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It's true that there's a natural map $BO(1)^n \to BO(n)$, which you should think of as taking the direct sum of $n$ real line bundles to get an $n$-dimensional real vector bundle, inducing a quotient map $BO(1)^n / S_n \to BO(n)$, although we need to be a bit careful about what we mean by that quotient: properly speaking it's a homotopy quotient. It is not ...

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It is easy to construct examples of isotopies on the $n$-torus $T^n = (S^1)^n$; all we have to do is work in the natural coordinate system induced by the identification of $S^1$ with the unimodular complex numbers. Indeed, representing points of $S^1$ as $e^{i\theta}$ where $0 \le \theta < 2\pi$, we see that for any real $\alpha$, the map $\phi_\alpha: ... 1 It might be useful to compare the proof given in May's book to, say, the more concrete proof of the classical van Kampen in, say, Hatcher's book. The proof given is fairly straightforward, except that it's dressed up in the language of category theory (which is not a bad thing at all, especially in a field like algebraic topology that actively uses the ... 0 You are asking to compute the map$i_2 : \mathbb{Z} = H_1(U \cap V) \to H_1(V) = \mathbb{Z}$induced by the inclusion$U \cap V \subset V$. As you say, both of the spaces$U \cap V$and$V$are homotopy equivalent to circles. You can use this to good effect if you can prove a little bit more, namely, there exist homotopy equivalences$f : U \cap V \to S^1$... 1 The words you are missing are "homotopy equivalence". What you are saying in your second paragraph is that your space is homotopy equivalent to$S^1 \vee S^1$. This is indeed true. But in a sense that's making things too complicated. A more straightforward statement with the same outcome but which is easier to prove, is that the union of the six ... 0 Here is a simple visualization of the example that Pece is talking about. The top map is a 2-fold covering but the bottom map has infinite fibers. The composition isn't locally trivial and is therefore not a covering map. However, it is still a semicovering map in the sense that it is a local homeomorphism which has the unique path lifting property. 0 My instructor informed me that the paper in question is by Prof. Robert Edwards of UCLA. I contacted him over e-mail about his paper and he was kind enough to correspond. He informed me that : ..there is no published paper, only his abstract in the AMS Abstracts of the May 1999 AMS meeting in Denton TX. He also mentioned that there is a clarification needed ... 0 It depends what the definition of "universal covering" is. If the definition is that a universal covering of$B$is a simply connected covering (as it is in Munkres), then this statement as you're writing it isn't quite true. If$B$is path-connected and simply connected but not locally path connected, then the identity function$B\to B$is the universal ... 0 Let$\gamma$be a loop whose unique lift starting at$x$ends at$x*\gamma$. The statement that$\pi_1(Y,y)$'s action on$p^{-1}(y)$is transitive means that, given an$x$, for any$x'$you pick there is a loop$\delta$such that$x*\delta=x'$. Obviously picking an element$x'$arbitrarily is the same as picking an arbitrary element from$p^{-1}(y)$. But ... 1 That notation doesn't mean much to me, either. But to try to help: I would guess that perhaps the prof has defined an action of$\pi_1(Y)$on the fiber$p^{-1}(y)$, an action denoted by "*". So this says that the path$\gamma$, acting on$x'$, gives$x$. (The definition of the action would be "lift the path to one in X that starts at$x'$, and see where it ... 3 Every closed oriented 3-manifold admits a Heegaard splitting, from which it follows that its fundamental group admits a presentation with$g$generators and$g$relations for some$g$. (To get a corresponding condition for the not-necessarily-oriented case look at the orientation double cover.) Such a presentation is called balanced, and most groups don't ... 0 Let$\pi\colon A\to B$be a universal covering of$B$. Let$b\in B$and$U\ni b$open. We want to show that$b$has a path-connected neighbourhood$U'\subseteq U$. Let$a\in A$with$\pi(a)=b$and$V$the connected component of$\pi^{-1}(U)$that contains$A$. I.e.,$V$contains all points that are reachable from$a$via a path in$\pi^{-1}(U)$. But then ... 0 In his book Knots and Links, Rolfsen shows that, written Antoine's necklace$A$as an intersection of chains of tori$\bigcap\limits_{n \geq 0} C_n$as usual, the inclusions$\mathbb{R}^3 \backslash C_n \hookrightarrow \mathbb{R}^3 \backslash C_{n+1}$are$\pi_1$-injective, so that the fundamental group $$\pi_1( \mathbb{R}^3 \backslash A) = ... 