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I have proven the injectivity of the Hurewicz map $h \colon \pi_n(x) \to MO_*(X)$ where $X$ is (n-1)-connected for $n \geq 2$. The basic idea is that if you have a bordism $(M,g)$ between two maps $f_1,f_2\colon S^n \to X$ you can construct this bordism as a cylinder with various handles attached. And because of transversality you can also arrange that the ...

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There is of course an obvious counter-example when $N=M$ and $i$ is the identity. In this case, $w(\nu)=1$ (since $\nu$ has rank $0$), but $i^*(1 + \sigma) = 1 + \sigma$. In the case you are considering, they are likely using the fact that the pullback of the tangent bundle of $M$ splits as $TN \oplus \nu = i^* TM$. Hence, the Whitney product formula and ...

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Assume $p$ is a quotient map: As any open set $U \subset Y$ is open in $Y \iff p^{-1}(U)$ is open in $X$, we have that $p$ is continuous. For any saturated open subset, $V \subset X$, consider $p(V)$. As $V$ is saturated, we can write $$V = \cup_{y \in p(V)}p^{-1}(y)$$ That is: $$V = p^{-1}(p(V))$$ Thus as $p^{-1}(p(V))$ is open, $p(V)$ must be too, and ...

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This is true in complete generality by just checking the universal property of the colimit (which is in your case also called a coequalizer, as already mentioned by Ronnie Brown): A cocone $f$ over the diagram of your question gives a family $\{f_i\colon U_i\rightarrow Y\}_{i\in I}$ of continuous maps, such that that the restrictions $f_i|_{U_i\cap ... 2 The Mayer-Vietoris sequence goes$\cdots \rightarrow H^{n - 1}(U) \oplus H^{n - 1}(V) \rightarrow H^{n-1}(U \cap V) \rightarrow H^{n}(M) \rightarrow H^{n}(U) \oplus H^{n}(V) \rightarrow 0$. Note that since$V \cong \mathbb{R}^{n}$we have$H^{n}(V) = 0$. We also already know that$H^{n-1}(U \cap V)$and$H^{n}(M)$are both$\mathbb{R}$, so in particular they ... 2 The nice concept to use here is that of coequaliser. Thus the diagram $$\coprod_{(i,j)\in I\times I} U_i\cap U_j \rightrightarrows^a_b\coprod_{i\in I} U_i \to^c X$$ is a coequaliser in the sense that a map$f: X \to Y$is completely determined by maps$f_i: U_i \to Y, i \in I$such that$f_ia=f_ib$for all$i$. The condition of an open cover is ... 3 The Mobius band cannot be extended to a closed surface. Added: To answer the additional question, suppose that$\Sigma \subset \mathbb{R}^3$is the given orientable surface. Then there is an embedding$f : \Sigma \times [0,1] \to \mathbb{R}^3$such that$\Sigma = f(\Sigma \times 0)$. So then$\Sigma$extends to a closed surface $$f\biggl(\bigl(\Sigma ... 0 You have the right idea, but you're line of thought is not entirely correct. The map$$S^{n-1} \hookrightarrow U \setminus \{ x \} \hookrightarrow \Bbb{R}^{n} \setminus \{ x \}$$is indeed a homotopy equivalence, so the composition$$H^{n-1}_{dR}(S^{n-1}) \rightarrow H^{n-1}_{dR}(U \setminus \{ x \}) \rightarrow H^{n-1}_{dR}(\Bbb{R}^{n} \setminus \{ x \})$$... 1 Start the ordinary construction of the mobius band: gluing a rectangle R with left side of R identified to right side of R (with a flip). Next, cut the rectangle into pieces, lets say R_1,R_2,R_3. Next, reconstruct the Mobius band by doing all the identifications at one time: the right side of R_1 is glued to the left side of R_2 (with no ... 3 I like the idea of using an algebraic model of a covering map, and one which has long been available is that of a covering morphism q: H \to G of groupoids, dating back independently to C. Ehresmann and P.A. Smith, see also P.J. Higgins (1964) and his Categories and Groupoids, and the book by P. Gabriel and M. Zisman. The condition for this is that for ... 0 This actually can be done by some basic advanced calculus technique. In particular the Stokes Theorem. First by a translation and a scaling, assume that x is the origin and the unit sphere S is in U. Let$$\eta = \sum_{i=1}^n (-1)^{i-1} x^i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots dx^n,\ \ \ \ \omega = \frac{1}{r^n} \eta,$$where r = ... 