Two functions are homotopic, if one of them can by continuously deformed to another. This gives rise to an equivalence relation. A group called homotopy group can be obtained from the equivalence classes. The simplest homotopy group is fundamental group. Homotopy groups are important invariants in ...

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graphs and homotopy extension property

If $T$ is a spanning tree of a graph $X$. How to prove that the pair $(X,T)$ has the homotopy extension property, without using the definition of CW complexes? I mean I don't need the general case ...
5
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1answer
58 views

The action of the group of deck transformation on the higher homotopy groups

This is for homework. I'm supposed to do exercise 4.1.4 in Hatchers "Algebraic Topology", which is to show that given a universal covering $p: \tilde{X} \to X$ of a path-connected space $X$, the ...
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19 views

Is there a fibration whose base, total and fibre is a Moore space?

I want to find a example which base, total of fiber spaces is a Moore space. I don't want trivial fibration. For example the projection $p : M(G,n) \times M(H,n) \to M(G,n)$ and path fibration ...
6
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1answer
49 views

Can the rank of the homology group of an abstract simplicial complex be computed in polynomial time?

I want to write a function that does the following: Input: An integer $n$ A function $f$ that maps nonempty subsets of $\{1, \dots, n\}$ to "yes" or "no" such that (a) every singleton set gets ...
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67 views

Homotopy, retraction related question

How to solve the following problem: $A$ is a strong deformation retract of $X$ such that exists a continuous function $α:X\to I$, $\alpha^{−1}({0})=A$. (a) If $H:X\times I\to X$ is a homotopy (rel ...
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Visualize Fourth Homotopy Group of $S^2$

I know $\pi_4(S^2)$ is $\mathbb{Z}_2$. However, I don't know how to visualize it. For example, it is well known that $\pi_3(S^2)=\mathbb{Z}$ can be understood by Hopf Fibration. Elements in ...
4
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1answer
66 views

What's the point of spectra?

I'm familiar with the definition of a spectrum, the one due to Adams, however, I'm not really sure why someone would want to define such a thing. I know they allow one to generalize homology and ...
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25 views

Naive question: Why $B:Mon\rightarrow Top^{*}, \Omega: Top^{*}\rightarrow Mon$ is an adjoint functor?

It is not clear to me why we have a bijection of the form $$Mor_{Top^{*}}(BY,X)\rightarrow Mor_{Mon}(Y,\Omega X)$$where $Mon$ is the category of topological monoids and $Top^{*}$ the based topological ...
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1answer
42 views

Covering infinite sheeted covering of torus

Suppose I have subgroup $H=\operatorname{span}\langle (a,b)\rangle\subset \pi_1(\mathbb{T}^2)=\mathbb{Z}^2$, where $a,b$ are integers where $(a,b)\neq(0,0)$. I know the covering space is $S^1\times ...
5
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1answer
95 views

What is the homotopy colimit of the Cech nerve as a bi-simplical set?

Let $f:Y_\bullet\to X_\bullet$ be an epimorphism of simplicial sets and define the bi-simplicial set $$ F_{\bullet\bullet}=\ldots Y\times_X Y\times_X Y\underset{\to}{\underset{\to}{\to}}Y\times_X Y ...
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1answer
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$SU(n)$ is simply connected (proof without fibrations, $n>2$)

How to show that $SU(n)$ is simply connected for $n>2$ if I don't know about fibrations yet? For $SU(2) \cong S^3$ the fact is said to be known. For any matrix $A \in SU(n)$ there is a matrix $S ...
5
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2answers
32 views

elementary reference for Hopf Fibrations

I am looking for a good introductory treatment of Hopf Fibrations and I am wondering whether there is a popular, well regarded accessible book. ( I should probably say that I am just starting to learn ...
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1answer
27 views

Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference?

I'm looking for a reference for the following result: If $G$ is a compact and simply connected Lie group and $\Sigma$ is a compact orientable surface, then every principal $G$-bundle over $\Sigma$ ...
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23 views

Mathematica code for solving MHD viscoelastic fluid flow [migrated]

Where might I find Mathematica code for solving MHD viscoelastic fluid flow using Homotopy Analysis method?
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0answers
37 views

How to show $\mathbb{S}^{2n-1}$ is a $\mathbb{S}^{n-1}$ bundle over $\mathbb{S}^{n}$ then $n=1,2$ or divisible by $4$?

I was asked to show if $$\mathbb{S}^{n-1}\rightarrow \mathbb{S}^{2n-1}\rightarrow \mathbb{S}^{n}$$ holds. Then $n=1,2$ or divisible by $4$. The $n=1,2$ cases in the converse are verfied. I am ...

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