Consider the set of all functions $\mathbb R^2\to\mathbb R^2$ defined by $f(\vec v)=\cos(\vec v\cdot\hat u+c)$ where $\hat u$ is a unit vector and $c$ is a constant
The question I uploaded yesterday, i.e. The complement of a closed discrete subspace (S) of $\Bbb R^n$ is simply connected if $n\geq 3$. Using the set $S$ is discrete, for each $x_i\in S$, we can find small n-1 sphere $S_i$ centered at $x_i$ such that $S_i$'s are disjoint.
How can I fill the rest of the space in $\Bbb R^n$? i.e $\Bbb R^n-\bigcup_i S_i$
Question ask to use proposition 1.26 so I tried to use 1.26(b) by attaching $n$-cell to $S_i$'s to make $\Bbb R^n-\bigcup_i B_i$
Using the fact that the space deformation retract to $S_i$ union some line segments is homotopy equivalent to wedge sum of $S_i$'s and $\pi_1(S_i)=0$ so using van Kampen, the whole space is simply connected. This is what you're saying right?
I am looking at the question : Check the convergence of the sequence $a_n=\left (\frac{n+2000}{n-2000}\right)^n$, $n>1$. If it converges calculate the limit.
We have $$a_n=\left (1+\frac{4000}{n-2000}\right) ^{n-2000}\left (1+\frac{4000}{n-2000}\right) ^{2000}\to e^{4000}$$ Having found the limit means that the sequence is also convergent, right? But could we have shown the convergence also in an other way?
@MaryStar You can also look at the numerator and denominator separately after cancelling $n$: $\lim\limits_{n\to\infty}\left(1+\frac{2000}n\right)^n=e^{2000}$ and $\lim\limits_{n\to\infty}\left(1-\frac{2000}n\right)^n=e^{-2000}$
Exercise one: a) All homotopy, homology, cohomology groups of klein bottle times 2-torus. Hell idk how to prove the fundamental group of the klein bottle on the spot, higher homotopies are easy though. b) show reduced zeroth homology of S^n minus smooth closed embedded smaller-dimensional submanifold is equal to n-1th reduced cohomology of the manifold, so the k=0 case of alexander duality c) asks whether K^2 times T^2 embeds into R^5
Exercise 2: a) show that klein bottle is an S^1-bundle over S^1. This is easy and was on the exercises but I wasted too much time doing this carefully until …
last week there were still lectures, I have 3 other exams in the coming weeks and all this intersection form stuff we only covered in the week before last week
Hi, why is this true? Let $r_i \geq 0 $ such that $\sum_{i=1}^k r_i = 1$. Then if $\{z_i \}_{i=1}^k \subset S^1 \subset \mathbb{C}$, and $\sum_{i=1}^k z_i r_i = 1$, all $z_i = 1$ necessarily
Or rather, for all $i$ such that $r_i > 0$, $z_i = 1$
Hi, I have a question. Let $G$ be a group acting on a topologiacl space $X$ and $H_1,H_2$ be two infinite cyclic subgroups of $G$ generated by $\varphi_1,\varphi_2$, respectively. Define $f:X/H_1\to X/H_1$ as $[x]_1\longmapsto[x]_2$. Is $f$ a well-defined homeomorphism?
Does anyone have examples of questions where I have to find the equation of a tangent to a curve, GIVEN AN EXTERNAL point? I am having trouble finding worksheets or papers of this kind. many thanks
probability brain doesn't work out nicely, one can in principle view this as a real random variable with finitely many values and $\varphi_X(1) = \Bbb E[e^{iX}] = 1$, and then try to prove it must have values in $2\pi \Bbb Z$. Wouldn't know how to do that though
Your map is clearly not usually well-defined which you'd see if you wrote out what it means for it to be well-defined
If the actions are conjugate (so the generators are conjugate elements in $G$) then this is true but that's the best you're going to get
Otherwise the actions are basically unrelated...