1 For convenience, we may assume that the maps \phi_i are inclusions of closed subspaces X_i\subseteq Y_i. These maps induce a closed inclusion \phi:X_1\times X_2\times I\hookrightarrow Y_1\times Y_2\times I. Seeing X_1*X_2 as a quotient of X_1\times X_2\times I via the quotient map q_X identifying (x_1,x_2,0)\sim(x_1,x'_2,0) as well as ... 0 SO(3) acts not only transitively but without fixed points on the tangent circle bundle of S^2. This action is the differential of its action on S^2 by rotations. So the rotation group is diffeomorphic to the tangent circle bundle of the 2-sphere. The fiber above a point is the subgroup of rotations that fix that point. 0 Just so this has an answer: Fixing a Kähler class on a compact Riemann surface amounts to fixing the overall area, with some fine print. (Similarly, fixing a Kähler class on a manifold of higher dimension fixes the areas of 2-dimensional homology classes.) In more detail, a torus T[1] admits a unique flat Kähler metric g of unit area.[2] Let ... 3 In general it really depends on how well you understand f and its action on the cohomology of X. Here are two nice special cases where you only need to understand the action of f on a single cohomology group and then everything else is determined by the cup product. Tori Let \Gamma be a lattice in \mathbb{R}^n. (You can take \Gamma = ... 4 I really like this proof because it uses a pretty neat trick: Declare an equivalence relation on B by saying a \sim b if there exists a bijection between p^{-1}(a) and p^{-1}(b). Try to prove that the equivalence class [x] of x \in B is both open and closed. Since B is path-connected (and hence connected), this implies that [x] = B, which ... 3 f : S^1 \to S^1 be a map without fixed points. Consider the homotopy$$H(s, t) = \frac{(1-t)f(s) - ts}{||(1-t)f(s)-ts||}$$between f and the anitpodal map -\text{id}, which is well-defined since f(x) \neq x for all x. Compose this with the homotopy$$H(s, t) = e^{i\pi (1-t)} s$$between -\text{id} and the identity map \text{id}. Thus, f \sim ... 2 Let us denote the orbit space obtained from the S_n-action on X \times X \times \cdots \times X as SP^n(X). Given a continuous map f : X \to Y, SP^n induces a map SP^n(f) : SP^n(X) \to SP^n(Y). In particular, SP^n : \mathbf{Top} \to \mathbf{Top} is a functor. It can be proved that SP^n takes homotopic maps to homotopic maps, which means if ... 0 Since the question asks only for hints I refer to the following paper (available here) R. Brown, P/R. Heath, and K.H. Kamps, Groupoids and the Mayer-Vietoris sequence'', J. Pure Appl. Alg. 30 (1983) 109-129. which deals with the more general question of components and vertex groups of the pullback by a morphism f: A \to C of a groupoid ... 1 You are right about the first part. If p\omega is nullhomotopic in A, then the homotopy to the constant loop lifts to a homotopy between \omega and the constant loop. This is because of the lifting properties of covering maps. Note that a covering map p:\tilde X\to X restricts to a covering map p^{-1}(A)\to A for every subset A of X. Regarding ... -1 The Statement is not true. Für example X=\overline{X}={\mathbb R}^2, which is simply connected, p the Identity, and A=S^1. The Fundamental Group of A is nontrivial, so i_\sharp is not injective, but p^{-1}A=A is connected. By the way, the Statement as above has no meaning anyway. The Path component is path connected anyway, you must mean something ... 2 The first relation: take any vector space with subspaces C, Z\supseteq B. The natural surjection Z/B \to (Z+C)/(B+C) has kernel (Z\cap C +B)/B which is isomorphic to Z\cap C /B\cap C so the isomorphism you want follows from the first isomorphism theorem. \partial^0 is the map formed as the direct sum of the maps C_{dp}/C_{d,p-1}\to ... 0 @ Sadok Kallel: The book Topology and Groupoids has Chapter 11 on "Orbit spaces, orbit groupoids". The main result is to give circumstances under which the morphism of fundamental groupoids \pi_1(X) \to \pi_1(X/G) presents the latter as an orbit groupoid. As a consequence, 11.5.4 calculates the fundamental group of a symmetric square. 0 For the exactness at the left, you can use the long exact sequence in K theory. Then every map involving (suspensions) of i will split (why). Then the long exact sequence breaks down in short exact sequences. 