ and a canonical map $X\rightarrow Y\cup_fX$. the homeomorphism $X/A\cong Y\cup_fX$ commutes with these maps.
"not from a serious college math student perspective."
Those wouldn't want to self-study ahead, or how would their notions differ?
you often care about these maps as much as you care about the quotient/adjunction space, so this can matter
e.g. you might wanna know when $X\rightarrow X/A$ is a homotopy equivalence and this is the case iff $X\rightarrow Y\cup_fX$ is a homotopy equivalence, for this reason
@TedShifrin its definitely far faster than high school courses, that's one reason why I liked it
We're not talking about high school here. We're talking about substantive university courses.
Anyhow, if someone can do that course, or — say — Artin's algebra course, on his own and really master all the material, then why take the course?!
My experience has generally been that students who think they "know" stuff really know very little.
That does not apply to the truly exceptional.
But this is not Lake Wobegon. I've only had a handful of truly exceptional students in 40 years of teaching.
@TedShifrin dunning-kruger effect. That's what you discovered
6:03 PM
Well, to varying extents, yes. I never knew that term until a few years ago. I think I learned it in the Tromp era.
the more you know, the more you know how little you know; or however the saying went
When I was in college and graduate school, I never finished a course thinking I had mastered it ... even if I got an A. Only studying for qualifying exams and then teaching did I start to really master courses.
That said I would hate myself for getting less than an A if I ever did
I got less than an A several times. I mentioned in here earlier that one B was in the material of the multivariable math book; I think I'm relatively good at that stuff :P
"hate" is a strong word
6:08 PM
I agree. Hate is not helpful.
my perspective on even the most simple concepts shifts the more I advance into more complicated areas
@ペガサスSeiya I would be very hard pressed to hire a job applicant with straight As.
@user726941 most of my family is compromised of people that are exceptional at math, the pressure to perform up to that par is a lot
less is more; or however the saying went
Often straight As are a sign that you were avoiding challenges and trying to get just As
6:09 PM
When I see a transcript with nothing but As, my immediate assumption is that this person never stretched themselves. They never took a class outside of their interests, and never took a "hard" class.
I don't trust 4.0s.
@XanderHenderson well, I'm taking a lot of math classes. The only ones I am not taking as far as I know are statistics
How about 3.999... :p
@user726941 That is a person who took too many classes, and got an A- once. Same problem. :P
Only worse, because they took 10 years to finish a 4 year degree.
@ペガサスSeiya Right, you aren't taking classes outside of your core interest.
Take an anthropology class.
@XanderHenderson I mean, I don't like seeing a "C- in World History" on my report card
You need to be more "well rounded."
6:12 PM
In much of the world, Xander, college is effectively grad school. No gen ed.
I get hammered badly in non-math courses I'll admit
Fix that.
not taking classes outside your core intereste is fine
@ペガサスSeiya When I am on hiring committees, I much prefer a transcript that has a C- in World History (but otherwise As and A-s) over a transcript with nothing but As.
Its not easy. Words are harder for me to understand properly. I generally have to work A LOT harder than most to pass the history class
6:14 PM
Can’t do that in the US, Thor.
@TedShifrin Oh, I get that. But in much of the rest of the world, folk are never given a classical liberal education. I want to hire people with classical liberal educations. :D
well, can't quite do it here either
I double majored in math and French. 🤷‍♂️
have to do a minor after all
@XanderHenderson just curious. Why not hire someone who specializes at that job as opposed to someone "well rounded"?
@TedShifrin Bonjour
6:15 PM
@ペガサスSeiya I want a specialist who has a well-rounded background.
Which is what a typical American PhD (or even masters grad) is going to have.
@XanderHenderson what are you building? A team of super humans?
With respect to my current position, I am at a community college. A lot of the students we teach are severely disadvantaged, and I want to know that anyone we hire has something in their background which is going to help them empathize.
empathy wasn't part of my curriculum!
Maybe some sociology or anthropology classes, maybe a few Cs---but something that is going to help folk relate beyond their area of specialization.
Not a part of mine either
@XanderHenderson video games?
6:18 PM
I will note that, contrary to many of the CCs in Arizona, we require two semesters of writing for any degree, including things like "welding" and "cosmetology".
And the broader community here loves us for it---our graduates are often complemented for being able to write and communicate well.
@TedShifrin Are you referring to your Multivariable Calculus and Linear Algebra course? I'm doing self-study on it right now. Thanks for putting it up! Just finished Day 1.
@XanderHenderson "write and communicate well." That's my kryptonite
@ペガサスSeiya I wouldn't want to hire you, then. You would be a bad fit for this institution.
writing and communication are important
@XanderHenderson That's too bad, I would've loved to pull pranks on you
6:21 PM
@GratefulDisciple Yeah, I was referring to that. Still, without the course/book one does not have the exercises, and to me that is the major part of the course (together with feedback thereon).
By the way, if you do get the book, you can find my homework assignments hidden away on the web (unless they're finally gone).
[Illegal copy]
One day, Ted, one day I'll convince you to let me buy the legal copy
Ha ha.
@TedShifrin I do plan to study it with your book and I found the syllabus and homework assignments here and here. So I plan to do the exercises too.
OK, cool. The only comment I'll make is that the assignments are mostly theoretical/proof stuff, because all the computational problems were done on WebWork (and graded automatically)!
@TedShifrin I see. Thankfully, there are answers to the selected exercises in your book. I'm taking this for math background for machine learning (I'm in Computer Science).
6:27 PM
Geometry is fun
There are plenty in the book, @user726941. I mostly modeled the WebWork problems (which literally took me hundreds of hours to code and troubleshoot) on problems in the book. There are a few exceptions. I still have the WeBWork problem code, but you would need access to WeBWork to enable it.
Can we say that $D^4=D^2\times D^2$?
D is the unit closed disk.
@TedShifrin thanks, sounds like a labour of love. Don't let it go to waste professor
I need this to show that $S^3$, S= solid torus can be written as a union of two solid tori.
yesterday, by Koro
because in general I'm getting: $\partial (A\times B)= (\partial A\times \bar B)\cup (\bar A\times \partial B)$.
@Koro if $=$ means homeomorphic, then sure
it's worth thinking about why
6:31 PM
hmm, thinking.
@user726941 Too late for that. I am retired and removed from it all. I think the people teaching the course at UGA have quit using my WebWork.
"then why take the course?!"
I'm not familiar with how the course-system is yet, but don't you need to take some courses to receive a degree? Or is there a way to take some test that proves that you know the material?
@Koro That is definitely not equal. I would say that $I^4 = I^2\times I^2$, but I would write $D^4\cong D^2\times D^2$ :D
@ILikeMathematics My point is that very few students know the material as well as they think. We had a course challenge system in place at UGA (not at all levels). Almost every challenge that happened while I was associate department head (8 years) was a failure. I think out of many tries, one student got credit for the course that he thought he knew so well.
I think people would usually write $D^4=D^2\times D^2$ in the context of smooth manifolds (where it is understood that products of manifolds with boundaries are automatically smoothed!), but not in other contexts
I studied Quotient topology today from various sources including Loring Tu's. I feel like I understood many things. I can now create/think about examples of quotient maps which are not open/closed :-). There was also an exercise on 'sphere with a hair' is not a manifold. I understood it :-).
6:35 PM
that's good
quotient constructions are of universal importance in topology
But I still feel I lack basics for algebraic topology. $D^2\times D^2= D^4$ should not be something to get stuck upon but I'm stuck here.
I suggest thinking about replacing $D$ with $I$
and whether that's allowed
with I, it is okay and we get the equality.
@Thorgott I was in the topological category. I am not in the custom of presuming manifolds with corners have them smoothed.
that's by definition of cartesian product.
with D^4, it's not the equality for sure.
6:37 PM
@TedShifrin are they still trying this challenge?
right, so are $D$ and $I$ homeomorphic
I have no idea. I've been gone 8 years, and was not associate department head past 2013 (I think).
D and I are not homeomorphic.
I even stopped giving credit for linear algebra courses taken at community colleges, because they tended to be completely computational and without proofs.
6:39 PM
because if I remove an interior point from I, connectedness is gone.
What are $D$ and $I$ without superscripts?
Hold it. Be slightly pedantic and put in superscripts.
I claim $D^1 = I^1$.
do imagine a superscript (the same on both!), I was too lazy to write it
@TedShifrin Ohh
And is $D$ an open ball or a closed ball?
then of course D=I.
D is closed unit ball.
6:40 PM
$D$ is closed, $B$ is open, if you ask me
OK. Then that's what Thor was talking about with corners.
My RC plane model is almost complete. Just need to modify the wing design slightly to allow for better control authority at higher angles of attack
(never ignore the power of removing points from spaces, don't be afraid of it)
@Thorgott yes, with superscripts.
now put all these observations together and you'll be done
I have a question. In computer science, we have an algorithm that works for a some subset (lets say A) and it has been proven that will take this subset as input and returns an output.

