# Tag Info

## New answers tagged algebraic-topology

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If $E=\{1,2\}$, $B=\{3\}$ and $p(x)=3$ for both $x\in E$, then $p^*\colon H^0(B) \to H^0(E)\colon\mathbb{Z}\to\mathbb{Z}^2$ is the map $n\mapsto (n,n)$ which is not surjective.

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@ vrume21: Assuming, by the tags and some other hints on your question, that the kind of network you are talking about is related to the classical definition of "graph" (non-empty collection of vertices together with a subset of the relation of the cartesian product of the vertices, meaning, the set of connected vertices), and by some definitions in ...

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When I analysed this proof in the 1960s I found it, and the notion of change of base point, could be generalised nicely and this can be found in Chapter 7 of the book Topology and Groupoids (T&G). The point is that the fundamental group $\pi_1(Y,y)$ is given by homotopy classes rel base point of maps $(S^1,1) \to (Y,y)$. So why not replace $(S^1,1)$, ...

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To elaborate on Qiaochu's comment: Suppose $H:A \times I \to A$ is a homotopy from the identity to the constant map on $x_2$. Then $\lambda=H(x_1,-)$ is a path from $x_1$ to $x_2$. I claim that $P_{0 \to x_1, 1 \to x_2}(A)$ is homotopy equivalent to $\Omega A$, the based loops at $x_1$, which is contractible. Consider the maps $$f:P_{0 \to x_1, 1 \to ... 2 With regard to intuition, imagine tying together two broom handles A,B by a rope going once round A and then once round B; this is not the same as, i.e. cannot be deformed into, going once round B and then round A. With regard to proof, I prefer to use the van Kampen theorem for the fundamental groupoid \pi_1(X,S) on a set S of base points found in ... 1 Let us assume X and Y are Hausdorff (otherwise, I'm not sure what "the one point compactification" is supposed to mean). Then X\wedge Y is compact Hausdorff, being the quotient of the compact Hausdorff space X\times Y by a closed equivalence relation (the equivalence relation is closed because it is the union of the diagonal in (X\times Y)^2 and ... 1 You're right that \iota_{\mathbb{F}_2,0} is not the multiplicative unit. However, it still is its own square. Indeed, for any space X, every element of H^0(X,\mathbb{F}_2) is its own square. Even more strongly, this holds on the chain level: any singular 0-cochain on a space with coefficients in \mathbb{F}_2 is its own square. Indeed, a ... 1 For a homotopist's proof see the comment I made above. There is a simpler proof however. Take a component X of \Omega S^1. It corresponds to some integer n because \pi_0(\Omega S^1)\cong \Bbb{Z}. That means that X is the subspace of loops going n times around the circle. By the general theory of universal covers, you can prove that X is ... 3 The whole Section 1.2 of Hatcher's book is dedicated to van Kampen's theorem, where nonabelian fundamental groups are abundant. The more-or-less canonical example of the wedge sum S^1 \vee S^1 is in particular described, and I find the explanation in the introduction of the section rather clear. You have two loops, a and b, and it's not possible to ... 0 What exactly is not working out? Let B_x be the interior of a closed ball around x, and let Z = M - B_x, and A = M - \{x\}. Then the excision isomorphism is exactly what you're asking for:$$H_i(\Bbb R^n, \Bbb R^n \setminus \{0\}) \cong H_i(B_x, B_x - \{x\}) = H_i(M-Z,A-Z) \cong H_i(M,M - \{x\}).$$(The first isomorphism is because the open ball is ... 1 How does the uct arise? Because we have the natural epimorphism H^k(M;G) \twoheadrightarrow Hom(H_k(M),G). UCT tells you that this splits, hence for G=\mathbb Z, as the kernel is torsion, you obtain for k=3 that this epimorphism is non-trivial. But as H_3(M;\mathbb Z) is a finitely generated abelian group, 0\neq Hom(H_3(M),\mathbb Z) \to Hom(H_3, ... 1 In order to avoid a long discussion, let me clarify this: the first proof is incomplete. Let \alpha is a path from f(x_0) to g(x_0), then your H is a homotopy from f to a constant map x\mapsto g(x_0). On the other hand, g\circ F is another homotopy from g to the constant map x\mapsto g(x_0). Then you use the fact that being homotopic is an ... 