Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know the definition of homotopy equivalence:

Let $f,g:X \to Y$ be continuous between topological spaces.

We say $f$ is homotopic to $g$ if there is a continuous map $H : X \times I \to Y \ \ $ satisying

$H(x,0) = f(x)\ and \ H(x,1) = g(x)$.

Now, if $\gamma:I \to X $ is a closed curve such that $\gamma(0) = p =\gamma(1)$ for some $\gamma \in X$.

If we define $H : I \times I \to Y \ $ by $H(s,t) = \gamma (st)$ , then $H(s,1) = \gamma(s)\ and \ H(s,0) = \gamma(0) = p$

So we have that any closed curve is homotopic to a constant map, but I know that is not true! For example not all closed curve can be contract to a point on a torus.

So, where is the main point? is $H(s,t) = \gamma (st)$ dis-continuous ? or I misunderstand the definition between homotopy and homotopy equivalence?

share|cite|improve this question
up vote 3 down vote accepted

Usually when people speak of curves being homotopic, they mean homotopic relative endpoints, which is the additional requirement that the $H(0,t)$ and $H(1,t)$ be constant. This fails in your example unless $\gamma$ is a constant curve.

share|cite|improve this answer
Oh! I see, so I need to fix curves st endpoints $0$ and $1$. Thank you very much! – Peter Hu Mar 14 '12 at 9:38

Your Answer


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.