# Tag Info

10

Any group (including $\mathbb{Q}$ and $\mathbb{R}$) can be realized as the fundamental group of some space - in fact, for every group $G$, there is a two-dimensional CW complex with fundamental group $G$. See Proposition 1.28 of Hatcher here.

9

For any connected manifold $M$, there is a homomorphism $\pi_1(M)\to\mathbb{Z}/2$ which sends a loop to $0$ if going around the loop preserves orientation and sends the loop to $1$ if going around the loop reverses orientation. This homomorphism is trivial iff $M$ is orientable. Since $\mathbb{Z}/2$ is abelian, this homomorphism factors through the ...

7

The very first thing we need is that $H_*(pt) = \Bbb Z$ in degree zero and is zero elsewhere. Your definition, as stated, does not have this. Clearly $C_*(pt) = \Bbb Z$ in all degrees, but the boundary map is always identically zero (as opposed to the simplicial case, where the boundary map alternates between being the identity and being zero). So the ...

6

Following Mike's nice hint, note that $\pi_{1}(\mathbb{R}\mathbb{P}^{3}) = \mathbb{Z}/2\mathbb{Z}$, and $\pi_{1}(S^{2} \times S^{1}) \cong \pi_{1}(S^{2}) \times \pi_{1}(S^{1}) \cong \mathbb{Z}$, so the induced map $\pi_{1}(\mathbb{R}\mathbb{P}^{3}) \to \pi_{1}(S^{2} \times S^{1})$ must be the zero map. Recalling that $p \colon S^{2} \times \mathbb{R} \to S^{... 5 The cohomology ring of$\Bbb{CP}^n$is$\Bbb Z[x]/(x^{n+1})$, with$|x|=2$. Any map$f: \Bbb{CP}^n \to S^2$induces zero on cohomology when$n>1$, because if$z \in H^2(S^2)$is a generator, then$0 = f^*(z^2) = f^*(z)^2$, so$f^*(z)$must be zero. Now use the fact that the universal coefficient theorem is natural to see that the induced map on$H_2$is ... 5 You trivialize$E\times_B E\to E$using exactly the nullhomotopy of$f\circ p:E\to BG$. For this you need to know that principal$G$-bundles over$X$, up to isomorphism, are in bijection with maps$X\to BG$, up to homotopy, and that given$f:X\to BG$classifying$q:Z\to X$and a map$g:Y\to X,f\circ g$classifies$Z\times_X Y\to Y$. The first fact is proven ... 4 It's not always possible to visualize things perfectly! But in this case, the spaces involved are quite reasonable. A useful fact here is that a CW pair$(X,A)$yields a quotient space$X/A$which inherits a cell structure from$X$(p.8 in Hatcher). The cells of$X/A$are the cells of$X -A$plus one new$0$-cell, the image of$A$in$X/A$. You get the ... 4 CW-complexes are a generalization of the much more readily visualized simplicial complexes: https://en.wikipedia.org/wiki/Simplicial_complex. Besides purely topological research topics involving these complexes (e.g., homotopy theory), simplicial complexes are used in many other ways. One way not mentioned in the above Wikipedia article is the Finite ... 4 Let$X=[0,1]$and$A=\{0,1,1/2,1/3,\cdots\}$. Then the quotient$X/A$is homeomorphic to the Hawaiian earring which has uncountable$H_1$(try to prove this). On the other hand,$H_1(X,A)$is isomorphic to$\bigoplus_{i=1}^\infty \Bbb Z$, which is countable. 3$X=Y=D^2$(closed 2-disc). Since$D^2$is contractible, any map$f:D^2\to D^2$is homotopic to constant map. Now consider$g_1:D^2\to D^2$be a constant map and consider a surjective map$g:D^2\to S^1$and define$g_2= i\circ g:D^2\to D^2$where$i$define the inclusion map of$S^1$into$D^2$. Then image of$g_1$is a point and image of$g_2$is circle which ... 3 Endow$Q$with the discrete topology$Q_d$and consider the classifying space corresponding to$Q_d$. Same construction with$R$. 3 Pick any non-identity involution$i$of$\Bbb{RP}^2$. It has two lifts to a map$S^2 \to S^2$; pick the one whose square is the antipodal map. Because the antipodal map has no fixed points and is order 2, your lift must have no fixed points and be order 4. Note that necessarily the involution$i$downstairs must have fixed points; there is no surface with ... 3 A quick proof using Stiefel–Whitney classes: a manifold$M$is orientable iff the first SW class$w_1(M) \in H^1(M;\mathbb{Z}/2\mathbb{Z})$is zero. But by the universal coefficient theorem, $$H^1(M;\mathbb{Z}/2\mathbb{Z}) = \operatorname{Hom}_\mathbb{Z}(H_1(M;\mathbb{Z}), \mathbb{Z}/2\mathbb{Z}) = 0.$$ Of course under the hood I don't think there's ... 2 First, note that$\pi: E \to M$is a submersion, so in particular is transverse to any submanifold$S \subset M$. So$\pi^{-1}(S)$has a canonical manifold structure. Similarly for$\pi^{-1}(U \cap S)$. Any smooth map, restricted to a submanifold, is automatically smooth, so$\phi$is smooth, as is its inverse. 2 The key theorem you need from elementary topology is that any surjective map from a compact space to a Hausdorff space is a quotient map. To apply that theorem you must set up the appropriate quotient maps. This is a purely topological problem, requiring you to guess the correct formulas for the appropriate quotient maps. You have described constructions 1 ... 2 Thinking first of$\mathbb{C}^2$, the action of the group$G = \mathbb{Z}/3\mathbb{Z}$is generated by the linear transformation $$\begin{pmatrix}0 & 1 \\ -1 & 1 \end{pmatrix} \cdot \begin{pmatrix}w \\ z \end{pmatrix} = \begin{pmatrix}z \\ -w+z \end{pmatrix}$$ Under the projection map$\mathbb{C}^2 - \{0\} \to P^1$, all the points in a single "... 2 1) As is shown here, if$M$is a compact connected orientable$n$-dimensional manifold which is a co-$H$-space, then$M$is simply connected and has the same integral homology as a$S^n$(i.e.$M$is a$\mathbb{Z}$-homology sphere). By a version of Whitehead's Theorem (see Hatcher's Algebraic Topology, Corollary$4.33$),$M$has the same homotopy groups as$...