0 I myself am struggling too with this but I think I understand it atleast intuitively: We are looking for the incidence number: deg(p_{\sigma_{i-1}}\circ f_{\partial \sigma_i}) Let the attachment map of a point of \partial B_{\sigma_i}^{i} be the map that identifies the point with the antipodal point and maps both on the same point y on K^{(i-1)}. Now ... 1 There is in fact a number concerned with the minimal amount of open subsets to cover M which satisfy contractability in M --- the Lusternik Schnirelmann category cat(M). This gives you (at least with my definitiong of a chart)$$ cat(M)\leq \eta(M). $$There are a lot of interesting techniques presented in the literature for this, which might be ... 2 If the composition$$f\circ \phi:\mathbb{S}^n\to X_n\to Y$$is nulhomotopic, then it extends to the disk, i.e., there exists \bar f:D^{n+1}\to Y such that if i:\mathbb{S}^n\to D^{n+1} denotes the inclusion, the obvious diagram commutes: f\phi = \bar fi. Now recall that X_{n+1}=X_n \cup_\phi D^{n+1} is the pushout of the diagram \begin{array} ... 2 The answer is yes for the following reason: \omega_n is the Euler class mod 2 Now use e.g. Theorem 4.7 here which says e(\nu_N)([N]) counts number of intersections, or you argue that the Thom class of \nu_N in M is the Poincaré dual of N (follows from this exercise). Hence, by pulling back to the cohomology of N the result follows. You ... 0 Hints: If F is a field then it follows from the universal coefficient theorem (for homology and co-homology) that H^*(X;F)= Hom_{F-mods}(H_*(X;F);F). This implies if f_* is surjective then f^* is injective. 1 Perhaps try thinking about matrices. You know from linear algebra that the set of invertible n \times n matrices is a group with group operation matrix multiplication. If you write out the functions that express the entries of the product of two matrices in terms of the entries of the matrices you can see that they are continuous (since they involve just ... 1 For your second paragraph, it might be helpful to see an example of a topological space with a discontinuous group operation. The topological space is X = \mathbb{R}. I will pick a discontinuous bijection of \mathbb{R}: f(x) = x if x \ne 0,1, f(0)=1, f(1)=0. Using this bijection, I will define a group operation \oplus on X, having identity ... 1 It might be easier to think of a topological group as a topological space with a group structure rather than a group with a topology. Let X be a topological space. This means that certain subsets of X are in a family \tau\subset\mathcal{P}(X) such that the usual axioms apply. Now assuming the axiom of choice, there is a group structure on X. That ... 1 The Solution is correct but more complicated than necessary. One can just show that X and \partial D are not homotopy equivalent by Computing their fundamental group or First homology. Namely, \partial D is homeomorphic to S^1, so \pi_1=H_1=Z. But X=T^2-D^2 is homotopy equivalent to the wedge of 2 circles, so \pi_1=Z*Z and H_1=Z\oplus Z. 2 The first Chern class of the tautological bundle over P(E_p) is a generator of H^2(P(E_p)). Lets call this a=-i^{*}(x) with i:P(E_p)\rightarrow P(E) the inclusion of the fiber. The cohomology ring of P(E_p) is H^*(P(E_p))=Z[a]/a^{n} (note that P(E_p)\cong \mathbb{C}P^{n-1}). So an additive basis of the cohomology of the fiber is ... 1 The inclusion * \times X \subset S^1 \times X is a cofibration (the singleton * is a sub-CW-complex of S^1, hence * \subset S^1 is a cofibration, and by this question the product of a cofibration with an identity map is a cofibration), thus:$$H^*(S^1 \times X, * \times X) \cong \tilde{H}^*\bigl( (S^1 \times X) / (* \times X) \bigr)$$(this is a ... 1 "I mean for anyone who trusts the eye theorem is obvious." How can you trust your eye when you are unable to fully visualize general continuous curve, e.g. a fractal? If a point z_0 lies on a fractal that is a Jordan curve, then there is no way to tell in what direction the "inside" is. For this reason we must "be blind" and resort in our proof only to ... 0 [Note that you should write something like \mathbb{Z}\{x_0,x_1\} rather than \mathbb{Z}[x_0,x_1], since the latter usually means the polynomial ring over \mathbb{Z} in x_0 and x_1.] The definition of the reduced homology is not the quotient of \mathbb{Z}\{x_0,x_1\} by the subgroup generated by x_0-x_1; rather, it is the subgroup of ... 