# Tag Info

8

You're almost there. Try using the long exact sequence for homotopy to find $\pi_{k-1}(Fiber)$, and then use the Hurewicz theorem. I'll see you after class on Wednesday, Kevin.

8

Assume that we already know that the action is transitive. Then what is the stabilizer of the point $v=(0,...,0,1) \in S^n?$ I claim that if we let $O(n)$ be identified with the subgroup of $O(n+1)$ of all matrices of the form $$\begin{pmatrix} A & \vdots \\ & 0\\ \dots 0 & 1 \end{pmatrix}$$ where $A \in O(n)$, then $O(n)$ is the stabilizer ...

7

I claim that there do exist sequences $X\rightarrow Y\rightarrow Z$ of simply connected spaces (even CW complexes) which are both fibration and cofibration sequences. Here is my example. For an abelian group $A$ and an integer $n\geq2$ we denote by $M(A,n)$ the degree $n$ Moore space, characterised by the property that it is a simply connected CW complex ...

7

Write $X=S^3\vee S^6$ and note that this space admits at least one comultiplication, since it is a suspension. Denote the suspension comultiplication $$c:X\rightarrow X\vee X$$ and observe that it is coassociative, cocommutative and counital, since $X$ is a double suspension. Thus for each space $Y$, the comultiplication $c$ furnishes the homotopy set $[X,Y]$...

6

First, note that $\emptyset \hookrightarrow \Delta^n$ is a cofibration (because any monomorphism is a cofibration in $\textbf{sSet}$), so the right lifting property of an acyclic fibration $p : X \to Y$ implies $p_n : X_n \to Y_n$ must be a surjection. In general, if you know a Kan fibration $p : X \to Y$ restricts to a surjection $p_0 : X_0 \to Y_0$, then ...

6

I think there's a typo in that MO answer: $BG$ is modeled by $EG \times_G E(G/H)$ since this is just $BG \times E(G/H)$ and $E(G/H)$ is contractible. Now we have a natural map $EG \times_G E(G/H) \rightarrow B(G/H)$ which is a fibration with fiber $(EG)/H \cong BH$. For a great reference on all this stuff written in a very friendly style see: http://www....

6

To be honest, I have trouble finding out what is the needed argument, but I'll try to give another one. I will consistently use radial coordinates on the disk, ie. $(r, \phi) \in D^{k}$, where $r$ is the distance from the centre and $\phi \in S^{k-1}$. Let $\tilde{F}_{n-1}: S^{i} \times I \rightarrow X_{n-1}$ given by $\tilde{F}_{n-1}(\phi, t) = F_{n-1}(t-1,... 6 If the fibration is smooth and the spaces involved are compact, then it is a fiber bundle. Just to make clear what I mean by smooth fibration: Definition. A smooth map$p\colon E \to B$is said to satisfy the homotopy lifting property in the smooth category if given the following commutative diagram where all maps are smooth: there exists an smooth map$\...

6

Consider $\mathbb{R}\hookrightarrow\mathbb{R}^3\to\mathbb{R}^2$ where the projection map is given by projection onto the first two coordinates. I'm going to define connections in terms of one-forms. Since one-forms annihilate a two-dimensional subspace, the kernel of the one-form will give the horizontal subbundle of the tangent bundle of the total space. ...

6

Yes, it is always the case. If you just need a weak homotopy equivalence, look at the long exact sequence in homotopy: $$\require{cancel} \dots \to \cancel{\pi_{n+1}(B)} \to \pi_n(F) \to \pi_n(E) \to \cancel{\pi_n(B)} \to \dots$$ to see that the inclusion $F \to E$ induces an isomorphism on all homotopy groups. More generally, if you want a full-on homotopy ...

6

Let $p:E\rightarrow B$ the projection and $i:F\hookrightarrow E$ the fibre inclusion. Choose a common basepoint $e_0\in F\subseteq E$ for both spaces and let $b_0=p(e_0)\in B$ be the basepoint of $B$. Let $[\beta]\in\pi_1(E,e_0)$ be the homotopy class of a loop $\beta:I\rightarrow E$ and consider the following commutative diagram $\require{AMScd}$ \begin{...

5

It's very important not to confuse the "topological" words used to describe the higher-groupoidal aspects of types in HoTT with the sets and topological spaces that can be defined internally to HoTT (just as they can in any sort of mathematics). This problem of terminology arises because homotopy theory studies objects that are properly called ∞-...