0 Yes there is. The following is a classical example (I think it is in Hatcher for example). Take H to be the hawaiian earings space, that the subspace of \mathbb C which is the union of the circle C_n\, (n\geq 1) of center \frac 1 {2n} and radius \frac 1 {2n}. Then you can get a composition of covering maps which is not a covering map as follow : ... 2 Hint. Suppose there is an homotopy H \colon Y \times [0,1] \to Y such that H(\cdot, 0) = \mathrm{id}_Y. For y\in Y, H(y,\cdot) is a path in Y with starting point y : what are then the allowed values for the ending point (that is H(y,1)) ? 0 The short answer is that \mathrm{im}\partial_1 is contained in the subgroup B of C_0 given by \{\sum \lambda_i v_i : \sum \lambda_i = 0, \lambda_i \in \mathbb{Z}, v_i \in K_0\} This subgroup is proper as soon as K_0 is non-empty: v \notin B for v \in K_0. The long story is that you can slightly change your simplicial complex by adding a ... 1 "Minus the zero section" means you remove the zero section from the total space. For example, the trivial bundle X \times \mathbb{R}^n over X, minus the zero section, is X \times (\mathbb{R}^n \setminus \{ 0 \}). 1 Let U be X with the top half of L and the north hemisphere of S^2 removed. Let V be similar but bottom and south. Use Van Kampen's theorem and the result that you already mentioned for S^2\cup L. 1 Yes, this is correct. Notice that your hom sheaf is the internal hom in the category of étale sheaves on S resp. the subcategory of finite étale schemes over S. Recall that the category of finite \pi_1(S,s)-sets is equivalent to the category of finite étale schemes over S. Using the equivalence, the problem becomes simply: If X,Y are finite ... 0 Consider the matrix I + E_{ij} where E_{ij} has a 1 in the ij slot and a 0 elsewhere. Then, for any matrix X, the product (I+E_{ij})(X) is what you get by adding row j of X to row i. For i\neq j, form a path \gamma(t) = (I+tE_{ij})X. Notice first that \det(\gamma(t)) = \det(I+tE_{ij})\det (X) = 1\det(X)\neq 0, so \gamma(t) is ... 0 An idea: "squeeze" the disk to the point \;(0,0,0)\; , so that you get two tangent spheres joined at this point and with a "stick" (the line \;\{(0,0,z)\;:\;-1\le z\le 1\}\;) as common diameter through their tangency point... 0 Either method works. They essentially the same homotopy, just different expressions: The angle between \vec{n}_t and \vec{n}_1 is the angle in the second homotopy. They fix the endpoint p because all planes H_t pass p and q. Note: You may need to pay attention to the orientation of the paths C_1 and C_2, as the above homotopy may send ... 1 Let S^0 = \{1,-1\}. Then S^0 \times I = \{1,-1\}\times I = \{1\} \times I \cup \{-1\} \times I is basically a disjoint union of two copies of the unit interval I. Hence a homotopy H between two maps f,g \colon S^0 \to X "consists" of two paths in X, one, t\mapsto H(1,t), connecting f(1) and g(1), the other, t \mapsto H(-1,t), connecting ... 2 Here is some intuition as to why D^n/S^{n-1} \simeq S^n is true. Observe that the boundary of D^n is homeomorphic to S^{n-1}, so what this relationship is expressing is that if you glue the boundary of an n-dimensional ball to a point, what you get is the n-dimensional sphere. This can be nicely visualized in the case of n=1 and n=2. In the ... 3 \pi_1 preserves arbitrary products (not just finite ones); this is easy to check. \pi_1 does not preserve coproducts in general. See math:SE/320812. Seifert van Kampen's Theorem only applies under certain assumptions. \pi_1 does typically not preserve pushouts. For example S^1 is the pushout of two open intervals which have trivial \pi_1, but ... 3 Look at the definition. U \subseteq Y is open if and only if f^{-1}(U) \subseteq X is open. If f isn't surjective (don't use the word projection), then let y \in Y not in f(X). Then f^{-1}(\{y\}) = \varnothing \subseteq X which is open. Therefore you get the quotient topology on f(X) and the discrete topology on its complement. 2 There is a fibre bundle$$O(n)\to V_k(\mathbb{R}^\infty)\to G_k(\mathbb{R}^\infty)$$where V_k(\mathbb{R}^\infty)=\cup_nV_k(\mathbb{R}^n) with V_k(\mathbb{R}^n))=\{k \text{ frames in } \mathbb{R}^n\}. In particular, the total space is contractible, so the long exact sequence in homotopy groups gives us isomorphisms$$\pi_i(O(n))\cong ... 0 This is not an answer, but I had trouble posting it as a comment. Here's a special case that already seems problematic: let$f : [0,1] \to [0,1]$be monotone increasing, fixing 0 and 1.$f \simeq \mathrm{id}_{[0,1]}$. Now if$\mathscr{O}$is a open cover of$[0,1]$, can you find a sequence of homotopies between$f$and$\mathrm{id}_{[0,1]}$such that each ... 0 I know this is an old question, but here's an answer anyway: let$\pi_1(U) = \langle a,d | a^p = d^q \rangle$and the map$f: \langle c,d | cd = dc \rangle \to \pi_1(U)$be given by$f(c) = e, f(d) = d$. The pushout of$\pi_1(U)$and$\pi_1(V)$over$\pi_1(N)$is isomorphic to the$\pi_1(X)$above. Alternatively$\pi_1(U) = \langle a,d,c | a^p = b^q, cd = ...

