/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Define \(f: S^{1} \times I \righ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Define \(f: S^{1} \times I \rightarrow S^{1} \times I\) by \(f(\theta, s)=(\theta+2 \pi s, s),\) so \(f\) restricts to the identity on the two boundary circles of \(S^{1} \times I .\) Show that \(f\) is homotopic to the identity by a homotopy \(f_{t}\) that is stationary on one of the boundary circles, but not by any homotopy \(f_{t}\) that is stationary on both boundary circles. [Consider what \(f\) does to the path \(\left.s \mapsto\left(\theta_{0}, s\right) \text { for fixed } \theta_{0} \in S^{1} .\right]\)

Short Answer

Expert verified
f is homotopic to the identity via \( f_t(\theta, s) = (\theta + 2\pi s(1-t), s) \) stationary on one boundary, not both.

Step by step solution

01

Understand the Map

The function \( f \) is defined on the cylindrical space \( S^1 \times I \) where \( S^1 \) represents the circle and \( I \) is the interval \([0, 1]\). The map \( f(\theta, s) = (\theta + 2\pi s, s) \) changes the angle \( \theta \) by a value proportional to \( s \), while keeping \( s \) fixed.
02

Explore the Identity Map

The identity map, denoted as \( id(\theta, s) = (\theta, s) \), leaves every point unchanged. Our goal is to establish a continuous deformation (homotopy) from \( f \) to \( id \) without affecting one of the boundary circles.
03

Define the Homotopy

To show that \( f \) is homotopic to \( id \), let us define a homotopy \( f_t : S^1 \times I \rightarrow S^1 \times I \) from \( f \) to \( id \) by \( f_t(\theta, s) = (\theta + 2\pi s(1-t), s) \). Here, \( t \in [0, 1] \) is the parameter of the homotopy.
04

Verify Stationarity on One Boundary

Check if \( f_t \) leaves one boundary circle unchanged. For the boundary where \( s=0 \), we have \( f_t(\theta, 0) = (\theta, 0) \), which is stationary. At the other boundary \( s=1 \), \( f_t(\theta, 1) = (\theta + 2\pi(1-t), 1) \), which is not stationary except at \( t = 1 \).
05

Investigate Path Behavior

Consider the path from \( s=0 \) to \( s=1 \) given by \( s \mapsto (\theta_0, s) \). The map \( f \) transforms this path to a helix, increasing \( \theta \) by \( 2\pi \) as \( s \) goes from 0 to 1.
06

Analyze Stationarity on Both Boundaries

If a homotopy \( f_t \) were stationary on both boundary circles, the path \( (\theta_0, s) \) would need to return exactly to its starting angle \( \theta_0 \) at \( s=1 \), meaning \( 2\pi s(1-t) \to 0 \) would yield a contradiction along the boundary \( s=1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Identity Map
The identity map is a fundamental concept in mathematics, especially when discussing homotopy. In the context of a geometric space like a cylinder, denoted as \( S^1 \times I \), the identity map is defined as \( id(\theta, s) = (\theta, s) \). This means that every point on the space stays exactly as it is.

Imagine having a transparent overlay on top of a circle, and you let it move but want every point to stay the same. That's what an identity map achieves. It is both a beginning and reference point for understanding transformations like homotopies.

When considering homotopies, the objective is often to find a way to continuously transform one map into another while sometimes preserving certain features, like the behavior on boundary circles.
Cylindrical Space
Cylindrical space is a mathematical abstraction that represents a cylinder without the inside being filled. In mathematical terms, it is denoted as \( S^1 \times I \) where \( S^1 \) represents a circle, and \( I \) signifies an interval, typically \([0, 1])\).

Picture this as the surface of a tin can. It has two ends (the boundaries of the cylinder), represented mathematically by \( s=0 \) and \( s=1 \). These are often referred to as boundary circles.

This space is useful for exploring transformations and paths. It helps mathematicians understand how different functions act on these spaces, such as twisting or stretching, like in the example map \( f(\theta, s) = (\theta + 2\pi s, s) \). This function subtly alters angles as it progresses along the cylinder.
Boundary Circles
In cylindrical space, boundary circles play a crucial role. They are the edge points where the cylinder meets barriers, showing where transformations might be restricted. In a cylindrical space \( S^1 \times I \), these are found at \( s=0 \) and \( s=1 \).

In terms of homotopy, operating on these boundary circles is akin to ensuring that the transformations respect these limits. For instance, in the provided exercise, the map \( f_t(\theta, s) = (\theta + 2\pi s(1-t), s) \) ensures that one of the boundary circles remains unchanged, specifically the one where \( s=0 \).

This selective immovability is key in studying homotopies, as it shows how transformations can be controlled or restricted to specific parts of the space, reflecting a deeper understanding of function behavior within given constraints.
Continuous Deformation
Continuous deformation refers to a transformation process where a function or shape changes smoothly over time. In mathematics, this is often described using homotopy, which shows how one function can continuously transform into another.

In our scenario, the homotopy \( f_t \) transforms function \( f \) to the identity, over a time parameter \( t \) which ranges from 0 to 1. This gradual transformation modifies parts of the original map, ensuring specific features stay unchanged.

Continuous deformation is powerful because it preserves the intrinsic properties of the space while allowing specific portions to shift or move. It is like watching a video slowly change frames without jumping out of sequence, maintaining a smooth visual progression.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(a\) and \(b\) be the generators of \(\pi_{1}\left(s^{1} \vee s^{1}\right)\) corresponding to the two \(S^{1}\) summands. Draw a picture of the covering space of \(S^{1} \vee S^{1}\) corresponding to the normal subgroup generated by \(a^{2}, b^{2},\) and \((a b)^{4},\) and prove that this covering space is indeed the correct one.

Let \(X \subset \mathbb{R}^{3}\) be the union of \(n\) lines through the origin. Compute \(\pi_{1}\left(\mathbb{R}^{3}-X\right)\)

If \(X\) is a finite graph and \(Y\) is a subgraph homeomorphic to \(S^{1}\) and containing the base point \(x_{0},\) show that \(\pi_{1}\left(X, x_{0}\right)\) has a basis in which one element is represented by the loop \(Y\).

Let \(X\) be the wedge sum of \(n\) circles, with its natural graph structure, and let \(\tilde{X} \rightarrow X\) be a covering space with \(Y \subset \tilde{X}\) a finite connected subgraph. Show there is a finite graph \(Z \supset Y\) having the same vertices as \(Y,\) such that the projection \(Y \rightarrow X\) extends to a covering space \(Z \rightarrow X\).

Let \(\varphi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation \(\varphi(x, y)=(2 x, y / 2) .\) This generates an action of \(Z\) on \(X=\mathbb{R}^{2}-\\{0\\} .\) Show this action is a covering space action and compute \(\pi_{1}(X / Z)\). Show the orbit space \(X / \mathbb{Z}\) is non- Hausdorff, and describe how it is a union of four subspaces homeomorphic to \(S^{1} \times \mathbb{R},\) coming from the complementary components of the \(x\) -axis and the \(y\) -axis.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.