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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 ... 1 Read Conway's zip proof at Conway. 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 ...

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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:

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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 ...

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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).

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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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