What do you call an object that represents the identity functor? I'm pretty sure these had a name I've heard. For instance hom(C[G],V)~V in the cat of G-Reps, and hom(G,X)~X in the cat of G-sets.
Munkres wants me to prove that if $p : E \to B$ is a covering map, $B$ is connected and $p^{-1}(b_0)$ for some $b_0$ in $B$ has $k$ elements, then so has $p^{-1}(b)$ for any $b \in B$.
Here's what I have tried : Consider an evenly covered neighborhood $U_0$ of $b_0$. Take a point $x \in U_0$. Assume $p^{-1}(x)$ has more than $k$ elements. But as $p^{-1}(b_0)$ has exactly $k$ elements, and $p^{-1}(U)$ is union of disjoint ${V_\alpha}$s, there is a $V_i$ which contains an elt of $p^{-1}(x)$ but not of $p^{-1}(b_0)$
This gives us a contradiction to the fact that $p$ is a covering space. Now as $B$ is connected, $U_0$ has nonempty boundary. So take a boundary point $b_1$, and we know that every nbhd of $b_1$ lying entirely in $U_0$ has $k$ fibers over it. Consider an evenly covered nbhd $U_1$ of $b_1$, and we have similarly than $f^{-1}(x)$ has k elements for any $x \in U_1$. Continue this way to cover all of $B$.
sorry i got disconnected
@anon that's an interesting idea. but i seriously doubt about continuity.
@BalarkaSen pick $k\in\Bbb N$ and let $U$ be the set of $x\in B$ such that $|p^{-1}(x)|=k$. picking any $x$ in $U$, we know there exists a $V_x\ni x$ open in $B$ (using a subscript cuz it will depend on $x$...) for which $p^{-1}(V_x)$ is $k$ disjoint open sets mapping homeomorphically onto $V_x$. Thus, $V_x\subseteq U$. And then we know that $U=\bigcup_{x\in U}V_x$ is open.
since the fibers of $B\to\Bbb N$ are open and $\Bbb N$ is discrete, this means $B\to\Bbb N$ is continuous
and since $B$ is connected, it must have a connected image, hence a singleton
@Committingtoachallenge Do you see the discussion i had @anon earlier about x/y vs y graphs. Well, I wanted to ask him something about it but I can't put it into words.
I just wanted intuition of the behavior of such plots. When the two functions are proportional to each other the plot is a straight line ("y = x" type )but what if their relationship wasn't linear? What if we were plotting $y^2$ against $\sqrt{x}$
Somebody somewhere on this site mentioned in the comments somewhere that this is equivalent to plotting $y^2 = \sqrt{x}$ on the normal cartesian plane.
@Nick I've looked at a few of these. I can show the determinant is zero for linearly dependent vectors, it's showing that the determinant is greater than zero for linearly independent vectors.
@DanielFischer @robjohn How does one prove this $$\int_0^\infty\frac{\cos ax -\cos bx}{x}\,dx=\ln\left(\frac{b}{a}\right)$$Frullani's theorem came across to mind but the problem is how does one justify $\displaystyle\lim_{x\to\infty}\cos x=0$? W|A says $\displaystyle\lim_{x\to\infty}\cos x=-1\text{ to }1$.
@Venus Try writing it as $$\lim_{h\to0^+}\left[\int_h^\infty\frac{\cos(ax)}{x}\mathrm{d}x -\int_h^\infty\frac{\cos(bx)}{x}\mathrm{d}x\right]$$ change variables on the right integral so that you have $\int_{bh/a}^\infty\frac{\cos(ax)}{x}\mathrm{d}x$ then you are left with $$\lim_{h\to0^+}\int_h^{bh/a}\frac{\cos(ax)}{x}\mathrm{d}x$$
@Venus I use the same method in this answer. After another change of variables, you end up with $$\lim_{h\to0^+}\int_1^{b/a}\frac{\cos(ahx)}{x}\mathrm{d}x$$ I hope that answers your question. I have to take a whining dog to the park.
Anyway, @Balarka, the construction is of a cell/CW complex. If you understand how they're constructed in general, you'll understand the special case. And examples help for understanding stuff in general.
@MikeMiller A ok.. I want to show that the algebraic set $V$ of $K^n$ is irreducible iff $I(V)$ is a prime ideal. That's what I have tried:
We suppose that $I(V)$ is a prime ideal and $V$ is reducible. That means that there are two non-empty subsets $V_1,V_2$ of $V$ such that $V=V_1 \cup V_2$. Then, $I(V)=I(V_1 \cup V_2)=I(V_1) \cap I(V_2)$. We have that $V_1 \subset V \Rightarrow I(V) \subset I(V_1)$ and $V_2 \subset V \Rightarrow I(V) \subset I(V_2)$ How can we find a contradiction?