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For some reason, the problem computes the cohomology for $n\geq 2$, but the author starts the base case at $n=1$! It's not the "otherwise" that's the problem. This can't possibly refer to $n<2$: that's simply not how scope of variables works in any kind of mathematical writing. It must refer, instead, to $d\notin\{0,n-1\}$, since those are the two cases ...

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Showing that a potentially short exact sequence is exact without knowing what the maps are is a fruitless endeavor. If the writer did not explicitly give the maps, its likely they just meant the most obvious ones (in this case where the first map is the map induced from the inclusion and the second map is induced from $d^i$).

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I don't think your question has anything to do with topology. This is just a linear algebra question. You have inclusion map $\ker(d^i)\to A^i$. Compose it with the projection $A^i\to A^i/Im(d^{i-1})$. By definition, the kernel of the resulting map is $\ker(d^i)\cap Im(d^{i-1})$. But since $d^id^{i-1}=0$, you know that $Im(d^{i-1})\subset \ker(d^i)$, and ...

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Let X be a mapping cylinder of the obvious map between the subspaces of the real line $\{n | n \in N\} \rightarrow {0}\cup\{1/n |n \in N^+\}$. Then the obvious map $X \rightarrow ({0}\cup\{1/n |n \in N^+\})\times I$ meets the requirement. Denote $({0}\cup\{1/n |n \in N^+\})\times I$ by $Y$, and ${0}\cup\{1/n |n \in N^+\}$ by $Z$. So we have $f : X ... 2 Let me denote by$i : B^p \to B$and by$j : X^p \to X$the given inclusions, and by$q : X^p \to B^q$the base change of$p$. Start with$\alpha : S^n \to X^p$, nullhomotopic in$X$and let$s : \mathrm{Cone}(S^n) \to X$be a nullhomotopy for$\alpha$. The composition$ps : \mathrm{Cone}(S^n) \to B$is a nullhomotopy for$q \alpha$. Choose a homotopy$h : ...

