Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Well I tried to imagine all this paths in my mind ( actually I used a glass and wool) to define informally the homotopy between these four paths, I think that they are homotopic, but how can I prove this formally (maybe I'm wrong and some of them are not)

Let's consider the space $ X = S^1$ x $[0,1] $ with it's topology as a subspace of the euclidean space $ \Bbb R^3$. The paths are the following:

$i) f(t)=(1,0,t)$

$ii) g(t)= (\cos(2\pi t) , \sin(2\pi t) , t^3 ) $

$iii) h(t) =(\cos(4\pi t) , \sin(4\pi t) , t^3 ) $

$iv) (f*j)(t)$ where $j(t)=(\cos(2\pi t) , \sin(2\pi t) , 0 )$

And where $f,g,h,j : [0,1] \to X$

share|improve this question
    
What does the star in part (iv) mean? And are these homotopies supposed to, say, keep the endpoints fixed (a detail which may change the answer to the question)? If you can visualize, try to turn your visualization into an explicit formula, i.e. a function from $[0,1] \times [0,1]$ to $X$ that starts at one of the paths and ends at the other. –  Paul VanKoughnett Feb 27 '13 at 3:32
    
$f*g$ denote the product of paths, first we travel around the path $f$ with double speed, then $g$ (that is only for having defined again the new path in the $[0,1]$ –  Eustass Feb 27 '13 at 3:38

1 Answer 1

Recall that if $Y$ is a contracticle space, and $X$ is path-connected, then the set of classes of homotopic maps $Y\rightarrow X$ is trivial. That is, every map is null-homotopic and all null-homotopic maps are homotopic to each other (because $X$ is path-connected). Given that $[0,1]$ is contractible, and your cylinder $X$ is path-connected, what does this tell you about $f,g,h,j$?


The question asker has mentioned that he is using homotopy relative to a subspace, in which case the above doesn't hold. Note that the Cylinder $S^1\times [0,1]$ is homotopy equivalent to the circle $S^1$ via the standard projection map. This projection map also has the nice property that each of your paths become loops. That is, $p\circ f(0)=p\circ f(1)$ and similarly for $g,h$ and $f*j$ where $p$ is the homotopy equivalence given by projection.

Now let $X$ and $Y$ be homotopy equivalent spaces and $f_1\colon X\rightarrow Y$ and $f_2\colon Y\rightarrow X$ be such that $f_1f_2\simeq 1_Y$ and $f_2f_1\simeq 1_X$. A path $\gamma\colon [0,1]\rightarrow X$ is homotopic to a path $\lambda\colon [0,1]\rightarrow X$ relative to end points if and only if $f_1\circ\gamma$ is homotopic to $f_1\circ\lambda$ relative to end points.

In your case, $Y$ is a circle, $f_1$ is projection on to the base circle and so your paths are only homotopic relative to end points if the induced loops on the circle are homotopic loops relative to base point. Hopefully you know how to test if loops in the circle are homotopic relative to base point (look at the winding number) and so you should be able to tell that only $g$ and $f*j$ are homotopic paths relative to base point.

share|improve this answer
    
But I'm talking about homotopy of paths , i.e $ F(0,t)=x_0 , F(1,u)= x_1 $ –  Eustass Feb 28 '13 at 2:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.