@Alizter I am only saying that I can understand someone deprecating their own answer and saying that someone else's is better, especially if their answer has gotten a lot of upvotes that they don't think they deserve.
No, he's not. He's saying the genus $g$ surface is a cell complex.
You start with a 0-dimensional cell complex (a point, in this case); you attach 1-cells to get something that looks like the boundary of the fundamental domain
Now you attach a 2-cell to the 1-skeleton to get the surface itself.
Give it some thinking. I've got to do some work. These are very important (so very important to understand), and they're what's used in your $\pi_1$ construction.
Hey @robjohn!!! I want to show that the algebraic set $V$ of $K^n$ is irreducible iff $I(V)$ is a prime ideal. That's what I have tried:
We suppose that $I(V)$ is a prime ideal and $V$ is reducible. That means that there are two non-empty subsets $V_1,V_2$ of $V$ such that $V=V_1 \cup V_2$. Then, $I(V)=I(V_1 \cup V_2)=I(V_1) \cap I(V_2)$. We have that $V_1 \subset V \Rightarrow I(V) \subset I(V_1)$ and $V_2 \subset V \Rightarrow I(V) \subset I(V_2)$ How can we find a contradiction?
I only know auto and iso so far, and I've only seen that phrase used for isomorphism but I don't know if it's used elsewhere as I don't know what it means
@UserX I can't possibly answer to your question without knowing the context. But for example, upto isomorphism means considering groups/rings/fields upto identification by isomorphism
@Balarka it deals with transforming information efficently, yet with as little chance as possible for it to be recieved wrongly after going through noise
@Studentmath: It isn't going to be pretty, but I've written my last exam for probability and am working on the final. Did conditional variance formula, have a few more examples and the Central Limit Theorem to discuss. That's about the end.
@Ted He means the construction of a CW-complex with a particular fundamental group, by taking the wedge of $n$ circles and gluing on discs to kill the words in the presentation.
@Mike @Ted it took me long to actually understand these things (not that I do now), until then it could possibly make some sense to me, when it was just used to solve things.
One of my students thought of a totally wrong solution to a problem turned in yesterday. It gives the right answer, but I can't for the life of me understand why.
@Studentmath: In the time I've known you in this chat, you've made a tremendous amount of progress on mathematics.
-No series, @Ted. Numerical integration, Taylor polynomials, and error bounds for those. Couple problems on convergence of sequences (using epsilon-M definition... but the problem asked them to find an M for epsilon=1/1000, say).
interesting exercise from munkres : $p : E \to B$ be a covering space with $p^{-1}(b)$ finite for each $b \in B$, $B$ being compact, then $E$ is also compact.
@TedShifrin The most interesting stuff I have encoutered up until now is probably the fact that for any path connected $X$ and $G$ acting on $X$ properly discontinuously and freely, there is a short exact seq $1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$. this mildly resembles the short exact seq in galois groups, imho.
So, @Studentmath, the perplexing question was this. Let $X$ denote the number of tosses of a die it takes to get a $1$, and let $Y$ denote the number it takes to get a $2$. What is $E[X|Y=5]$? My student's answer was: $(1-(4/5)^4)\cdot 5 + (4/5)^4\cdot 7$, as $5$ is the expected number it takes given no $2$'s, and $7$ is the number it takes if you toss a $2$ on the first try.
Yes, @Balarka, there's a total correspondence between the Fund Thm of Covering Space Theory and the Fund Thm of Galois Theory.
Two of my students kept saying it made intuitive sense, but as soon as they actually tried to explain it (or derive it with conditional expectation), they got very silent. @Studentmath :)
@TedShifrin That reminds me, one of the exam problems was quite tricky (since it made them think a teeny bit). It gave them the value of $f(0)$, $f'(0)$, and a bound on $|f''(x)|$, and told them to prove $|f(1)| < 2$.
@Ted because there is some intutition in it indeed, and it gives the right answer, I think it should be deriveable somehow.. but then again, certainly not the way presented
@TedShifrin The idea was to use the Taylor polynomial bound... the first taylor polynomial is just 0, so we get $|f(1)| \leq 4*1^2/2!$. But this was a little tricky.
@Ted I think he might've went with the formula of the expectation of geometric RV, and he said - given no 2's, the probability of getting a 1 is 1/5, thus E[X]=5, and given a 2 on the first time, the probability of getting 1 is 1/6 on the latter times, so the expectaion is 1+6=7.
