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Hot answers tagged algebraic-topology

10

The reason that $<$concept$>$ is important is that their are enough examples of $<$concept$>$ that justifies making the definition. So I will answer this by giving some examples. The first example that we come across is the real numbers, $(\mathbb{R}, +)$ (by extension all the $\mathbb{R}^n$). We see that the map $sub:\mathbb{R}^2\to \mathbb{R}$ ...

7

To a topologist, topological groups are interesting in their own right. The group structure actually gives us interesting topological structure, too! One interesting fact is that the fundamental group (an important topological invariant) of a topological group is Abelian, a fact that spectacularly fails to be true in general - any group can be the ...

5

Topological groups also arise out of purely algebraic situations. Here are just two of the most common examples. In number theory one considers the rings of $p$-adic numbers $\mathbb Z_p$ for primes $p$. I won't get into their construction except to note that their topology is built out of purely algebraic considerations, and that $\mathbb Z_p$ is in fact a ...

4

There are a number of different definitions of dimension, depending on context. You are correct about the definition of the dimension of a vector space. Similarly, we often define the dimension of a manifold (something like a torus, a sphere, etc.) which locally looks like $\mathbb{R}^n$ to be $n$ for the same reason. If every point "locally requires $n$ ...

3

Converted from comment to answer: Topological groups are used to study continuous symmetries, and are used very frequently in modern theoretical physics. Local symmetries form the basis for gauge theories - QED is an abelian gauge theory with the symmetry group $U(1)$, quantum chromodynamics, is a gauge theory with the action of the $SU(3)$ group, and the ...

3

$\DeclareMathOperator{\cls}{cls}$Let $\epsilon : C_0(X) \to \mathbb{Z}$ be defined by $\epsilon(\sum m_i x_i) = \sum m_i$. This is clearly a homomorphism, and it vanishes on boundaries (easy check), so it yields $\epsilon : H_0(X) \to \mathbb{Z}$. Suppose that $\cls(\gamma)$ is a generator of $H_0(X)$ and let $n = \epsilon(\cls(\gamma))$. Let $x \in X$ be ...

2

In order to prove that $\tilde f^*(E)$ is isomorphic to $S^1 \times \mathbb{R}$ you need to construct a map $\tilde f' : S^1 \times \mathbb{R} \to E$ having the property that its restriction to the fiber $(S^1 \times \mathbb{R})_z$ is a linear isomorphism between $(S^1 \times \mathbb{R})_z$ and $E_{z^3}$. For this purpose you cannot use the exact same ...

2

Am I right in saying that any two lifts of $A$ to an endomorphism of $F_2$ differ by some inner automorphism of $F_2$? No, using a free basis $F_2 = \langle a,b \rangle$, the map given by $f(a)=a$, $f(b) = a b^3 a^3 b^{-1} a^{-4} b^{-1}$ is not an inner automorphism but it induces the identity on $\mathbb{Z}^2$, as does the identity automorphism ...

2

You can indeed use other objects instead of a closed interval. The important thing to make things work is when considering a homotopy between two morphisms $f,g:X\to Y$ is to replace $Y$ by a cylinder object for it. A cylinder object for $Y$ is an a space (and more generally the situation can be considered in any Quillen model category, so in fact the ...

2

I think the answer is yes: if $X$ is a connected finite 2-dimensional CW-complex with $\pi_1(X)$ free, then $X$ is homotopy equivalent to a wedge of 1-spheres and 2-spheres. This is stated in the paragraph spanning the first and second pages of Trees of Homotopy Types of Two-Dimensional CW-Complexes by Dyer and Sieradski, available here: ...