1 Yes. The condition on homotopy groups (vanishing in higher degree, isomorphism in lower degree) is satisfied for the products. And the product of two fibrations is a fibration, so all involved maps are fibrations. 4 For a simple counterexample, take X=S^1 and Z to be a point. Then \chi(X\setminus Z)=1\neq \chi(X)-\chi(Z)=-1. Morally, what's going wrong here is that you should consider \chi(X\setminus Z) to be -1 rather than 1, since it is made up of only a single (open) 1-cell and no other cells. But if you take the standard homological definition of ... 4 This is most probably the pullback of the covering map \mathbb{R} \to S^1. In a more general context, suppose that p : E \to B is some covering space, and f : X \to B is any continuous map. Then the linked Q&A shows that$$f^*E := E \times_B X = \{ (e,x) \in E \times X \mid p(e) = f(x) \}$$is a covering space via the projection q : f^* E \to X, ... 1 It seems to me that you have all the correct ideas about topology. Topology works best as a mixture of intuition and precision. The fuzzy intuitive dreaming is fun, but it needs to interact correctly with the precise and rigid part. For instance, dreamy intuition can be a helpful step in formulating a precise mathematical description of what is true and what ... 2 It sounds like the question arises from two points of confusion. 1) If you think of * as an element of X_0, then you need to remember that an element of X_0 gives rise, via the degeneracy maps, to an element in each X_n. (Play around with the simplicial identities to see why you get a well-defined element in X_n, regardless of which degeneracy maps ... 2 Personally, the way I think of this is to look at the following decomposition of \mathbb{C}^2 into two sets, and see how they fit together into the quotient. Let U = \{(x, y) \in \mathbb{C}^2 \mid y \neq 0\} and let P = \{(x, 0) \in \mathbb{C}^2\}. Note that these are disjoint and their union is all of \mathbb{C}^2. When we quotient P by the ... 3 Suppose that you have a homotopy equivalence f : X \to Y with homotopy inverse g : Y \to X, where Y is some discrete space (of any cardinality, finite or infinite). The space X is compact, thus f(X) \subset Y is compact; but the only compact subspaces of a discrete space are finite, so f(X) is finite. Since f and g are inverse homotopy ... 1 You know, since \alpha is a cocycle, that \alpha \circ d^A_n = 0. However, you do not know that \hat\alpha \circ d^X_n = 0! So \hat\alpha is not even a cocycle. What you know is that, if i_* : C_*(A) \to C_*(X) is the inclusion, then$$\hat\alpha \circ d^X_n \circ i_{n+1} = \hat\alpha \circ i_n \circ d^A_n = \alpha \circ d^A_n = 0,$$and so by ... 3 Take C_*=0\to \mathbb{Z}\stackrel{2}\to\mathbb{Z}\to 0 (with the \mathbb{Z}s in degree 0 and 1 and every other term in the complex 0) and let D_*=0\to\mathbb{Z}/2\mathbb{Z}\to 0 (with the \mathbb{Z}/2\mathbb{Z} in degree 0 and every other term in the complex 0). Then there is a unique nonzero chain map f:C_*\to D_*, and this induces an ... 1 If the point (r(t)x, t) is to lie on S^{n+1}, then r(t)^2 + t^2 = 1, so, since r(t) is non-negative, one must have r(t) = \sqrt{1 - t^2}. But this function does not satisfy r(0) = 0. The way to fix this is consider I = [-1,1] instead, and carry the proof as you did, just changing 0 and 1 to -1 and 1, respectively. Of course, the ... 3 If it is not surjective, it factors over the sphere minus a point, which is contractible. Consequently, the map is null homotopic and hence has degree zero. 0 Just so this question is answered, the surface you are describing looks like this: It is referred to by the following names: double torus, 2-holed torus, \mathbb{T}^2 \# \mathbb{T}^2, or "a sphere with two handles". 3 If you think of a torus by taking a cylinder and identifying its ends, then it is clear that cutting along this circle gives you a cylinder (a Möbius strip with zero half twists). 1 We claim for any sphere S^n, there is a homotopy between the identity and the double reflection$$(x_0, x_1, x_2, \dots, x_n) \mapsto (-x_0, -x_1, x_2, \dots, x_n).$$To see this, define$$\ell: S^n \to \mathbb{R},\text{ }r(x_0, \dots, x_n) = \sqrt{x_0^2 + x_1^2},$$and$$\theta:S^n \to [0, 2\pi),\text{ }\theta(x_0, \dots, x_n) = \text{Arg}(x_0 + ...

