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## Hot answers tagged fundamental-groups

7

Taking the stereographic projection originating from one of the points in $S^m$, we have an homeomorphism between $S^n- S^m$ and $\mathbb{R}^n-\mathbb{R}^m$. The latter is homotopy equivalent to $\mathbb{R}^{n-m} \backslash \{0\},$ which in turn is homotopy equivalent to $S^{n-m-1}$. Hence, $S^n-S^m \stackrel{hmtp}{\simeq} S^{n-m-1}$. It follows that if $n-... 5 Not possible. Alexander duality is an obstruction. Check Corollary 3.45 Hatcher. If$X\subset R^3$, then$H_1(X)$has to be torsion-free. But$H_1(X)= \pi_1(X)= Z_2$is torsion (and non-trivial) for your case. 5 Let$F\to E\to B$be a fibration. Then we have the following long exact sequence of homotopy groups $$\ldots \to \pi_{n+1}(B)\to \pi_n(F)\to \pi_n(E)\to \pi_n(B)\to \pi_{n-1}(F)\to\ldots.$$ In particular, for the Hopf fibration$S^1\to S^3\to S^2$, we obtain $$\ldots \to \pi_{n}(S^1)\to \pi_n(S^3)\to \pi_n(S^2)\to \pi_{n-1}(S^2)\to \ldots.\ \ \ \ \ (1)$$ ... 4 Are you asking why these describe the same space? You can match them up by matching up the two dots in the bottom diagram with, say, the lower left and upper right corners in the top diagram. 4 Let's start from the begining.$S^1$is given and another, distinct$D^2$is given. The boundary$\partial D^2$of$D^2$is$S^1$as well, but since it is distinct I will denote it as$\partial D^2$. 1- I do not understand the statement: "by attaching a disk$D^2$along the boundary circle" what do the question mean by$along the boundary$? does it mean ... 3 That is correct. Note that$\pi_1(X)$has a much simpler presentation: $$\pi_1(X)=\langle \alpha,\beta,\gamma,\delta\mid\delta=\alpha\beta\gamma^{-1}\rangle=\langle\alpha,\beta,\gamma\rangle=\mathbb F_3$$ This can also be seen by noting that the original space$X$is homotopy equivalent to$S^1\vee S^1\vee S^1$(a bouquet of three circles), and applying van-... 2 The key idea is that since$X$is Hausdorff, any compact subspace of it is closed, and so$A$and$B$are both closed in$X$. This gives you a handle on the topology of$X$and not just the topologies of$A$and$B$. In particular, you can cover$X$with open sets which deformation-retract to$A$and$B$(and whose intersection is contractible), which is ... 2 The desired fundamental group (1) is$\mathbb{Z} \times F_{r-1}$and the desired space (2) is$S^1$cross a graph. Also, the sentence "If they are not linked, I can find a deformation retraction from..." is not correct. If the circles are not linked then the space deformation retracts to a one point union of a disjoint collection of circles and two-... 2 Suppose we have a closed disc$\bar{D}^2$whose boundary$\partial D^2$is attached to a circle$S^1$by a map$\gamma:\partial D^2\to S^1$that wraps$\partial D^2$in total$n$times around$S^1$. We call the resulting space$Y_n$. Let$U$and$V$be open subsets of$Y_n$defined as follows. The set$U$is given by$(U\cap D^2)\cup S^1$, where$D^2$... 2 In the figure, the space$X$is given by the green ball with the blue line on the left. Now I let$U$be the red shell and the blue line on the right, and$V$the orange shell. Observe that$U$is homotopic to$S^1$and$V$is contractible, while$U\cap V$is homotopic to$S^1$. By van Kampen's theorem,$$\pi_1(X)\cong \pi_1(U)\underset{\pi_1(U\cap V)}{*}\... 1 Unfortunately, this is not possible. I would assume that Lee Mosher misspoke there, and as is pointed out, was probably just thinking of the trefoil. To show there is no map from$\pi_1 = \pi_1(S^3 - 4_1)$to the dihedral group$D_6$, we will use the idea of$n$-coloring. A knot$n$-coloring is an assignment of$n$colors,$\{0,1, \ldots, n-1\}$to the ... 1 A base for the space is the open sets of the sphere that do not contain the poles the open sets of the diameter that do not contain the poles the open sets of the sphere that contain just one pole and an open ended line segment from the pole along the diameter. 1 First consider$V:=\{(z_0,\ldots,z_n)\in\mathbb{C}^{n+1}:z_0^2+\cdots+z_n^2=1\}$and verify that the standard projection restricted to$V$becomes a 2-sheeted covering map (actually it is sufficient to notice that$p\mid_V$is a quotient map, but in this particular case it can be easily seen by noticing that$V\$ is a cover). To be specific, let an arbitary ...

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