Paths and loops in a topological space

From Maths
Revision as of 19:23, 16 April 2015 by Alec (Talk | contribs) (Created page with "* This article aims towards defining the Fundamental group * The name of the page was chosen to make it distinct from paths and loops, which are terms in graph theory ==P...")

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
  • This article aims towards defining the Fundamental group
  • The name of the page was chosen to make it distinct from paths and loops, which are terms in graph theory

Path in a topological space

A path in [ilmath]X[/ilmath] is any continuous map [ilmath]p:[0,1]\rightarrow X[/ilmath][1].

Loop in a topological space

A path [ilmath]p[/ilmath] is a loop if [ilmath]p(0)=p(1)[/ilmath]

Loop based at

If [ilmath]p[/ilmath] is a loop based at [ilmath]x_0[/ilmath] if [ilmath]p(0)=p(1)=x_0[/ilmath]

Concatenating paths

Given two paths [ilmath]p_0[/ilmath] and [ilmath]p_1[/ilmath] in a topological space [ilmath]X[/ilmath] with [ilmath]p_0(1)=p_1(1)[/ilmath] we can obtain a new path by performing [ilmath]p_0[/ilmath] first, followed by [ilmath]p_1[/ilmath] in the same time by moving at double speed, this new path is called [ilmath]p_0*p_1[/ilmath] and is defined as:

  • [math](p_0*p_1)(t)=\left\{\begin{array}{lr} p_0(2t) & \text{if }t\in[0,\tfrac{1}{2}] \\ p_1(2t-1) & \text{if }t\in[\tfrac{1}{2},1] \end{array}\right.[/math]
    this is fine to do as the functions agree when restricted to the intersection, that is to say when [ilmath]t=\tfrac{1}{2}[/ilmath] both "halves" agree

Note that:

  • If [ilmath]p_0[/ilmath] and [ilmath]p_1[/ilmath] are loops based at [ilmath]x_0[/ilmath] then [ilmath]p_0*p_1[/ilmath] is always defined and is itself a loop based at [ilmath]x_0[/ilmath]
  • [ilmath]p_0*p_1\ne p_1*p_0[/ilmath] usually - which is really easy to see.
  • We can't define a group yet.
    • Note that [ilmath]a*(b*c)[/ilmath] performs [ilmath]b[/ilmath] and [ilmath]c[/ilmath] at 4 times their normal speed and [ilmath]a[/ilmath] at just double whereas:
    • [ilmath](a*b)*c[/ilmath] performs [ilmath]a[/ilmath] and [ilmath]b[/ilmath] at 4x their normal speed and [ilmath]c[/ilmath] at just double - these are clearly different paths - so we don't even have associativity

References

  1. Introduction to topology - lecture notes nov 2013 - David Mond