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The mapping cone of $f$ is the quotient of $S^1\times I\cup S^1$ by $(x,0)\simeq (x',0)$ and $(x,1)\simeq f(x)$. The quotient of $S^1\times I$ by $(x,0)\simeq (x',0)$ is $D^2$ and to obtain the mapping cone here, you make the quotient of $D^2$ by $e^{it}\rightarrow e^{2it}$. The fibres of this map are the antipodal points. So the resulting quotient is ...

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You can calculate explicitly the intersection between the line defined by $x$, $f(x)$ and the border of the ball. The line in parametric form is $$t\longmapsto f(x)+(x-f(x))t.$$ The condition "the point is in the border" is $$\|x+(f(x)-x)t\| = 1$$ (or any other radius $R$).

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Start with the double torus case. We have the following octagon: Start by cutting it along the diagonal depicted in the picture by a red dotted line. You'll get two squares with an "edge", i.e., the red line. Once one identifies both of those pieces together one gets two punctured torii: that is, $\Bbb T^2$ with a small disk removed from each. The ...

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Let $\alpha$ be an $n$-cell of $X$ and $\beta$ be an $n$-cell of Y. Then $f(\alpha)=\sum_{\beta\in J_n'} y_{\alpha \beta} \beta$. Here $J_n'$ denotes the set of $n$-cells of $Y$. We wish to determine the values of $y_{\alpha \beta}\in \mathbb Z$. Let $\varphi_\alpha:S^{n-1}\to X^{n-1}$ be the attaching map of $\alpha$. Let $\overline{f}:X^n/X^{n-1}\to ... 1 Restricting$x$to the unit sphere and projecting all the vectors$f_i(x)$onto the orthogonal complement of$x$, this is equivalent to asking whether the sphere$S^{n-1}$is parallelizable (since the$f_i(x)$will be a basis for the tangent space at$x$for each$x\in S^{n-1}$). By a famous deep theorem (see, for instance, section 2.3 of these notes by ... 3 It seems you're having trouble in the definition of quotient spaces, not in the definition of CW-complexes. If$X$is a topological space,$\sim$an equivalence relation on$X$,$X/\sim$is defined to be the space with underlying set consisting of the$\sim$-equivalence classes of$X$and topology coming from the final topology on the quotient map$q : X \to ...

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Hint: $G \to Aut(S^n) \to \{1,-1\}$.Compose!

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In higher dimensions this is very difficult and, I think, there is no simple characterization along the lines of the ones in low dimensions. The main difficulty comes from characterization of compact manifolds among, say, compact metrizable topological spaces. The characterizations of spheres in dimensions 1 and 2 are (implicitly) based on the fact that in ...

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This is not true for every affine subvariety since a point of $C^2$ is a compact subset. But it is true if the affine subvariety is defined by a function $f$. A compact subset of $C^2$ is bounded. Suppose that the affine variety $V$ defined by $f\in K[x,y]$ is compact, there exists $M$ such that for every $x\in V$, $\|x\|<M/2$. But $f(M,y)$ has a ...

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In 2-dimensions you might want the Kline sphere characterization theorem.

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Ok it seems I've found what I was looking for. It is a consequence of the axioms of a multiplicative cohomology theory. The question I've written may seem a little confused, but I think there is a splendid explanation on Tammo tom Dieck's Algebraic Topology at page 413, Proposition 17.3.1 clearly, $\sigma$ is the suspension isomorphism he defined ...

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