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18

Yes. We can consider a sequence of disks $D_1,D_2,\ldots$ around $(\sqrt 2,\pi)$ such that the $n$th disk has precisely $n$ lattice points in its interior (and none on its boundary). Let $z_n$ be the lattice point in $D_n\setminus D_{n-1}$. Construct accordingly a sequence $C_1, C_2, \ldots$ of simply connected compact sets with smooth boundary such that ...


6

Let $\mathcal S$ be a compact surface, possibly with boundary. Let $\text{Homeo}^+(\mathcal S)$ refer to the group of orientation-preserving homeomorphisms that fix the boundary pointwise with the compact-open topology. Throwing an $n$ in there means we add $n$ marked points in the interior, which I prefer over deleting points. (I'd be worried about the ...


5

A nontrivial wedge sum $M \vee N$ of closed manifolds fails to satisfy Poincaré duality, and so cannot be homotopy equivalent to another closed manifold $X$. First, if $M$ and $N$ have the same dimension, then top (co)homology distinguishes $M \vee N$ from a closed manifold. So assume WLOG that $M$ has strictly smaller dimension than $N$. Recall that ...


4

The four non-orientable non-compact Euclidean 3-manifolds are: $M\times \mathbb{R}$, $M\times S^1$, $K\times \mathbb{R}$, and $K \times S^1$, where $M$ denotes the open Mobius band (i.e. the quotient of $\mathbb{R}^2$ by a glide reflection) and $K$ denotes the Klein bottle. Note that these are non-homeomorphic, since their fundamental groups are ...


3

Here's an idea on how to construct a solution for the bijection you describe. It appears feasible by a diffeomorphism of the plane. First we solve the following problem : extend the bijection \begin{array}{RCL} \phi:\Bbb Z\times\lbrace-1,0\rbrace & \longrightarrow & \Bbb Z\times\lbrace-1,0\rbrace\\ (n,\epsilon) & \longmapsto & \begin{cases} ...


3

Here's a sort of diagrammatic argument from an old homework assignment: Construct $\mathbb{R}P^3$ as the quotient of $B^3 \subset \mathbb{R}^3$ under the antipodal map $a: \partial B^3 \to \partial B^3$. Let $K$ be the knot in $\mathbb{R}P^3$ obtained as the quotient of the vertical segment $V=\{(0,0,z) \in \mathbb{R}^3 : -1 \leq z \leq 1\}$ in $B^3$. As ...


2

Your idea of using a triangulation each of whose simplices lies in one of the $U_i$ is a good one. There exists such a covering of the form $V_0 \cup \ldots \cup V_D$ where $D$ is the dimension of $M$, such that if $M^{(k)}$ denotes the $k$-skeleton of the triangulaion then $$M^{(k)} \subset V_0 \cup \ldots \cup V_k \quad\text{for each $k=1,\ldots,D$.} $$ ...


2

I think it is path-connected. Let $S = \{ D_n \, | \, n \in \mathbb{N} \}$ be the set of closed disjoint disks under consideration. For any integer $m \ge 1$, let $N_m = \{ n \in \mathbb{N} \, | \, \mathrm{radius}(D_n) \in [1/m, 1/m-1) \}$ (with the convention $1/0 = \infty$) and $S_m = \{ D_n \, | \, n \in N_m \}$. Let $P, Q \in \mathbb{R}^2 \backslash S$ ...


2

Contrary to the request of the OP, this is still a sketch. I put it here because it points to some relevant literature, hoping that somebody more expert than me will help to clarify it. The question reminded me of this paper by Fischer and Zastrow: the relevant part is the end of p. 12. See also this answer by George Lowther. You can assume that $\cup D_n$ ...


1

So here is my attempt to clean up and clarify the other answer. We pick $A,B \in \Bbb R^2 \setminus S$ For some disks $D_n = B(c_n,r_n)$ we're going to find a "thickening" of $D_n$, $T_n$, such that its border $B_n$ is disjoint from every disk, and find a continuous projecti on $f_n : T_n \setminus \{ c_n\} \to B_n$. We also want the $T_n$ to be disjoint ...


1

Jørgensen's inequality says that if two elements $A, B \in SL_2(\mathbb{C})$ generate a non-elementary discrete group, then $$ |\mathrm{Tr}(A)^2-4| + |\mathrm{Tr}(ABA^{-1}B^{-1})-2| \ge 1 $$ Assuming that $A=\begin{bmatrix} 1& 1 \\ 0 & 1\end{bmatrix}$ and $B=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (with $a,b,c,d \in \mathbb{R}$ and ...


1

Here is another solution, somewhat in the same spirit as Olivier Bégassat's answer. In fact, we will prove a stronger result : any bijection $f : \mathbb{Z}^2 \to \mathbb{Z}^2$ comes from a smooth isotopy $h : \mathbb{R}^2 \times [0,1] \to \mathbb{R}^2$ such that $h(x,0) = x$ and $h(-,1) = f : \mathbb{Z}^2 \to \mathbb{Z}^2$. As in Hagen von Eitzen's answer, ...


1

Yes, these points are described by the open set $h^{-1}(0, \infty)$, where $h(x,y)= g(x,y)-f(x,y)$, and $h$ is smooth ( though continuous would be enough).


1

The expression you wrote does not define a Riemannian metric for coordinates on $S^3$, because when $\eta=0$ or $\pi/2$ the metric degenerates. It does define coordinates for a portion of $S^3$, namely the portion where the coordinates satisfy the strict inequalities $0 < \eta < \pi/2$, $0 < \xi_1, \xi_2 < 2 \pi$. So perhaps you meant to use ...



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