Now I created a new algorithm that takes a larger subset (lets say B such that A is a subset of B) and it works. But how to prove compatiblity with the previous algorithm?

I think I need to say the new algorithm is well-formed and extends the previous algorithm.

What is this process called in Math and is there an example for it?
6:45 PM
$I^2\simeq D^2$. So $I^2\times I^2\simeq D^2\times D^2$
$I^4\simeq D^2\times D^2$
f: A -> R
f': B -> R'

but if we pass A into f' we also get R

How to prove f' extends f
Have you ever taught any neurodivergent students that you knew of? @TedShifrin
@TedShifrin It's great you still have the WeBWorK files. If you like, I would be more than happy to try to host a WeBWorK system, since it appears that WeBWorK is an open source project. That way, the students of your YouTube lecture recordings can have their homework graded automatically.
@TedShifrin did any of them show signs of struggling to communicate?
Is that a problem in classes?
6:56 PM
All different issues. Plenty of ADHD. Some with dyslexia. Several with autism and Aspergers. They tended to be a bit disruptive.
@GratefulDisciple Interesting. I wonder if the various updates in WeBWork over the past decade will cause compilation issues with my code. The problem is that a course needs to be set up with an instructor, who then gets bombarded with questions from students working the homework. I'm not willing to be that generous :D
@TedShifrin Yeah, I happen to be an aspie
@TedShifrin And after you've worked on the challenge problems and feel like you understood everything from the course, would you then study ahead the courses you will take next?
How busy were your office hours with that particular class @TedShifrin
02:00 - 19:0019:00 - 00:00

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