3 First of all notice that 1-dimensional manifolds generically embedded in n-manifolds are unlinked for n\geq4. As you already noticed your space A := S^5 \setminus (S_1 \cup S_2 \cup S_3) is homeomorphic to \mathbb{R}^5 \setminus (\mathbb{R} \cup S_2 \cup S_3). This space is homotopically equivalent to (\mathbb{R}^5 \setminus \mathbb{R}) \vee ... 0 As you said, an exact sequence 0 \to M \to N \to 0 forces M = N, since the map M \to N has kernel \operatorname{im} (0 \to M) = 0 and image \ker (N \to 0) = N. Since all the reduced homology groups \tilde H_n(X_1) \oplus \tilde H_n(X_2) vanish, the Mayer-Vietoris sequence gives isomorphisms H_n(X) = H_{n-1}(X_1 \cap X_2) for each n. 0 The standard generators of H_1(T) are the latitudinal and longitudinal circles. Call them \alpha=\{x,y\in [0,1]^2\mid x=0\} and \beta=\{x,y\in [0,1]^2\mid y=0\} and write an element a[\alpha] + b[\beta]\in H_1(T) as (a,b). The matrix A=\left(\begin{smallmatrix} 1 & 1\\ 0 & 1 \end{smallmatrix}\right) then acts on H_1(T) by ... 0 The question is not that trivial. Though it's pretty obvious that the composition$$Y\hookrightarrow S^1\times[0,1]\sqcup Y\xrightarrow q C_f$$is injective, you still need to show that it's a homeomorphism onto its image, i.e. an embedding. To this end, let C\subseteq Y be closed. Then \bar C=f^{-1}(C)\times\{0\}\sqcup C is closed in ... 4 Related to QiaochuYuan's comment. It might seem a little tricky but maybe you can use something out of it. If you do not want to go through orbifold maybe this discussion could help you : \pi_1 and H_1 of Symmetric Product of surfaces Instead of looking topologically \mathbb{T}^n/S_n you could say it is an orbifold (i.e. almost everywhere a manifold ... 1 With field coefficients you have that cohomology is the dual of homology. By linear algebra, we get an isomorphism. Proof for my statement is the fact that Ext_{R}(H_i(M;R),R)=0 for a field R (because dualizing is exact for vector spaces), which you apply to the universal coefficient theorem. About the reasoning in the first part, you were completely ... 0 Let y\in Y_2 and let U be a neighborhood of p_2(y) which is canonical for both p_2 and p_2\alpha (where canonical means the set is open and path connected and each path component of its preimage is open and is mapped homeomorphically onto that set). Let V be the path component of p_2^{-1}(U) containing y. If p_2^{-1}(U)=\bigsqcup_i V_i, ... 0 Here, you have to show with given informations that α is a surjective open map that is locally a homeomorphism, meaning that each point in X has a neighborhood that is the same after mapping f in Y. Tell me where exactly is your problem, and I will try to help you with the informations you give me. 0 The below is the content of Leon's comments above, which give a full answer to this question. A sketch of the proof for the first equality is given in 4-Manifolds and Kirby Calculus, Gompf, Stipsicz, 1999 AMS GSM 20, page 125, Proposition 4.5.11. The last question: if we cut out the unknot S^1\times B^2 = thickened z-axis from ... 5 You can't recover a sheaf from its stalks alone. For instance, all vector bundles \mathcal{E} (of fixed rank r) on a variety X have the same stalk at any given point x \in X:$$\mathcal{E}_{X, x} \cong \mathcal{O}_{X, x}^{\oplus r}$$Stalks provide local information, whereas a sheaf encompasses not only local information but also information about ... 3 The use of Tietze transformations to prove that \langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2 is correct. There are no mistakes in that. The mistake is to claim that \langle a,b;(ab)^2\rangle is a presentation of the projective plane. In fact, the relation \langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2 together with the Classification ... 2 You more or less have the idea. Since every point of X \subseteq Y is collapsed to a point, and in Y every point of S^{n-1} = \partial D^n is mapped to X, the quotient Y/X is precisely D^n / S^{n-1}, which is S^n. More precisely, you can prove that Y/X = (X \sqcup D^n) / {\approx} where t \approx x for all t \in S^{n-1} and all x \in ... 