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6

First of all, $y \mapsto y - x$ is a homeomorphism from $\mathbb{R}^n\setminus\{x\}$ to $\mathbb{R}^n\setminus\{0\}$. Note that $\mathbb{R}^n\setminus\{0\}$ is homeomorphic to $S^{n-1}\times(0, \infty)$ via the map $y \mapsto \left(\frac{y}{\|y\|}, \|y\|\right)$. As $(0, \infty)$ is homeomorphic to $\mathbb{R}$ via the homeomorphism $z \mapsto \ln z$, we ...

5

Consider the composite map $$(D.\partial D) \rightarrow (M,M-D) \rightarrow (S^2, x_0)$$ It suffices to show that $(D.\partial D) \rightarrow (S^2, x_0)$ is not nullhomotopic. For example it induces an isomorphism in homology $H_2(D.\partial D) \rightarrow H_2(S^2, x_0)$.

4

The answer depends on what you mean by a solid genus-2 handlebody, and the trouble is that this is ambiguous and requires some interpretation. Had you asked about "the complement of the solid genus-2 handlebody then I would assume you meant this and the answer would be yes, as in the answer of @QuangHoang. But since you asked about "the complement of a ...

4

You can compute a double homotopy colimit as an iterated homotopy colimit, that is, $\mathrm{hocolim}_{I \times J} F(i,j) \cong \mathrm{hocolim}_I \mathrm{hocolim}_J F(i,j)$. For your case, take the vertical homotopy colimits first, to get the diagram $\ast \leftarrow \Sigma X \to \ast$ (recall that for any $Y$ you have $\mathrm{hocolim}(\ast \leftarrow Y ... 4 It is helpful to break the argument down into two steps. First, show that$\mathbb{R}$is homeomorphic to$\mathbb{R}_{>0}$, the latter denoting the positive real numbers. Then show that$\mathbb{R}^n \setminus \{x\}$is homeomorphic to$S^{n-1} \times \mathbb{R}_{> 0}$. For notational convenience, I will assume that$x = (0,0,\dots,0)$, as it is ... 4 I quickly threw this animation together to demonstrate that the "figure-8" immersion of the Klein bottle admits a decomposition into two Möbius bands with the same "apparent handedness," whatever that means. 4 They are only in two version if you consider the Mobius strip as being an object embedded in three-dimensional ambient space. This becomes pretty moot when you consider the Klein bottle can't be embedded into$3$-space, so instead we usually only consider the Mobius band and the Klein bottle as topological spaces, in which case they don't come with ... 3 This is a lemma due to S.Kinoshita, from his paper "Notes on Covering Transformation Groups", Proceeding of AMS, volume 19, 1968. Lemma itself is on page 422. The paper is freely available from the Jstor. Edit. Here is the detailed argument. Let$p: \tilde{X}\to X$is a covering with the group of automorphisms$G$. First of all, since$X$is Hausdorff, ... 3 For$(1)$take$V_1=\Bbb R^2\backslash X$and$V_2=U$, by the Jordan Curve Theorem$V_1$is connected. On the other hand$V_1\cup V_2=\Bbb R^2$hence$H^1(V_1\cup V_2)=0$. As$V_1$and$V_2$are connected and$H^1(V_1\cup V_2)=0$by using the exercise5.8 it follow that$V_1\cap V_2$is connected, but$V_1\cap V_2=U\backslash X$. For$(2)$use same argument. ... 3 Because$i_*$is injective, the images of the boundary maps$\partial$are zero (by exactness of the sequence given in axiom 4). The exactness of the sequence also means that$j_*$is surjective. Hence the long exact sequence given by axiom 4 breaks up into short exact sequences $$0\overset{\partial}{\to} ... 3 Let f : M \to M be a diffeomorphism, and (to be explicit) define the mapping torus to be$$T_f = M \times [0,1] \,\, / \,\, (x,1) \sim (f(x),0) $$Pick a triangulation \tau of M. Perturb f by a small isotopy so that the triangulations f(\tau) and \tau are in general position with respect to each other. It follows that there exists a ... 3 Recall that if \Delta^n = [v_0, v_1, \ldots, v_{n-1}, v_n] then for a map \sigma\colon\Delta^n\to X we define the boundary operator \partial\colon C_n(X)\to C_{n-1}(X) by$$\partial(\sigma) = \sum_{j=0}^n (-1)^{j} \sigma \mid [v_0,v_1, \ldots, \hat{v}_j,\ldots, v_{n-1},v_n]$$and if, for an arbitrary chain \alpha, we have ... 2 Draw a picture. If you know how to visualize the universal cover of S^1 \times S^1, then it will become very clear how to visualize the homotopy. The universal cover is \mathbb{R}\times \mathbb{R}. Each lift of the path a is a horizontal path from (m,n) to (m+1,n) for some integers m,n. Similarly each lift of the path b is a vertical path from ... 2 The problem is, "the same way" isn't well-defined. To specify a transition map for an SO(n) bundle is to specify an element of SO(n) to multiply by at each fiber in the intersection. In order to lift it to a Spin(n) bundle, you have to pick which of the two preimages of that element you want (at one point, and so everywhere in that intersection by ... 2 Fulton's Algebraic topology: A First Course contains quite some material on the topology of surfaces. In particular, this includes a fairly detailed computation of their fundamental groups. 