@AsafKaragila Now factor out the action by the subgroup. The covering projection of the universal covering factors through this projection and what you want to show is then more or less a tautology.
@HenningMakholm ah, if you wanna cap, go for LHF!!
Those fruit hung so low that they already have roots...
Now, what I don't understand is, from here Bartle and Sherbert claim that, the union of the intervals in $\dot P_1$, say $\mathcal I$ is such that $$[1,3-\delta]\subseteq \mathcal I \subseteq [1,3+\delta]$$ @tb
Oh, you thought that I meant with his personal life? Well, while I am indeed curious about the personal lives of a few people in this chatroom I am not that interested...
No, I start with a partition $P$ (dots dropped). Now consider the partition that you can complete for $[1,3]$ from only the points of $P$ and adding possibly the end points. I call this $P_1$. @tb
@KannappanSampath Yes, like taking a hot shower and massaging the painful area under hot water. I just need to finish this first... Unless you mean something like suicide, in which case I'm not sure it would actually solve the problem as it will take a quotient by the subgroup of complaints... :-)
or: put another way, if one is going to use integrals of step functions to approximate integrals of other functions, the step functions themselves ought to be integrable
and the point with the riemann integral, is we are going to let the "mesh" of the partition go to 0, which will force the "deltas" to go to 0 as well (they have to be smaller than the mesh)
if not, then one of the sub-intervals stops at 3.
the difference between the darboux and the riemann integral, is that in the riemann integral, we don't specify "which" element of the subinterval we tag. any one will do.
@N3buchadnezzar I would use polar coordinates but I would not translate. You should get an integral of the form $$\int_{0}^4 \int_{\text{a function of } r}^{\text{another function of }r} r\,d\varphi\,dr$$
$D$ is a circle of radius $2$ centered at $(2,0)$.
@N3buchadnezzar Sorry I should have written $r \cdot r \,d\varphi\,dr$ in the integrand (one $r$ is your function and the other $r$ from the change of variables into polar coordinates).
I choose not to reply to derogatory comments, but this is getting like a personal attack. To recount, @Gigili, you claimed that you did not want to talk to me [which you're free to, without announcing] not because I did not get your point [which I have no obligation to] but because I am an Indian. [This if you think is very reasonable, well, you MUST learn what is called a chat room] .
If I have not made myself clear, as in, if I see more derogatory comments, I have this same message on a .txt file which I'll painlessly reproduce here. [cf. with all the efforts you'll have to put in to type things.] : ) @Gigili
@N3buchadnezzar Again, I'm not translating the coordinate system. Since $D$ is a disk of radius $2$ centered at $(2,0)$ the radius $r$ must go from $0$ to $4$. Now you have to determine the angle of the two intersections of the circle of radius $2$ around $(2,0)$ with the circle of radius $r$ around $(0,0)$. This should give you something like $\pm \arctan{(\text{something with a square root})}$ for the upper and lower bounds of the inner integral.
(What I'm trying to exploit is that $f(r\cos\varphi,r\sin\varphi) = r$ but I haven't done the computation.)
@DavidWheeler It started when I had my picture as avatar and this guy kept addressing me as "he" and adding "I don't care [you're female]". Like I do care he's male or all the other people here are as well.
Well, @Gigili, Can you present the true facts. When on hell, did I say, "I don't care [you're female]"? Yes, I'd say, I don't care but never have I said, you're a female. Like you announced you were not going to me, you should have announced, well this is my real self. How am I supposed to know people whose avatar is a female is a female? @Gigili. I would definitely need you to clarify this point.
@GraceNote Hi Grace! I've seen the flag and I think Kannappan and Gigili have a misunderstanding that keeps going on and on and on. But it should be possible to sort it out without a major uproar.
I don't even know what Obsequiousness is, haha. A blue user on chat is a chat moderator (in other words, anyone who is a moderator on any site on Stack Exchange).
I personally am a Community Coordinator working for Stack Exchange. Nice to meet you. ♪
@MattN My mentor loaded me with a bunch more of exercises on the Power set ring. He said that ring is "mildly powerful" that you'll learn a lot of ring theory playing with them. : )
@KannappanSampath Alright. Cool with me. Btw, I've been thinking that it would probably be better for you if we do the exercises in the morning (my morning) then it's not so late for you. So what do you say if we do that tomorrow (i.e. about 11 hours from now)?
