# Tag Info

6

Your idea is OK, using homotopy groups. Suppose $\mathbb{Z} = G \times G$ for some (Abelian) group $G$, which must be infinite. Then note that $\{0\} \times G$ and $G \times \{0\}$ are infinite subgroups of $G \times G$ that intersect only in the unit element $\{(0,0)\}$. By the isomorphism that supposedly exists, such subgroups should also exist in the ...

6

Some points. 1) An oriented manifold $M$ which occurs as the boundary of an oriented manifold $W$ necessarily has $\chi(M)$ even. This means that not only are all of your surfaces not going to work, but neither is anything that's the boundary of a manifold. 2) Every odd-dimensional manifold has $\chi(M) = 0$ (even non-oriented ones). So you can't try that. ...

5

By the classification of surfaces, every closed oriented surface is a connected sum of tori, so you won't be able to find a $2$-dimensional example. By Poincare duality, any closed odd-dimensional manifold has Euler characteristic $0$, so the smallest dimension where you can find an example is $4$. To find a $4$-dimensional example, you can start with ...

4

We can use the same approach that works for the covering $\Bbb R \to \Bbb S^1$. By definition, a deck transformation $\phi : \Bbb S^1 \to \Bbb S^1$ must preserve the covering map $\pi : z \mapsto z^n$, that is $\pi = \pi \circ \phi$. Substituting gives $$z^n = \phi(z)^n ,$$ and rearranging gives $(z^{-1} \phi(z))^n = 1$, and hence for all $n$ $$\phi(z) = ... 4 I think that when he says f is continuous from E to B. he intends to mean that the important part which was justified was: (..) to B. and assumes continuity is a trivial matter (which is, since it is division by \frac{1}{1+\Vert x \Vert }. Since the norm is continuous, multiplication by scalar is continuous, inversion is continuous in ... 4 Two R-modules M and N are said to be "stably isomorphic" (see Stacks for example) if M \oplus R^n and N \oplus R^n are isomorphic for some n \in \mathbb{N}. It is possible for two non-isomorphic modules to be stably isomorphic. If M' and N' are such modules, let n be the smallest (necessarily positive) integer such that M' \oplus R^n \cong ... 4 There's a little abuse of notation. First, M has to be compact for all this to make sense; it's probably written somewhere before in the paper. When M is compact, orientable, and connected, H^{2d}(M;\mathbb{R}) is isomorphic to \mathbb{R}, and the choice of an orientation basically amounts to the choice of an isomorphism H^{2d}(M;\mathbb{R}) \cong ... 3 EDIT: Sigh... the torus is compact, but T\backslash \{x_0\} isn't. \blacksquare Nevertheless, I'm gonna keep the answer below. If A \subset B is a retract of B, then the induced map (by inclusion) on the fundamental group is injective. This is seen readily from functoriality. Note that the torus with a point removed has the free abelian group ... 3 As an alternative to Travis' nice answer (+1), you can also note that firstly all rotations by multiples of 2\pi/n are Deck transformations, and that secondly they together act transitively on each fiber. Since a Deck transformation is determined by its action on a single point, it follows that they are already all. 2 The long exact sequence of homology (which indeed comes from the snake lemma) reads$$H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})\to H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$$where the maps H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z}) and H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z}) are given by multiplication by n ... 2 So f\circ g \equiv Id_Y i.e H:Y\times I\to Y is a continuous map and H(y,0)=f\circ g (y) and H(y,1)=y. Img(f\circ g) \subset U (lets say). Then consider the path \gamma (t)=H(v,t). This connect U with a point in V. But U,V are disjoint open sets so they cannot be connected by a path. (Contradiction). 2 Every reasonable path connected subspace of \mathbb{R}^2 has fundamental group a free group, which is either trivial or infinite. "Reasonable" means homotopy equivalent to a tubular neighborhood of it in \mathbb{R}^2, and in particular means homotopy equivalent to an open subset of \mathbb{R}^2. Now, an open subset of \mathbb{R}^2 is a noncompact ... 2 Your question was answered in affirmative by George Lowther here: The Fundamental group of every subset of \mathbb{R^2} is torsion free? More precisely, he proves that if X is a subset of any closed surface S, for instance, S=S^2, and x\in X, then \pi_1(X,x) is torsion free. Note that no assumption is made about the nature of the subset X, ... 2 To give an answer along the lines suggested by OP, one needs only remark that for a field F we have a vector space isomorphism$$Hom_F (H_n(X;F),F)\cong H^n(X;F)$$Now H_n(X;F) is a vector space over F and if Hom_F (H_n(X;F),F)=0 then H_n(X;F)=0. 