5

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 $f : X \... 5 I assume that you mean a bundle where both fiber and the base have positive dimension and the manifold is closed. Then: In dimension 2 a manifold$M$fibers iff$\chi(M)=0$. In dimension 3 life is much more complicated. For aspherical manifolds fibering with circle or surface fiber (allowing for Seifert fibrations which are generalizations of the ordinary ... 5 Good news as long as we only care for CW complexes. In D. Barnes, The simplicial bundle of a CW Fibration (jstor link), to every fibration with base and fibres being CW, there is associated a fibre-homotopic (simplicial) fibre bundle. 5 The path space in the path space fibration is not the universal cover, even up to homotopy, unless$X$is aspherical. Its fiber, rather than being the fundamental group, is the based loop space of$X$. But it is reasonable to think of it as a higher analogue of the universal cover. 5 This definition of a "fibration" is pretty much useless except for pedagogical purposes, in order to ease you into the correct definition. In fairness, two pages later Schwarz gives the standard definition of a "locally trivial fibration", this is the one that he (and you, most other people) will use. Here are few exampless to consider which show how bad ... 5 The group$SU(n)$has a natural action on$\Bbb C^n$and also on the unit ball within, the set$\{z\in\Bbb C^n:\|z\|=1\}$which we can identify with$S^{2n-1}$. Then$SU(n)$acts transitively on$S^{2n-1}$. For a suitable point$z_0$of$S^{2n-1}$the stabiliser of$z_0$is$SU(n-1)$. Therefore we may identify$S^{2n-1}$with the coset space$SU(n)/SU(n-1)$. 5 (I will use "connected" to mean "path-connected" throughout) If it is a fibration, then you have a long exact sequence of homotopy groups$\pi_2(T^2)\to \pi_1(F)\to \pi_1(M)\to \pi_1(T^2)\to \pi_0(F)\to \pi_0(M)$, where$F$is the fiber. As$M$is connected and$\pi_2(T^2) = 0$you get$0\to \pi_1(F)\to \pi_1(M)\to \pi_1(T^2)\to \pi_0(F)\to \{*\}$Now ... 5 Let$F\to E\to B$be a fibration. Then we have the following long exact sequence of homotopy groups $$\ldots \to \pi_{n+1}(B)\to \pi_n(F)\to \pi_n(E)\to \pi_n(B)\to \pi_{n-1}(F)\to\ldots.$$ In particular, for the Hopf fibration$S^1\to S^3\to S^2$, we obtain $$\ldots \to \pi_{n}(S^1)\to \pi_n(S^3)\to \pi_n(S^2)\to \pi_{n-1}(S^2)\to \ldots.\ \ \ \ \ (1)$$ ... 5 As Tyrone points out there is the well-known Hopf fibration$S^1 \to S^3 \to S^2$, which is actually just the restriction of the fibration$S^1 \to S^{\infty} \to \mathbb{CP}^\infty$to the$2$-skeleton$\mathbb{CP}^\infty_{(2)} = \mathbb{CP}^1 \cong S^2$. From this fibration you find that$\pi_i S^2 \cong \pi_i S^3$for$i\geq 3$. This result will make your ... 4 It's actually much easier than that. Recall that adjoints are unique up to unique isomorphism, and that the adjoint of a composite is the composite of the adjoints. So if we have a pullback square $$\begin{array}{ccc} \bullet & \stackrel{z}{\to} & \bullet \\ {x} \downarrow & & \downarrow {y} \\ \bullet & \... 4 Yes, this is true and to put it a broader context, it can be seen as a special example of a more general principle: Suppose we have a topological group G and a subgroup H, then we get a fibre bundle$$G/H\rightarrow BH\rightarrow BG.$$One way to construct this bundle is to start with a model of the universal principal G-bundle$$G\rightarrow EG\... 4 You are right. If the fibration is smooth and the spaces involved are compact, then it is a fiber bundle. Just to make clear what I mean by smooth fibration: Definition. A smooth map$p\colon E \to B$is said to satisfy the homotopy lifting property in the smooth category if given the following commutative diagram where all maps are smooth: there exists an ... 4$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Modulo orientation, the complex line bundle coming from the Hopf map$\pi:S^{3} \to S^{2}$is the tautological line bundle$\mathcal{O}_{\Proj^{1}}(-1)$, whose total space is the blow-up of$\Cpx^{2}$at the origin. (The same real$2$-plane bundle with the opposite orientation is the hyperplane ... 4 Let me expand my comment. I need to assume that$X$and$B$are algebraic, compact and smooth. Then, the Hodge numbers of$X$coincide with the Hodge numbers of$F \times B$. For see this, consider the Grothendieck ring of varieties$K_0(\text{Var}/\Bbb C)$. This is the ring generated by varieties over$\Bbb C$, quotiented by the "scissor relation"$[X] = [...

4

In the notes you linked, $F(f)$ is defined as the pullback of the diagram $X \xrightarrow{f} Y \leftarrow PY$ where $PY$ is the path-space of $Y$. Thus the map $fp$ is the same as the map $F(f) \to PY \to Y$. But $PY$ is contractible, so the map is nullhomotopic. To see that $PY$ is contractible, note that we can define for a path $\gamma(s)$ the ...

4

Your descriptions indeed do not match, but that's because neither accurately describes what the product of topological spaces really is. Let's use $\hat\times$ to denote the operation you're describing: For a space $A$ and a pointed space $B$, we define $A \mathbin{\hat\times} B$ as the result of gluing a copy of $B$ to every point of $A$. Let's consider $\... 4 First, prove the following: A point$x$of a space$X$corresponds to a (continuous) function$\mathbf{x} : \ast \to X$that sends$*$to$X$. Paths$\gamma : I \to X$from$x$to$y$are in bijective correspondence to homotopies$H : \mathbf{x} \simeq \mathbf{y}$with$\mathbf{x},\mathbf{y}: \ast \to X$the functions associated to$x$and$y\$, where ...

Only top voted, non community-wiki answers of a minimum length are eligible