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Consider the parametrization of the 1-sphere $(\cos s, \sin s)$, $s \in [0,2\pi[$. I understand that the 1-sphere is not contractible.

However, why isn't the map $((\cos s, \sin s), t)$ to $(\cos ts, \sin ts)$ a homotopy of a constant map to the identity map?

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    Picture what your map is actually doing: immediately after time zero you're tearing apart the circle. – 2017-02-25
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    You are defining a contraction of $[0,2\pi)$, and the natural continuous bijection $[0,2\pi)\to S^1$ is not a homeomorphism. – 2017-03-01

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Your map is not continuous. Remember, $s = 0$ is supposed to be identified with $s = 2\pi$. But $(\cos(t0),\sin(t0))$ is not equal to $\cos(t2\pi),\sin(t2\pi))$ for all $t \in [0,1]$.

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    I have edited the question so that it is clear that $s \in [0,2\pi[$. This parametrization is just a way to make the map explicit – 2017-02-25
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    Well, if your $s$ is to parametrise a circle, then your map isn't continuous! – 2017-02-25
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It's not a homotopy because it's not a well-defined function.

We can compute the 'value' of $((1,0), 1/2)$ in two different ways:

$$ ((1,0), 1/2) = ((\cos(0), \sin(0)), 1/2) \mapsto (\cos(0), \sin(0)) = (1,0) $$ $$ ((1,0), 1/2) = ((\cos(2 \pi), \sin(2 \pi )), 1/2) \mapsto (\cos(\pi), \sin(\pi)) = (-1,0) $$

Since we get two different outputs for the same input, the formula you have given does not define a function.

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    Let's assume that $s = 2\pi$ is not included when defining the map, only $s=0$. – 2017-02-25