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7

This is $\mathbb{R}P^2\#\mathbb{R}P^2\#\mathbb{R}P^2\simeq\mathbb{R}P^2\#T^2\simeq \mathbb{R}P^2\#K$, where as usual $T^2$ is the torus, $\mathbb{R}P^2$ the real projective plane and $K$ the klein bottle. I will only give an intuitive argument for this. Consider the following sketch: To see this represents your manifold, note that this is a Möbius strip (...

6

Here I assume everything has a CW structure. If $X$ is path connected then $\sum X$ is simply connected. Now if we assume that $\sum^2 X$ is contractible then $\widetilde{H_n}(\sum ^2 X) = 0$ for all $n$. Again $\widetilde{H_{n+1}}(\sum^2 X)=\widetilde{H_n}(\sum X)$. This implies $\widetilde{H_n}(X)=0$ for all $n$. Thus Whitehead theorem implies that $\sum X$...

5

They are absolutely not all $B^4$. Given a smooth 4-manifold with boundary $S^3$, there is a unique way to cap that boundary component off with a copy of $B^4$, thus giving us a smooth closed 4-manifold. (Hidden in this statement is Cerf's quite nontrivial theorem that every diffeomorphism of $S^3$ extends over the 4-ball.) If the original manifold was a ...

4

To close the question, let me write my comment as an answer. $\chi(S^2)=2$, not $3$ as can be seen by computing the alternating sum of the Betti numbers. Alternatively, one may consider either of the two usual cell decompositions of the sphere - either the "one $0$-cell, one $2$-cell" or "one $0$-cell, one $1$-cell, two $2$-cells" description of the sphere -...

4

Since $S^n$ is simply-connected for $n>1$, any such map $f$ factors through $\mathbb{R}^n$, which is the universal cover of $T^n$ and has $n$-th degree trivial homology since its contractible. It follows that the $n$-th degree induced map of $f$ on homology is trivial by functoriality of homology. The key point here is the Lifting Theorem.

