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Chapter 8: Problem 145

Seien \(f, g: B^{\prime} \rightarrow B\) zwei homotope Abbildungen zwischen lokal wegzusammenh盲ngenden und semi-lokal einfach-zusammenh盲ngenden topologischen R盲umen. Zeigen Sie f眉r jede U'berlagerung \(p: X \rightarrow B\), dass die Basiswechsel von \(p\) entlang \(f\) und \(g\) isomorph zueinander sind.

Short Answer

Expert verified
The base changes of \(p\) along \(f\) and \(g\) are isomorphic because the homotopy ensures a lifted isomorphism.

Step by step solution

01

Understand the Concepts

We are dealing with continuous maps between topological spaces that are homotopic, which means they can be continuously transformed into one another. The spaces are locally path-connected and semi-locally simply connected, meaning every point has a path-connected neighborhood and small enough loops can be contracted to a point.
02

Define the Problem

You are given two maps, \(f\) and \(g\), which are homotopic: \(f hicksim g\). You need to show that for a covering map \(p: X \rightarrow B\), the pullback (or change of base) of \(p\) along \(f\) and \(g\) yields isomorphic covering spaces.
03

Use Homotopy Lifting Property

The homotopy lifting property states that if \(p: X \rightarrow B\) is a covering map and \(H: B' \times I \rightarrow B\) is a homotopy from \(f\) to \(g\), then there exists a unique lift \(\tilde{H}: B' \times I \rightarrow X\) such that \(\tilde{H}(x,0) = \tilde{f}(x)\), where \(\tilde{f}\) is a lift of \(f\).
04

Construct the Lifts

Given the homotopy \(H: B' \times I \rightarrow B\), construct the lift \(\tilde{H}\). Starting with a lift \(\tilde{f}: B' \rightarrow X\), lift the entire homotopy \(H\) to get \(\tilde{H}\). The endpoint \(H(-,1)\) gives us a lift of \(g\), \(\tilde{g}\), since \(g = H(-,1)\).
05

Show Isomorphism

The lifts \(\tilde{f}\) and \(\tilde{g}\) are related by \(\tilde{H}\), showing that both changes of base along \(f\) and \(g\) pull back the covering space \(X\) to isomorphic covering spaces \(f^*(X) \cong g^*(X)\) over \(B'\). The lifts are homotopic and hence isomorphic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homotopy
In topology, homotopy is a concept that helps us understand when two mathematical structures can be continuously transformed into each other. Imagine you have a rubber sheet (or space) and you're trying to mold it from one shape into another without tearing or gluing. If you can do this, the two shapes are homotopic.

Formally, two functions, say \(f\) and \(g\), are homotopic if there is a continuous transformation called homotopy \(H\) that smoothly morphs \(f\) into \(g\). This transformation is often expressed as \(H: B' \times I \to B\), where \(I\) is the interval \([0,1]\). At \(t=0\), \(H\) behaves like \(f\), and at \(t=1\), it transforms into \(g\).
  • Homotopy is a fundamental concept in algebraic topology.
  • It allows for flexibility in handling continuous functions.
  • Homotopy equivalence is a looser relation than homeomorphism, as it doesn't require the map to be invertible.
Covering Spaces
Covering spaces are an interesting structure in topology. They can be thought of as an expanded version of a given space, where each point is 'covered' by several corresponding points in a larger space. This idea is a bit like copying a point multiple times on a higher plane.

A formal covering space consists of a map \(p: X \to B\) where \(X\) is the covering space and \(B\) is the base space. Locally, this map looks like a simple projection, ensuring that every small enough neighborhood in \(B\) has a complete, disconnected inverse map in \(X\), spread out like a fan.
  • Covering spaces help simplify the analysis of spaces, especially in the presence of loops.
  • The concept is crucial for understanding phenomena in fundamental groups.
  • They allow leveraging global properties by examining local structures.
Path-Connected Spaces
Path-connected spaces are types of spaces where any point can be joined to any other point by a continuous path. Think of it as being able to draw a line connecting any two points without lifting your pen.

If you picture the surface of a ball, it's clear that you can draw such paths between any two points; hence, a ball's surface is path-connected. In contrast, a space split into two separate islands cannot have such paths linking them, so it isn't path-connected.
  • Ensures any two points in the space can be navigated continuously.
  • Helps in analyzing the space's overall structure and continuity.
  • It's a vital property when considering the existence of continuous mappings and transformations.
Simply Connected Spaces
Simply connected spaces are essentially souped-up versions of path-connected spaces. Not only can any two points be connected by a path, but any loop in the space can be shrunk to a single point. Imagine the space as a stretchy rubber band that you can shape or squish without it tearing or having holes.

Formally, a space is simply connected if it is path-connected and any loop within it can be continuously contracted to a point, removing concerns about holes or twists that prevent such shrinking. Consider the surface of a sphere: any loop on it can be pulled tight to a point, illustrating why it's simply connected.
  • Being simply connected implies that the fundamental group is trivial (consisting only of the identity element).
  • Such spaces are incredibly important in complex analysis and differential geometry.
  • They simplify the calculation of the space's properties, especially regarding its fundamental group.

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Most popular questions from this chapter

Geben Sie ein Beispiel f眉r einen surjektiven lokalen Hom枚omorphismus, der keine U'berlagerung ist.

Die Abbildung $$ ] 0, \infty\left[\longrightarrow S^{1}, t \mapsto \exp (2 \pi i t)\right. $$ ist keine U'berlagerung.

Gibt es eine Polynomfunktion \(p: \mathbb{C} \rightarrow \mathbb{C}\) mit \(\operatorname{grad}(p) \geqslant 2\) welche eine U?berlagerung ist?

Sei \(p: X \rightarrow Y\) eine U'berlagerung. Beweisen oder widerlegen Sie: (1) Ist \(Y\) kompakt, so auch \(X\). (2) Ist \(X\) kompakt, so auch \(Y\). (3) Ist \(Y\) ein Hausdorff-Raum, so auch \(X\). (4) Ist \(X\) ein Hausdorff-Raum, so auch \(Y\).

Sei \(p: X \rightarrow B\) eine 脷berlagerung mit endlicher Bl盲tterzahl \(n \geqslant 2\) und nicht leeren, wegzusammenh?ngenden R盲umen \(X\) und \(B\). Dann gibt es eine Schleife in \(B\), welche sich nicht zu \(X\) hochheben l盲sst. (Das entsprechende gruppentheoretische Resultat wurde von Jordan bewiesen, siehe (Ser03).)
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