2

It might be amusing to see how this works in the context of covering spaces. If $G$ is discrete, a $G$-bundle is specified by a representation $\rho: \pi_1(B) \to G$; this representation is determined by picking a basepoint in the total space $x \in E$, and sending a loop to the unique $g$ such that if you lift the loop to start at $x$, it ends at $gx$. This ...

2

They mean to have the locally path-connected assumption, presumably, but didn't feel the need to state it (often "space" can be taken to mean "non-terrible space"). As a counterexample in general, take the Warsaw circle $X$. This has trivial homotopy groups and homology groups (exercise), but collapsing the 'bad part' gives us a non-null map $X \to S^1 =: Y$...

2

Use the fact that $B^{n+1}\setminus\{p,q\}\cong S^n\times(0,1)$, and define your map $f\colon S^n\times(0,1)\to S^n$ to just be projection onto the first factor.

1

What Milnor calls the strong topology on the join $$X= X_1 * \cdots * X_n$$ is called the join topology in Topology and Groupoids (T&G) Section 5.7, and is the initial topology with respect to the functions he gives. More formally, this means that a function $f: Z \to X$ is continuous if and and only if its composites with all the "coordinate ...

1

It depends on what you mean by "fix." What you want to do is include the basepoint as part of what a covering map is. Basically, Hatcher means that there is a bijection between subgroups of $\pi_1(X,x_0)$ and based covering maps $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$. So if two based covering maps $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ and $q:(\tilde{Y},\... 1 For a "sufficiently nice" action of a group$G$on a simply-connected, based space$X$, we can lift a loop$\gamma:[0, 1] \to G/X$to a loop$\tilde \gamma$in$G$. The map$\gamma \to \tilde \gamma(1)$then induces a well-defined isomorphism$\pi_1(G/X) \to G$. In this particular case, "sufficiently nice" means "free and properly discontinuous," but you ... 1 Given a covering map$p: Y \to X$, the induced map$p_*: \pi_1(Y) \to \pi_1(X)$is injective, and the right cosets of$p_*(\pi_1(Y))$in$\pi_1(X)$are in bijection to the fiber$p^{-1}(x_0)$if$Y$is path-connected. The bijection is $$p_*(\pi_1(Y))a \mapsto \tilde a(1), \text{ where a is a loop at x_0 and \tilde a is its lift at \tilde x_0}$$ (... 1 This is not true. Take$X=\mathbb R, C_1=\{0\}, C_2=\{0,1\}$. Then$\text{int}(C_2)$is empty. 1 no, not necessary. "$f,g $are homotopic relative to$\{1\} $" is equivalent to "$[f]=[g]$in$\pi_1(X)$". "$f$is homotopic to$g $relative to$\emptyset$" is equivalent to "$[f]$and$[g]$lay in the same conjugacy class". it is easy to construct an example for arbitrary$X$with nonabelian$\pi_1$. 1 Yes, your proof and afterthought are correct. I'm not sure if there is a way to characterize all the orbit spaces$S^1/\mathbb{Z}_2$for various actions. Obviously, the resulting space will be compact and connected. Moreover, as a quotient of a locally compact Hausdorff space by a proper action, it will also be Hausdorff. Finally, if$\mathbb{Z}_2$acts ... 1 There is indeed a choice of homeomorphism$D^n/S^{n-1} \cong S^n$but as others have pointed out this won't affect any homology calculations as long as you consistently use this homeomorphism for each cell. However, the degree is affected by the choice of homeomorphism, so the cellular boundary formula as found in Hatcher is only right to a sign. In order ... 1 Note: This is an extension of my comment above. As I said there, I'm not very used to the definiton of manifolds as functionally structured spaces, but I think that the following works. The idea comes from the similar question found here. Any comments on possible mistakes are welcome. Answer: According to Bredon's definition 2.4, a diffeomorphism is defined ... 1 I've recently struggled with the same sort of things, so I'll try to explain how I understand it. This is all supposing you understand the technical tools such as the isomorphism$H_n(X,A) \cong H_n(X/A)\$, naturality, degree theory, and the likes. First, let me say that there are a variety of levels you can make this argument so "the right way to see it" ...

Only top voted, non community-wiki answers of a minimum length are eligible