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1

I am not sure If I understand the definition correctly, but if I do, then take a look at the maps $$ f : (0,1) \rightarrow [0,1], x \mapsto x$$ and $$ g : [0,1] \to (0,1), x \mapsto \frac{1}{3} x + \frac{1}{3}.$$ These are embeddings and composing these we get $$ f \circ g : [0,1] \rightarrow [0,1], x \mapsto \frac{1}{3} x + \frac{1}{3}$$ and $$ g \circ f ...


3

I will just deal with the case of adding in unknots with framing $+1$. Everything is entirely analogous in the $-1$ case. $\Bbb CP^2-\{pt\}$ is naturally an open $D^2$-bundle over $S^2=\Bbb CP^1$. Consider $p:\Bbb CP^2-[0:0:1] \rightarrow \{z=0\} $ with $p([x:y:z]) = [x:y:0]$. In words, $p([x:y:z])$ is the unique point on $\{z=0\}$ intersecting the line ...


8

The fundamental group of a (second-countable, Hausdorff) manifold is countable. See Lee, Smooth manifolds, proposition 1.16. The fundamental group of an orientable noncompact surface (in particular an open subset of $\mathbb R^2$) is free. See here. You can realize rather easily a free group on $n$ generators or $\mathbb N$ generators as the fundamental ...


4

Observe that: $F(S^m,2)/\mathbb{Z}_2$ is the space of all unordered pairs of points on $S^m$. $\mathbb{R}P^m$ is the space of all unordered pairs of antipodal points on $S^m$. The first space deformation retracts onto the second. Specifically, given an unordered pair $\{a,b\}$ of points on $S^m$ that are not antipodal, let $C$ be the great circle ...


3

Define $i:S^m\to F(S^m,2)$ by $i(a)=(a,-a)$. This map is equivariant with respect to the $\mathbb{Z}_2$-actions on both sides, and I claim it actually realizes $S^m$ as a $\mathbb{Z}_2$-equivariant deformation retract of $F(S^m,2)$. Indeed, given a point $(a,b)\in F(S^m,2)$, we can continuously move $a$ and $b$ in opposite directions from each other along ...


2

Theorem. There are no nontrivial homomorphisms $f$ (continuous or not) from $SU(2)$ to $Homeo(S^1)$. Proof. First of all, it is known that $SO(3)=SU(2)/(\pm 1)$ is a simple group (as an abstract group); a proof of this is a very nice exercise in elementary group theory and geometry. If you cannot prove it, see for instance Berger’s book “Geometry-I”, ...


1

EDIT: turns out the region is empty. Go Figure. the object in question is a convex pentagon in a 2-plane; The plane can be described by finding a favorite point $P$ in it in $\mathbb R^5;$ then find, say, an orthonormal basis $\vec{u}, \vec{v}$ for the plane given when all three right-hand sides are changed to $0.$ Your pentagon is then given as $$ P + s ...


4

A tubular neighborhood of $S^1$ in $\mathbb{R}^2$ is an annulus (which is topologically just a cylinder). Making the one-point compactification is just collapsing the top and bottom borders of the cylinder together into a point, which is equivalently a sphere that's had two points collapsed:


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 ...


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).


2

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 ...



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