I was quite surprised today. I was bracing for a lot of pain and suffering but Commutative Algebra is not so bad at all. Maybe I start to get the hang of maths a bit. But let's not count my chickens before they're hatched.
It was also quite reassuring to hear that guys around here thought about maths already when they were in high school. Made me feel less bad for lagging behind.
@KannappanSampath No. Not at all. But at the moment I'm bonged out. Maybe if you give me an hour of rest we can do some more exercises if that's not too late for you.
So someone asked: _what is the difference between "infinite moment" and "moments don't exist"?_ and I was gearing up for a discussion of Zeno's paradoxes ...
David, yes. However these are different spaces and the fundamental groups are different objects which are well defined. We can identify, but it's not the same (and we need that assumption).
So consider all the paths in $Y$ whose starting point is $y_0$, we say that $\gamma\sim_H\tau$ if $\gamma(1)=\tau(1)$ and $[\gamma\ast\tau^{-1}]\in H$, where $\cdot^{-1}$ is the inverse path and $\ast$ is concatenation.
We define $\rho:X\to Y$ by $\rho([\gamma]_H)=\gamma(1)$. This is well defined and all that shizzle.
Since $X$ has the homotopy lifting property we can lift $\gamma$ to a unique path in $X$ starting at $x_0$, and it ends in $x_0$ as well since $\gamma\in H$.
So we have that $H\subseteq\rho_\ast(\pi_1(X,x_0))$, which is fine. Now I need to show the other direction, that is take a loop in $\pi_1(X,x_0)$ and argue why its image is in $H$.
Hm. I think if we take $\gamma\notin H$ then it is not $\sim_H$ to $x_0$ so it has to end at a different point, thus not a loop in $X$.
Suppose the image is not in $H$, then you would not be able to submit the AT assignment. This <quote> for the love of god and all that is holy </quote> contradicts the fact that Asaf is the owner of this room.
i think a good X to think about is the torus, where we have, essentially, two different kinds of loops: the ones around the hole, and the ones through the hole and around.
the cover of one kind is a cylinder.
the universal cover is the plane, all the loops are trivial
when you quotient the plane (in one direction), you create non-trivial loops.
If I have a set C = {4, 2, 1, 3} it has a cardinality of 4, if I have a set of C = {4, 2, 1, 3, 1} is this still called a set and if so does it still have a cardinality of 4?
@t.b. I doing some calculations with derived torsion and derived completion. In the noetherian situation, you can compute them using tensoring or homing with the infinite Koszul complex (the Cech complex). This is not always true in the non-noetherian case, so I have to work hard to show that in my cases it is the case
@anonymous Sheesh, it's been a while and it's quite a bit outside my comfort zone... Have you checked whether the memoir by Beligiannis and Reiten is of any help?
@Ilya Well, there's nothing wrong about an accept rate in the fourties. It shows that the user knows that the accept button is there and what it's for; I'm not sure one could (or reasonably ought to) expect more. The SE people were told this, loudly, when the accept rate counter was introduced, but claimed that it probably wouldn't turn out to be a problem. Sigh.
well, I'm using `\documentclass[a4paper, 10pt]{amsart}` there and `\usepackage{amsthm, amsmath}` together with some custom specs for the theorem, definition and remark environments.
\newtheoremstyle{mythm}% hnamei {6pt}% hSpace abovei {6pt}% hSpace belowi {\itshape}% hBody fonti {}% hIndent amounti1 {\scshape}% hTheorem head fonti {.}% hPunctuation after theorem headi {.5em}% hSpace after theorem headi2 {}%
\newtheoremstyle{mydef}% hname {6pt}% hSpace abovei {6pt}% hSpace belowi {}% hBody fonti {}% hIndent amounti1 {\scshape}% hTheorem head fonti {.}% hPunctuation after theorem headi {.5em}% hSpace after theorem headi2 {}%
@MattN I would say everything up to 8 pages is very short. Normal papers in pure math are 10-20 pages, but occasionally they grow to 40 or fifty. More pages are the exception, I think.