There is an action of $\Bbb Z$ on the cylinder $S^1 \times \Bbb R$ whose quotient is the Klein bottle, but so that $2\Bbb Z$ just acts by a translation
@BalarkaSen I lovingly selected a small memedump for you as a celebration of a successful exam and then I got rekt. Now it's not appropriate anymore smh
yeh happens. Consolation is that even those who had a 5 page LaTeX written solution ready for every exercise session had massive problems, and I did not invest that level of effort
Shit is that I now have 3 other exams in the next 17 days and invested lots of time into the topology thing. I could probably have passed but that's not worth it for a possibly bad grade
Probably, so the cohomology ring of $K^2 \times K^2$ with $\Bbb Z_2$ coefficients is $\Bbb Z_2[x_1,x_2,y_1,y_2]/(x_1x_2,x_1^2-x_2^2,y_1y_2,y_1^2-y_2^2)$?
Hm but then these would not reach the fourth cohomology with any combination I see
$G_d$ is a space in the following sense. I know how to define distance between rooted finite graphs of degree at most $d$
If $\Gamma_1, \Gamma_2$ are two graphs with roots $o_1, o_2$, I define $d(\Gamma_1, \Gamma_2) = 1/k$ if $k$ is the maximum number for which the $k$-balls in $\Gamma_1$ and $\Gamma_2$ centered at $o_1$ and $o_2$ resp are isomorphic
This is the rooted distance
But how do you turn an arbitrary graph into a rooted graph? The root need not be god-given
Yeah but then it's a boring notion because small parts of the graph can have massive degree
You're only looking at that for a long time
I'll give you an example soon
Solution: Choose a root uniformly, cuz your graph is finite. That makes any graph in $G_d$ a random rooted graph, a distribution on the space $RG_d$ of rooted connected graphs of degree at most $d$.
$RG_d$ is the metric space, mind you, with the rooted distance. It's a compact metric space.
So finite graphs are just probability distributions on $RG_d$.
$d(P_n^k, P_n^{k'})=\max(1/(n-k), 1/k, 1/n-k', 1/k')$ so for every $\epsilon = {1\over K}$, we can separate the sum as $\sum_{k=1}^K f(P^k_n)+f(P^{n-k}_n) +\sum_{k=K}^{N-K}f(P_n^k)$
The left sum is bounded and the right sum is close to n times some limit
Hi, I'm happy with the definition of a sheaf of $\mathcal{O}_X$-modules, but it occurred to me that I don't know how these are defined as functors. In functorial language, perhaps atleast dealing with the presheaf part, it's a pair of functors $\mathcal{O}_X\times \mathcal{M}:\text{Open}(X)^{\text{op}}\to \text{CRing}\times \Bbb Z\text{-mod}$ and $\mathcal{M}:\text{Open}(X)^{\text{op}}\to \text{CRing}\times \Bbb Z\text{-mod}$, along with a natural transformation
If O is a presheaf of rings, then a presheaf F of O-modules ought to be a presheaf of abelian groups equipped with a map of presheaves O x F -> F satisfying the usual relations, yes?
I think you're looking for the notion of ring object in an abelian category, and then module object over a ring object in that category.
@MikeMiller I guess I was confused about the targets of those functors O x F -> F, and the natural transformation is a natural transformation between functors on which categories, that sort of thing
Btw, the only reason I was trying to do something like this is I want to take stalks on a complex of sheaves of O_X-modules, but wanted to state something about functoriality
Like you want to apply $\varinjlim_{U\ni x}: \mathcal{C}\to\mathcal{D}$ for whatever categories (in theory)
You could take them to land in Set, but ask that they factor through Ring, AbGrp etc I guess
Like $O:Open(X)^{op}\to CRing, F:Open(X)^{op}\to \Bbb Z-mod$, so the natural transformation doesn't immediately make sense to me functorially
say $M$ is a smooth manifold (with boundary, if that matters) and $f\colon M\rightarrow[0,1]$ is a submersion, can I always connect two points in different fibers of $f$ by a path tranverse to the fibers of $f$?