2 Yes, that's correct. To see why, one can think of S^3 as \Bbb R^3 with an infinity point whose a neighborhood is the complement of B^3 in \Bbb R^3. It is left to see that B^3-T\sharp T is equivalent to the punctured T\sharp T. Note 1: it may help to work with the solid torus first. Note 2: It is somewhat obvious in the view of Morse Theory, but ... 2 The inclusions C\subset X\subset A induce maps on the fundamental groups$$\pi_1(C)\to \pi_1(X)\to \pi_1(A).$$The composition of the two maps is an isomorphism since C is a deformation retract of A. It follows that \pi_1(C)\to \pi_1(X) is injective. 2 A map S^n\to X is null-homotopic if and only if it can be extended to the entire ball B^{n+1} (you should have proved this earlier in your course). Let \tilde{i}\colon B^{n+1}\to S^n be the extension of i to the ball and let f be the inclusion of S^n\hookrightarrow B^{n+1} and consider the composition \tilde{i}\circ f. What can you say about ... 2 \newcommand{\Z}{\mathbb{Z}}Here's a proof without reduced homology (which, by the way, makes the problem much simpler). To sum up what you've written (as far as I can tell it's correct), you're left with:$$0 \to \underbrace{H_1(U) \oplus H_1(V)}_{\Z^{e-v}} \to H_1(X) \to \underbrace{H_0(U \cap V)}_{\Z^2} \xrightarrow{g} \underbrace{H_0(U) \oplus ... 2 It is unnecessary to "replace 2. with the condition that$h$is linear", because you can always recover that condition after the fact. Assuming that this condition holds for all characteristic maps of simplices of dimension$\le i-1$, given an arbitrary characteristic map$f_c \mid \Delta^i \to X^{i-1}$satisfying conditions 1 and 2, you may precompose$f_c$... 2 The$f$and$g$appearing in the proof are not the same as the$f$and$g$appearing in the axiom. We have$g\circ f\sim\mbox{Id}_{X,A}\colon (X,A)\to (X,A)$and so axiom$5$implies that$(g\circ f)_*=(\mbox{Id}_{X,A})_*$, the other axioms then give us that$(g\circ f)_*=g_*\circ f_*$by funtoriality, and$(\mbox{Id}_{X,A})_*=\mbox{Id}_{H_k(X,A)}$also by ... 2 Unless I'm being silly, the torsion subgroup is contained in the kernel of the map$H^{j+1}(X,\Bbb Z)\to H^{j+1}(X,\Bbb C)$, so you're done by exactness of the long exact sequence. (For reasonable spaces, e.g., manifolds or simplicial complexes, the sheaf cohomology and singular cohomology agree.) 1 From the first definition we have that$X=\stackrel{\circ}{X-U}\cup \stackrel{\circ}{A}$because$\overline{U}\subset \stackrel{\circ}{A}$and$\stackrel{\circ}{X-U}=X-\overline{U}$then if i put$B=X-U$i obtain that$H_n(B,B\cap A)=H_n(X-U,A-U)$where$A$is the same and$X-U=B$in this case$X=\stackrel{\circ}{A}\cup \stackrel{\circ}{B}$so the two ... 1 See the Theorem$1$and$2$here, where there is equivalent. 1 If you are careful with your deformations when you draw topological examples, you can prove it by making a very crude drawing, like the one below. The cut occurs at the vertical plane of symmetry of the bottle. It is fairly clear that the final objects are Mobius bands. 1 At the request of Mike Miller, here is the (or an) explanation :) This picture illustrates the relationship between the functions$f,g,$and$h$. The left quadrangle is$f$, middle one$g$and right one$h$. We start out at the situation$[(f \cdot g) \cdot h]t$and end up at$[f \cdot (g \cdot h)]t$- initially$f$and$g$should have their$t$-interval to ... 1 I think it doesn't, which kinda makes sense since it is a composite of a left and a right adjoint. More precisely : takes a short exact sequence of abelian groups$0 \to A \to B \to C \to 0$, then you get an exact triangle of spectra$HA \to HB \to HC$. If you rotate, or suspend once you get$\Sigma HA \to \Sigma HB \to \Sigma HC$which is also exact, but ... 1 No, it is not. Recall that by the Chern character isomorphism, rational K-theory is periodic rational cohomology, so the cup product on odd rational K-theory doesn't vanish in any situation where the cup product on odd rational cohomology doesn't vanish; the simplest example is$S^1 \times S^1$. The stability argument is irrelevant because the cup product ... 1 I may state Kim's answer more precisely. If$f(x)\not=f(-x)$and$f$is not surjective, there's a point not in$Im f$. Then the map from$S^2\to S^2$can be seen as map from$S^2\to S^2\backslash\{0\}\cong \mathbb{R}^2$. By Borsuk-Ulam there's an antipodal point$f(x)=f(-x)$, but this contradicts$f(x)\not=f(-x)$. Thus$f$should be surjective. 1 wspin's counterexample: Identify the sphere with the inscribed cube and let$A$be the front, bottom, and back faces. Let$B$be the top, left, and right faces. Then each connected component of$A\cap A^\star$is disjoint from its image under the antipodal map, and similarly for$B \cap B^\star$. Assume$X \subset A$. Then$X\subset A\cap A^\star\$, so that ...

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