1 \newcommand\Z{\mathbb{Z}} Let P = \mathbb{CP}^2 \times \mathbb{CP}^2 be the cartesian product and S = \mathbb{CP}^2 \vee \mathbb{CP}^2 be the wedge sum. Then the smash product is \mathbb{CP}^2 \wedge \mathbb{CP}^2 = P / S. Since S \subset P is a sub-CW-complex, \tilde{H}^n(P/S) \cong H^n(P,S). It is well known that H^*(\mathbb{CP}^2) = ... -1 A CW-complex is built inductively, with cells of dimension n only allowed to be attached in the n-th step. A cell complex is similar but cells of any dimension may be attached in each step. \exists cell complexes that are not CW-complexes. 3 One can define characteristic classes in a more general context. For any topological group G, there is a universal principal G-bundle (henceforth just G-bundle) EG \to BG. BG is called the classifying space of G, because isomorphism classes of principle G-bundles over X correspond bijectively to homotopy classes of maps f: X \to BG, the ... 1 A handy way to see this is to consider the long exact sequence ...\to\pi_1(G)\to\pi_1(G/H)\to\pi_0(H)\to\pi_0(G)\to... you find \pi_1(G)=0 and \pi_0(G)=0 because G is simply connected and \pi_0(H)=H because H is discrete. This just give \pi_1(G/H)=H. 0 Here I am trying to give an intuitive idea...Let see how far I can make it clear... let U,V be two chart which cover these two spaces, U\cup V=X and U,V is homeomorphic to open disc \mathbb{D^2}...now since X is compact so boundary of U should properly contained in V similarly boundary of V contained inside U...now if I take a point u out ... 2 You can use the formula you mention, as long as you apply it correctly. Let's see how. One consequence of that formula is that if X \rightarrow Y is a map glueing together two points of X, then e(Y)=e(X)-1, since Y has a triangulation that is the same as that for X except that two vertices in X have been identified in Y. So now let F be a ... 2 Though not as easy as homological proofs, there is a known purely analytical proof that the sphere is not contractible. It follows from a purely analytical proof of Brouwer's fixed point theorem. http://people.ucsc.edu/~lewis/Math208/hairyball.pdf Once Brouwer's fixed point theorem is proved, it follows that the sphere is not a retract of the closed ball ... 1 We show that g is homotopic to g*e, by defining H:[0,1]\times[0,1]\to X by$$H(x,t)=\left\{\begin{array}{rl}g((t+1)x)&0\leq x\leq\frac{1}{t+1}\\g(1)&\frac{1}{t+1}\leq x\leq1\end{array}\right..$$It is easy to define a similar homotopy between g and e*g. Then, the two homotopies can be joined to one homotopy. 1 The covering space of S^1 \vee S^1 corresponding to the normal subgroup N is the Cayley graph \Gamma((\Bbb Z * \Bbb Z)/N) of (\Bbb Z * \Bbb Z)/N. That this is a covering space is seen by noting that \Gamma((\Bbb Z * \Bbb Z)/N) is homeomorphic to the orbit space \Gamma(\Bbb Z* \Bbb Z)/N, and the covering map p : \Gamma(\Bbb Z* \Bbb Z)/N \to S^1 ... 1 The two techniques you mentioned in your question: "the fundamental group of a wedge sum is the free product of each fundamental group" and "getting it by Van Kampen" are the same. Let X and Y be any nice spaces (we'll specify what nice means below) and let S be their wedge sum at a common point, which we'll call p. Then the copy of X in S ... 0 ***HINTS***try to prove that this picture is the required normal covering space...now you can see that this pic is symmetry so intuitively it is a normal covering... 1 You have the following two results in the book Simplicial homotopy theory of Goerss and Jardine: Corollary III.2.7. Suppose that A is a simplicial abelian group. Then there are isomorphisms$$\pi_n(A,0) \cong H_n(NA) \cong H_n(A),$$where$H_n(A)$is the$n$th homology group of the Moore complex associated to$A$. These isomorphisms are natural ... 2 A closed orientable surface embeds into$\Bbb R^3$; a closed non-orientable surface does not (codimension 1 connected closed manifolds in$\Bbb R^n$separate it into two pieces; one is compact, with boundary the closed manifold; the boundary of an oriented manifold is oriented). Because$X \# Y$is oriented iff$X$and$Y$are,$X \# X$embeds into$\Bbb ...

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An explicit example is the tangent bundle of $\mathbb{CP}^2$. More generally, the tangent bundle to $\mathbb{CP}^{2m}$ has non-zero second Stiefel-Whitney class as $\mathbb{CP}^k$ is spin if and only if $k$ is odd. Added Later: I thought I should add a few more details. As $T\mathbb{CP}^k$ is a complex bundle, its odd Stiefel-Whitney numbers are zero, in ...

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