## On the associative property of a binary operation of the fundamental group.

I was reading about the proof of associativity property of the operation on the fundamental group here. The book gives the following diagram then it says the reader should supply the elementary geometry necessary to derive the homotopy But I don’t know how to do that. Please show me how to derive the above homotopy … Read more

## Prove that exist bijection between inverse image of covering space [duplicate]

This question already has answers here: If h:Y→X is a covering map and Y is connected, then the cardinality of the fiber h−1(x) is independent of x∈X. (3 answers) Closed 2 years ago. Let B be path-connected and p:E→B covering map (with E as covering space). Prove that ∀a,b∈B exist 1-1 injection correspondence between p−1(a) … Read more

## Regular and non-regular covering spaces of S1∨S1∨S1 \Bbb{S}^{1} \vee \Bbb{S}^{1} \vee \Bbb{S}^{1} .

I tried to draw the regular and non-regular covering spaces of S1∨S1∨S1. I think the regular covering space is: Is it true? How do you draw the non-regular covering space of this one? Answer The example you drew is not a covering space of S1∨S1∨S1, because the unique vertex of S1∨S1∨S1 has valence 6, and … Read more

## Direct sum of nontrivial vector bundles?

When is it true that the direct sum (or whitney sum) of two nontrivial vector bundles is nontrivial? Also, if you have a direct sum of vector bundles, with a and b global sections respectively, does its direct sum have a+b sections? Answer If you know the characteristic classes of your vector bundles, you can … Read more

## Show H1dR(Sn)=0H_{dR}^1(S^n)=0 for n>1n>1 without de-Rham Theorem, and some similar questions.

Question Without using de Rham’s theorem, prove: (1) Show H1dR(Sn)=0 for n>1. (2) Use (1) to show H1dR(RPn)=0 (3) a n-form Ω is exact on Sn if and only if ∫SnΩ=0 (4) Use (3) to show that every smooth 2-form ω on RP2 has the form dα for some smooth 1-form α I know I … Read more

## \mathbb{R}^{2}\mathbb{R}^{2} and \mathbb{R} \times [0, +\infty]\mathbb{R} \times [0, +\infty] are homotopy equivalent, but not homeomorphic

So, let’s consider M=\mathbb{R}^{2} and N= \mathbb{R} \times [0, +\infty] – two topological spaces. Since \pi_{1}(M)=\pi_{1}(\mathbb{R}) \times \pi_{1} (\mathbb{R}) = \{0 \} (since \mathbb{R} is path-connected). This time, fundamental group of every convex subset of \mathbb{R} is also trivial, so it’s time to conclude that M is homotopy equivalent to N. But how to prove … Read more

## configuration-spaces and iterated loop-spaces

In the paper Configuration-Spaces and Iterated Loop-Spaces. Graeme Segal, page 213-214, it is obtained that the labelled configuration space Cn is homotopy equivalent to a topological monoid C′n. Let a homotopy equivalent map be ϕ:Cn→C′n. On the other hand, we have the path-connected components decomposition Cn=⨆k≥0F(Rn,k)/Σk as the disjoint union of k-th unordered configuration spaces. … Read more

## Bisection Theorem

I don’t know whether this is true or false But I did try to prove it as true using similar arguments as in the Bisection theorem Statement: Given a simple bounded region in R2 there exist at least two straight lines where each bisect it into two parts of equal area. But this one I … Read more

## Cohomology ring of RPn\mathbb RP^n with integral coefficient.

I know cup product structure on H∗(RPn;Z2)=Z2[α]/(αn+1). How to get H∗(RPn;Z) from this? I have two cochain complexes for two coefficient rings. Now my question is what will be the induced map between these two cochain complexes and what will be H∗(RPn;Z)? Answer I think, H∗(RP2n+1)=Z[α,β]/(2α,αn+1,αβ,β2), where α has degree 2 and β has degree … Read more

## What is the Wrong in this Triangulation of the Torus

On pg 133 of Roman’s Introduction to Algebraic Topology it is stated that one requires at least 14 triangles in any triangulation of the torus. Admittedly, I do not have a very good understanding of triangualations. From what I understand, the following seems like a perfectly valid triangulation of the torus: What the mistake in … Read more