Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20080704 11:27:46
PathsLet be defined by . Then p is a path in from (1,0) to (−1,0). In general: If a and b are points in a topological space X, a path in X from a to b is a continuous function p : [0,1] → X such that p(0) = a and p(1) = b. In our example, our path is the upper semicircle of the unit circle centred at the origin.Last edited by JaneFairfax (20080704 11:39:02) #2 20080704 20:35:25
Re: PathsLet where q(t) = −p(t) for each t ∈ [0,1] be another path in (the lower semicircle of the same unit circle). Suppose for each θ ∈ [0,1], we have a function defined by .Note that F_{θ} is a path from (1,0) to (−1,0) for each θ ∊ [0,1] and F_{0}(t) = p(t) and F_{1}(t) = q(t) for each t ∊ [0,1]. F_{θ} is a semiellipse with semimajor axis 1 and semiminor axis 1−2θ (except when θ = ½, when it’s a straightline path). Now, what happens when θ varies continuously from 0 to 1? Yes, the path p transforms continuously to q via the various intermediate paths F_{θ}. We say that the path p is homotopic to q in . #3 20080705 06:53:18
Re: PathsThe function F is called a homotopy from f to g. We can write to mean that F is a homotopy from f to g. #4 20080705 22:56:52
Re: PathsAnd we write . F can be seen as “fixing” the subset A while deforming f continuously into g. In particular, if p and q are paths, we can have a homotopy H which “fixes” the start and end points – i.e. . #5 20080706 06:04:17
Re: PathsI see you're starting to get into algebraic topology. Is this self study? What book are you using? "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #6 20080706 06:51:29
Re: PathsYes, it’s mainly selfstudy – but any help from anybody will be greatly appreciated. Last edited by JaneFairfax (20080706 06:58:34) #7 20080706 10:23:24
Re: PathsBy the way, MathsIsFun, if you need a page on algebraic topology on the website, I can write one up for you – up to and including the defintion of fundamental group. #8 20080714 00:06:05
Re: PathsLet a, b, c be points in a topological space X. Suppose are paths from a to b and are paths from b to c. First we define the product path of two paths, :Basically, is a path from a to c via b. Then we have this reault: It must be noted that it is important for the homotopies to be relative to {0,1}. If p and q (likewise r and s) are only homotopic, not homotopic relative to {0,1}, then the product paths and need not be homotopic! Here is an example to show why. Let and define paths as follows: Then p and q (likewise r and s) are only homotopic, not homotopic relative to {0,1} – and the product paths and are definitely not homotopic. The former is the unit circle centred at (0,0) and the latter the unit circle centred at (0,2) – as (0,0) is not in X, there is no way a loop enclosing (0,0) can be continuously deformed in the Cartesian plane into a loop that does not enclose (0.0). #9 20080714 07:47:22
Re: PathsLet . The null path at a is the constant path at a – i.e. the path where for all .#10 20080714 10:54:20
Re: PathsLet be a path from a to b. The inverse path of p is the path from b to a where for all .#11 20080724 19:04:31#12 20080727 02:41:27
#13 20080727 07:42:28
#14 20080808 05:08:23
