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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?

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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.

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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
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