TITLE: A little question about homotopy equivalence
QUESTION [3 upvotes]: 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?

REPLY [3 votes]: 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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