2

Consider any subset $E_m$ of your abelian group of size $m$. Then since the group is abelian, $|E_m^n|$ is no more than the number of ways you can choose $m$ non-negative integers to sum up to $n$, which is${{n+m-1} \choose {m-1}}$ which is no more than than $(n+m-1)^{m-1}$. Since the exponent is a constant, you will get $\lim_{n \to \infty}(n+m-1)^{(m-1)/n} ... 2 Do you know Weibel's note on the history of Homological Algebra? See page 17: The familiar "five-lemma" occurs for the first time on p. 16 of [ES]. (Its companion, the "snake lemma", first appeared in [CE].) Nobody who looked at Cartan-Eilenberg book has been disappointed. It's an evergreen, no matter if it is 60 years old! :) [ES]: S. Eilenberg and ... 1 Hint: The map $$\psi:\quad {\mathbb C}\to{\mathbb C}^*/G,\qquad w\mapsto e^w/_\sim$$ is a covering of your Riemann surface$R:={\mathbb C}^*/G$. Two points$w$,$w'$map onto the same point of$R$iff they are equivalent modulo the lattice generated by$\omega_1:=\log 2$,$\>\omega_2:=2\pi i$. 1 You should convince yourself with a drawing that the Klein bottle$K$minus a point (not your base point) deformation retracts onto the wedge of circles$S^1\vee S^1$generated by the two loops$\alpha$and$\beta$. If$\gamma$is a loop in$K$, it is based-homotopic to a loop that misses a point$\mathrm{pt}$(necessarily other than the base point). Use the ... 1 There is a way around using Van Kampen. In a sense it is a dirty cheat, since the core idea of the argument is essentially the same as the proof of the Van Kampen theorem. The idea is to use the supplied hint. By Sard's theorem, you can apply a small homotopy leaving the base-point fixed to any map$S^1 \to K^2$, ensuring the new map misses some point in ... 1 I'm assuming the ordinary Euclidean topology. First note that the boundary of$D^2$is (homeomorphic to)$S^1$(not$S^2$, mind you). If a point belongs to the interior of$D^2 $(which you can embed in$\mathbb R^2 $as the unitary disc) you may as well assume it is the origin; build up the map$ H : [0,1] \times D^2 \to D^2 $sending $$(t,x) \mapsto ... 1 I don't if I am right, but I will give it a try. This is what I know- (1) \displaystyle f_{*} (<\alpha>) = <f \circ \alpha> where \alpha is a loop based at 1 in S^1. (2) The exponential mapping \pi: \mathbb{R} \rightarrow S^1, x \mapsto e^{2\pi ix} and the path p_n(s) = ns, 0\leq s\leq1, n \in \mathbb{Z}, joining 0 to n in ... 1 Let U = S^n\setminus A and V = S^n\setminus B then by hypothesis U and V cover S^n so we can apply Mayer-Vietoris:$$ H_1(S^n) \overset{\partial_*}\longrightarrow H_0(U\cap V) \overset{(i_*,j_*)}{\longrightarrow} H_0(U)\oplus H_0(V) \overset{g}{\longrightarrow} H_0(S^n) \longrightarrow 0 $$If we assume that n\geq 2 then H_1(S^n) = \{0\}, ... 1 Note that the homomorphisms of \mathbb Z are all of the form n\mapsto kn for some k\in\mathbb Z, and are completely determined by the fact that 1 maps to k. You want to show that each of these can be realized as as the induced homomorphism of a map \varphi_k:S^1\to S^1. Intuitively, the isomorphism \pi_1(S^1)\to \mathbb Z counts how many times ... 1 They are just saying that if A=\bigcup_nA_n, then \ell^2(A)\simeq\bigoplus \ell^2(A_n). The direct sum has to be understood as a direct sum of Hilbert spaces, i.e. an \ell^2-sum. The left regular representation of \Lambda leaves each direct summand invariant. So we can think of it acting that way on \ell^2(\Gamma) (as a block-diagonal operator). ... 1 There is another way of putting this question. The cyclic group C_2 of order 2 acts on the circle S^1 by conjugation z \mapsto \bar{z}. The fundamental group of the circle at 1 is \mathbb Z and the induced action on \mathbb Z is n \mapsto -n; the quotient of \mathbb Z by this action is cyclic of order 2, which is the wrong answer, as ... 1 You could use the fact that a covering map with finitefibers is a perfect map, i.e. a surjective and closed map with compact fibers. Closedness of p is shown as follows: Let C\subset E be closed, b\notin p[C]. Then p^{-1}(b)=\{b_1,...,b_n\} is a finite subset of E-C. There is an evenly covered neighborhood U of b whose preimage is the union of ... 1 Think at \mathbb{R}P^2 as the quotient of the 2-sphere by the antipodal relation. Then think at a closed band around the equator (precisely, as if you think at the band between the tropics :)), under the previous identification this band gives rise to a Möbius Band, right? Then the complement in the 2-sphere are two open disks ("shortened hemispheres"), ... 1 In general, if two maps are homotopic, then their induced maps conincide: For a map f:(X,x_0)\rightarrow (Y,y_0) (that is a continuous map X\rightarrow Y with f(x_0)=y_0) we denote its induced map \pi_1(X,x_0)\rightarrow \pi_1(Y,y_0) by f_*. Claim: If f,g:(X,x_0)\rightarrow (Y,y_0) are homotopic relative to \{x_0\}, then f_*=g_*. ... 1 There are two finite 2-dimensional complexes A, B which are not homotopy-equivalent but are homology-equivalent, i.e., there exists a continuous map$$ f: A\to B$$inducing isomorphism of fundamental groups and homology groups. See the last paragraph (page 522) of this paper, the actual example is due to Dunwoody. Edit 1: The homology equivalence part ... 1 Your proof works fine, but you should say that$p\psi$is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy,$ψ$is path homotopic to a constant map, too, which implies$a=b$. With little more effort, we can show that if a loop$\phi$at$b_0$in$B$is in the image$p_*(\pi_1(E,e_0))\$ (which is ...

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