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You're really close. As you say, you now know that $[\sigma^{-1}\alpha\sigma]=[\sigma^{-1}][\alpha][\sigma]=[\sigma]^{-1}[\alpha][\sigma]$ and so now you just need to let $g=[\sigma]^{-1}$ so that $g^{-1}=([\sigma]^{-1})^{-1}=[\sigma]$, hence $[\sigma^{-1}\alpha\sigma]=g[\alpha]g^{-1}$. So the automorphism $\Phi_{\sigma}$ maps $[\alpha]\mapsto ... 0 Matthew Leingang gave the answer in a comment but I'll flesh it out a little. The quotient map$q\colon G\to G/H$is a fibration with fiber topologically isomorphic to$H$, and so we get a long exact sequence in homotopy groups $$\cdots \to \pi_{k+1}(G) \to \pi_{k+1}(G/H) \stackrel{\delta_k}{\to} \pi_k(H) \stackrel{i_*}{\to} \pi_k(G) \to \pi_k(G/H) ... 1 In the article you linked there is a proof at the bottom of the page (After “To prove these relations...”), but it's obfuscated by a typo. Whenever F\stackrel{i}{\rightarrow} E \stackrel{p}{\rightarrow} B is a fibration, there is a long exact sequence in homotopy, moving downward by degree. At the end of the sequence you have$$\pi_1(G) \longrightarrow ... 0 The space$\Omega^p(U) = \Gamma(\bigwedge^p T^*U)$, and$\bigwedge^p T^*U = 0$for$p > \dim T^*U = n$. Hence all such$\Omega^p(U)$vanish. The Mayer-Vietoris sequence does eventually terminate in zero modules, but it's not a short exact sequence (in general); that term refers specifically to a three-term sequence$0 \to L \to M \to N \to 0$. 1 If$p>n$, then$H^p(U)=0$since$\Omega^p(U)=0$(for open$U\subseteq \mathbb{R}^n$). If$U\subseteq \mathbb{R}^2$is an open set, then$H^1(U)$depends on$U$(but it is always finite dimensional and its dimension counts the number of "holes" in$U$, roughly speaking). The$\dots$just signifies that the sequence extends over all integer$p$. However, ... 0 I'll use two lemmas: Lemma 1: (Theorem 22.2 in Munkres) Let$p : X \to Y$be a quotient map. Let$Z$be a space and let$g : X \to Z$be a map that is constant on each set$p^{-1}(\{y\})$, for$y \in Y$. Then$g$induces a map$f : Y \to Z$such that$f \circ p = g$. The induced map$f$is continuous if and only if$g$is continuous;$f$is a quotient map ... 1 I just consider the case$h=3$, but the argument is completely general. From the classification of closed surfaces, we know that$N_3$is the connected sum of three projective spaces, so that$N_3$be the quotient of an hexagon$P$by identifying its sides as indicated by the following figure: Now let$U, V \subset N_3$be subspaces illustrated by the ... 3$\mathbb{CP}^{\infty} \times \mathbb{RP}^{\infty}$, being a product of Eilenberg-MacLane spaces$B^2 \mathbb{Z} \times B \mathbb{Z}_2$, is in particular a loop space and hence is homotopy equivalent to a topological group$G$(in fact a topological abelian group). As I mentioned in the other thread, for any topological group$G$the free loop space splits as ... 1 (Comment-turned-answer) I agree that it must be a typo, for the example you give is exactly the example usually given to show that$\mathbb{R}^2\setminus \{0\}$is homotopy equivalent to$S^1$. Although it is possible that the question have many different corrections, the most likely is the following: Prove that there is no continuous ... 1 There is another viewpoint different from the answer of @Travis, in which there is no ambiguity up to conjugacy. Namely, it is common to put the base point into the notation for the map$f$itself, like this: $$f : (X,x_0) \to (Y,y_0)$$ In this notation, one regards$f$not just a morphism in the category of topological spaces and continuous maps, but as ... 1 No, it is not true. An example is the sphere$S^2$: it is simply connected, but its loop space is not. Take a point$x \in S^2$and the loop of loops that has the following properties: the two endpoints are two loops which have constant value$x$; at every time it is a loop based on$x$; as the time passes, it "spans" the whole sphere. More formally, if ... 5 First let's look at the simpler case of the based loop space$\Omega X$(pick a basepoint). WLOG$X$is connected since$\Omega X$only sees the connected component containing the basepoint. This case is simpler because we know that $$\pi_i(\Omega X) \cong \pi_{i+1}(X)$$ so we conclude that$\Omega X$is connected iff$X$is simply connected, and$\Omega ...

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The connections between i), ii) and iii) are that the domains of the various homotopies are closely related to each other via quotient maps. Let me explain this by proving a couple of the directions. Item ii) means that for any $f : S^1 \to X$ there exists a map $H : S^1 \times [0,1] \to X$ such that $h(x,0)=f(x)$ and $H | S^1 \times 1$ is constant. To ...

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Let $f_t$ be a homotopy with $f_0(\theta)=f(\theta)$, $f_1(\theta)=x\in X$. Let $G(r\cos\theta,r\sin\theta)=f_{1-r}(\theta)$. This is a continuous map from $D$ to $X$.

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Firstly, you have computed the $H_0$; since $H_0 \cong \mathbb{Z}$ this tells you that the space is path connected. But anyway, you don't even need that from a formal point of view. Either (a) the space is path connected, and thus you can apply Hurewicz, or (b) the space isn't path connected, so in particular it's not simply connected. In both cases, you ...

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The problem is not in the morphisms … it is in the very definition of $\pi_1$ on objects: which group do you choose? There are, in principle, infinitely many groups $\pi_1(X,x)$ with $x\in X$, all different (because the underlying sets are different) but isomorphic. The isomorphism $\pi_1(X,x)\cong \pi_1(X,y)$ is not canonical, as it depends on the choice of ...

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