2 By simple computations, one checks AB=BA, AC=CA, CB=BCA is a presentation. So in the abelianization A=1, BC=CB the abelianized group is \bf Z^2. Note that dy, dz are invariant by G and form a base for the de Rham co-homology 2 For (a) you want to prove that every y\in Y has an evenly covered neighborhood in Y. Let's start with what we are given. We are given a continuous map f:Y\to X. So let x=f(y). Now we are also given a covering map p:T\to X. So this means x has an evenly covered neighborhood, call it U and let p^{-1}(U)=\bigsqcup_\alpha U_\alpha. Let ... 2 Here are some details on how to get the fibration in Qiaochu Yuan's answer:\newcommand{\Z}{\mathbb{Z}} Let H \curvearrowright E_H act freely on a weakly contractible space E_H, and let the exact sequence of groups 1 \to G \to G \rtimes H \xrightarrow{\phi} H \to 1 be given. Associated to the map \phi : G \rtimes H \to H, there is an action G ... 1$$ \require{AMScd} \begin{CD} I\times X @>\pi_\sim>>(I\times X)/{\sim} \\ @V\exp\times 1_X VV @VVh V \\ S^1 \times X @>>\pi_\wedge> \operatorname{S}X \end{CD} $$The map \exp is a perfect map, that is a closed map whose fibers are compact. For any space X, the map \exp\times 1_X is therefore a closed map. Now \pi_\sim identifies ... 1 I take "relative to A" to mean equivalence classes in \pi_1(X, A, x_0), that is paths \gamma: I \to X with \gamma(0) \in A and \gamma(1) = x_0 up to homotopy through other paths of the same form. The result then isn't too difficult to see. Let i: \pi_1(X, x_0) \to \pi_1(X, A, x_0) denote the inclusion. In one direction, if [i(\gamma_1)] = ... 1 The zero locus of a monic (in x) polynomial P(x,y)=0 in C^2 is always path-connected. The multiset of roots varies continuously in y for any suitable distance between multisets, such as edit distance based on a metric in C. If you want to trace the path of a particular x as a function of y this is not hard to extract from a given path of all ... 1 In general if a topological group G acts on a topological space X then each g\in G gives a homeomorphism \theta_g:X\to X defined by \theta_g(x)=g\cdot x. Let p:X'\to X be a covering map. We say that the action of G lifts to X' and is compatible with the action on X if there exists a map G\times X'\to X' such that the following diagram ... 1 So we can actually build up the answer inductively here: let's label the simplices of \Delta_2 as v_0, v_1, v_2, and the corresponding vertices of the top and bottom faces of \Delta_2 \times I as t_0, t_1, t_2, b_0, b_1, b_2. So let's define P on C_0(\Delta_2). Here \partial P + P\partial = \partial P (as \partial vanishes on 0-simplices) ... 1 Question 1 has been answered in principle and 2. is quite a nice question! One way of doing it is to realize S^2 as the union of two hemispheres E^2_+, E^2_- with intersection S^1. Then give S^1 the structure of a polygon with a set X of n vertices. Now each E^2_+. E^2_- can be given a cell structure with S^1 as the 1-skeleton, and one ... 1 Consider X=S^1\vee S^3 and its double cover X_2 i.e, attach two copy of S^3 one in north pole and one in south pole of S^1. Then \pi_1(X) =\mathbb{Z} = \pi_1(X_2). And covering map induced isomorphism in \pi_n for all n\geq 2. But they are not homotopically equivalent/ they have different homology groups since their Eular Characteristics are ... 1 Under certain conditions they do actually mean the existence of neighborhoods, containing the common point x_0 of the wedge, which do retract on x_0. With this condition we are able to choose suitable open path-connected sets A_i to compute the desired fundamental group. For instance if you take the n wedge of circles S^1 at x_0 you can ... 1 Suppose X=A_0\cup A_1 and we have x_0\in A_0\cap A_1. Now the Siefert-VanKampen Theorem states$$\pi_1(X,x_0)\cong\pi_1(A_0,x_0)\ast_{\pi_1(A_0,x_0)\cap\pi_1(A_1,x_0)}\pi_1(A_1,x_0) is an amalgam provided several assumptions are satisfied: 1) We must have that $A_0$ and $A_1$ are open in $X$. 2) We must have that $A_0$ and $A_1$ and $A_0\cap A_1$ ...

1

As homology with $\mathbb Z$ coefficients consists of abelian groups, you can think of $\mathbb Q$-homology as measuring the free part and $\mathbb Z/p$-homology as measuring the $p$-torsion part. Hence we have: $X$ is an integer homology point iff $X$ is a rational homology point and a $\mathbb Z/p$-homology point forall $p$. Next note that we always have ...

1

The reason that being $n$-simple is not enough is that, not only do we need to have a canonical identification of $\pi_n(F,*)$ for every choise of basepoint (which is where the $n$-simplicity condition shows up), we also need a canonical identification of the homotopy groups of different fibers $\pi_n(F,*) \cong \pi_n(F',*)$. Put another way, there's an ...

1

This is just the pasting lemma again: you are pasting two continuous maps, one on $\Omega Y\times \{(s,t)\in I\times I:0\leq s\leq\frac{t+1}{2}\}$ and one on $\Omega Y\times\{(s,t)\in I\times I:\frac{t+1}{2}\leq s\leq 1\}$. These are closed subsets of $\Omega Y\times I\times I$ whose union is the whole space, and the continuous maps agree on the ...

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