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.

Can someone explain me in descriptive words or maybe with an image, what homotopic actually means and what its relevance is?

Thank you for your time,

Chris

share|improve this question
11  
Which parts of the wikipedia article do you find vague? –  Tim Duff Jun 17 '12 at 2:37
add comment

1 Answer

up vote 1 down vote accepted

Let $X$ and $Y$ be topological spaces, and let $F_0, F_1:X\rightarrow Y$ be continuous maps. We say that $F_0$ and $F_1$ are homotopic and denoted by $F_0 \simeq F_1$ if there exists a continuous map $H: X \times I \rightarrow Y$ (which is called a homotopy from $F_0$ to $F_1$) such that for all $x \in X$,

$$H(x,0)=F_0(x),$$ $$H(x,1)=F_1(x).$$

If we think of the parameter $t$ as the time parameter which changes from 0 to 1, then the homotopy $H$ represents a continuous deformation of the map $F_0$ to the map $F_1.$

We know that a path in a topological space $X$ is a continuous map $f:I \rightarrow X$ such that $f(0)=x_0$ and $f(1)=x_1$. The points $x_0$ and $x_1$ are called initial point and terminal point of $f$, respectively. We say that $f$ is a path from $x_0$ to $x_1.$ For any two paths in the topological space $X$, we need a stronger relation between the paths in order to find the holes of the space $X.$

Let $f_0, f_1:I\rightarrow X$ be two paths in $X$. We say that $f_0$ and $f_1$ are path-homotopic and denoted by $F_0 \sim F_1$ if $f_0$ and $f_1$ have the same initial point $x_0$ and the same terminal point $x_1$ and there exists a continuous map $H: I \times I \rightarrow X$ (which is called a path-homotopy from $f_0$ to $f_1$) such that

$$ H(s,0)=f_0(s) $$ $$ H(s,1)=f_1(s) $$ $$ H(0,t)=x_0 $$ $$ H(1,t)=x_1 $$

for all $s \in I$ and for all $t \in I.$

In other words, the path-homotopy $H$ represents a continuous deformation of the path $f_0$ to the path $f_1$. Also end points remain fixed during the deformation.

share|improve this answer
add comment

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.