3

One can do this without covering spaces using the cup product in cohomology. Namely $H^n(T^n,\Bbb Z)=\Bbb Z\langle e_1\cup e_2\dots \cup e_n\rangle$ (this is supposed to denote integer multiples of $e_1\cup e_2\dots \cup e_n$), where $e_i$ are the generators of $H^1(T^n,\Bbb Z)$. Now if $f:S^n \to T^n$ has nonzero topological degree then $f^*( e_1\cup e_2\... 3 In this example, by drawing a picture you can easily guess what$\Delta(p)$is. You can then verify your guess formally using the method in the previous question. Here's the geometric picture to draw. The set$Y$is just the union of integer coordinate lines in$\mathbb{R}^2$: horizontal lines of the form$\mathbb{R} \times m$,$m \in \mathbb{Z}$; and ... 3 Cohomology is a ring when you take untwisted coefficients in a ring. (Untwisted means the$G$action is trivial.) As OmarAntolín-Camarena points out in the comments, you can still get a ring in the twisted case if the coefficients are a$G$-ring. However, this does explain the terminology in your reference, since group cohomology with untwisted coefficients ... 3 You have the covering space$p_G:EG\rightarrow BG$, let$Sing(BG)$be the singular complex of$BG$you can lift it to$EG$and obtain a complex$Sing(EG)$on which$G$acts, remark that the (co)homology of$Sing(EG)$is trivial since$EG$is contractible. Thus, endowed with the action of$G$,$Sing(EG)$defines a free resolution of$G$. 3 Recall that$\pi_1(X\times Y)\cong \pi_1(X)\times \pi_1(Y)$via the projections. Since$S^3$is simply connected, the projection map$p:S^1\times S^3\to S^1$induces an isomorphism$p_*:\pi_1(S^1\times S^3)\to \pi_1(S^1)$. But$p\circ i$is a homeomorphism${S^1}'\to S^1$, so$i_*p_*$is an isomorphism, and hence$i_*$is an isomorphism. 3 It looks like this: A closed orientable surface of genus$2$admits a hyperbolic metric, so corresponds to some quotient of the hyperbolic plane$\mathbb{H}$. The action of the fundamental group on$\mathbb{H}$can be used to draw a Cayley graph in$\mathbb{H}$. This is what this picture is depicting, in the Poincaré disk model of$\mathbb{H}$. 3 The projection$p:(x,y)\in T\to (x,x_0)\in A$onto the first factor has the property that$p\circ i$is the identity of$A$. It follows that the composition $$H_1(A)\xrightarrow{i_*} H_1(T)\xrightarrow{p_*}H_1(A)$$ is the identity of$H_1(A)$and, in particular, an injective map. Of course, it follows that$i_*$itself is injective. 3 You are looking for lens spaces. One can draw them: (The image comes from https://plus.maths.org/content/dont-judge-black-hole-its-area-2, and there you can find an explanation of what it means) 2 Your attempt at producing the homomorphism (which I would call the holonomy homomorphism, though I guess some would call the monodromy homomorphism) is perfectly correct. If$\gamma, \eta$are two loops based at$m$and$\bar \gamma$etc their lifts starting at$\bar m$, suppose$\bar \gamma(0) = h_1m$and$\bar \eta(0) = h_2m$. Then$\gamma * \eta$(meaning ... 2 Use$(r,\theta)\mapsto (\cos\theta\sin\pi r,\sin\theta\sin\pi r,\cos\pi r)$from the unit disk with polar coordinates to the unit sphere in$\Bbb{R}^3$. Clearly this is continuous, and when$r$is 1, all theta values map to$(0,0,-1)$. Hence this is a well defined map from$D^2/S^1\to S^2$. Since cosine has a continuous inverse on$[0,\pi]$, the map $$(x,y,... 2 The k-fold dunce cap (the generalization of the 3-fold dunce cap as seen in this question) is probably the simplest example. It's easy to draw faithfully, as long as you're comfortable with the idea of identifying edges: (Be warned: there are a few different spaces often called the "dunce cap"; only one of them has fundamental group \Bbb{Z}/k.) 2 Let X be a graph. There are two types of points in X: the points e interior to edges (I'll call them edge points) and the vertices v. Let's compute the local homology at each. To do this, we'll use the long exact sequence in homology:$$ \cdots\to H_{n+1}(X,A)\to H_n(A) \to H_n(X) \to H_n(X,A) \to H_n(A) \to \cdots $$(The computations are made even ... 2 There's no strong reason not to use G(F(x,t),t). One reason the author may not have is that they may have had in mind the general principle that you can concatenate homotopies, and that this often lets you build homotopies that "do multiple things" by doing each one separately and concatenating the pieces along the time coordinate. In this particular ... 2 Here are a couple more systematic ways to see this. First, you can notice that A is a retract of T, via the map r:S^1\times S^1\to S^1\times\{x_0\} defined by r(x,y)=(x,x_0). It follows by functoriality that the homology of A is a direct summand of the homology of T, and in particular the induced map is injective. Second, you can compute ... 2 Here is a comment from the author of the book you mention. I thought it might be relevant, however too long for a comment. The argument I had in mind was induction on the number of cells of X, but not explicitly using a cofiber sequence. Suppose X is obtained from a subcomplex X' by attaching an n-cell. Given a map f : X ---> Y, induction implies that ... 2 You are forgetting about H_0. Your long exact sequence is$$ \cdots\to H_n(U-x)\to H_n(U)\to H_n(U,U-x)\to \cdots\to H_0(U-x)\to H_0(U)\to H_0(U,U-x)\to 0 $$Now H_0(U-x)\simeq\mathbb{Z}\oplus \mathbb{Z} and H_0(U)\simeq\mathbb{Z}. If you do this explicitly you will see that kernel of H_0(U-x)\to H_0(U) is isomorphic to \mathbb{Z}. On the other ... 2 If \;x\in I\cap J\; , then \;G_x\; appears as free factor in both \;G_a,\,G_2\; , so it won't be true that \;G_r\cap G_s=1\; always, as one of the possible intersections will be \;G_x\cap G_x=G_x\; . The basic assumption is that the free products considered here are non-trivial, meaning: \;G_i\neq 1\; forall the groups considered. Yes, as ... 2 Whenever you have a subspace A\subseteq X and a deformation retraction H:X\times [0,1]\to X onto A (such that H(x,0)=x and H(x,1)\in A for all x\in X and H(a,1)=a for all a\in A), you get a homotopy equivalence as follows. Let f:X\to A be given by f(x)=H(x,1) and g:A\to X be the inclusion map. Then f\circ g=Id_A, and H is a ... 2 Morphisms of G-actions are the same as morphisms of G-sets. To clarify this, I need to introduce the necessary definitions. By right G-action (on set X) we usually mean a function \cdot\,\colon X\times G\to X such that x\cdot(gh) = (x\cdot g)\cdot h and x\cdot e_G = x. Morphism between G-actions on sets X and Y are defined as functions ... 2 Hint: Apply the Serre exact sequence since the fibre F is discrete, \pi_i(F)=1, i\geq 1, thus you have isomorphism between \pi_i(P) and \pi_i(B), i\geq 2 where P is the total space and B the base space. https://en.wikipedia.org/wiki/Fibration#Long_exact_sequence_in_homotopy_groups 1 This is just a notational error. It should say$$l'(\tau_0)+l'(\tau_2) = l'(\tau_0+\tau_2) = l'(\tau_0\ast\tau_2) = l'(\tau_1).$$To show that l'(\tau_0+\tau_2)=l'(\tau_0\ast\tau_2), you just directly use the definition of l':$$l'(\tau_0+\tau_2)=[u(x)\ast\tau_0\ast u(y)^{-1}\ast u(y)\ast\tau_2\ast u(z)^{-1}]=[u(x)\ast\tau_0\ast\tau_2\ast u(z)^{-1}]=l'(\... 1 I try to answer my question. I am pleased if you guys make some corrections. A mobius band can be think of a square$[0,1]×[0,1]$identified in both ends${0}×[0,1]$and${1}×[0,1]$by$(0,x)$~$(1,1-x)$. After seeing this picture, we can intuitively shrink the band from time to time to get a circle. So there is a deformation retract$f_t:M → M, t∈I$,$M$... 1 Yep, that's exactly what it means. 1 In topological spaces, a cell complex is usually the same thing as a CW-complex. The name "cell complex" comes from the fact that there exists generalizations to other categories, but if you're interested in topological spaces then for all intents and purposes "cell complex" = "CW-complex". A finite cell complex is a cell complex that has a finite number of ... 1 Recall what is the subgroup of$\pi_1(B)$associated to the covering$p$. To this end, recall this prop: Thm 1.38 [Hatcher page 67] Let$B$be path-connected, locally path-connected, and semilocally simply-connected. Then there is a bijection between the set of basepoint-preserving isomorphism classes of path-connected covering spaces$p \colon (\tilde{B} ,...

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