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I was solving an exercise and the following question came to me that I don't know how to attack:

It is true that two homotopic curves viewed in $C(S^1,X)$ are then also homotopic viewed as closed paths $C([0,1],X)$. Where $X$ is a topological space (Hausdorff if necessary).

Any ideas?

Zaragosa
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1 Answers1

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Both spaces are the same: $S^1$ arises from $[0,1]$ by identifying $0$ with $1$, so any continuous closed curve $\gamma : [0,1]\rightarrow X$ with $\gamma(0)=\gamma(1)$ becomes a continuous curve $\tilde\gamma : S^1\rightarrow X$. The converse is analogous, since "breaking up" $S^1$ at any point results again in $[0,1]$.

Thus, it doesn't really matter in which of these two spaces two curves are homotopic.

Symplectic Witch
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  • I understand the idea but could you formally detail what you mention? For example, the following function occurs to me between both spaces: $\bar{\alpha(t)}=\alpha(e^{2\pi it})$. But how do I prove the good definition and its bijection? – Zaragosa Sep 03 '22 at 03:51
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    @Zaragosa There isn't much more to it than what you have written in your answer. Essentially, there are just numerous ways to define $S^1$, which are all topologically equivalent. Just to name a few: $$S^1 = {e^{2\pi\ i t} : t\in [0,1]} = [0,1]/(0\sim 1) = \mathbb{R}/\mathbb{Z}$$ So, $\alpha : S^1\rightarrow X$ and $\alpha : [0,1]/(0\sim 1)\rightarrow X$ are two equivalent ways of expressing the same curve. These curves stand in 1-to-1 correspondence with curves $\bar\alpha : [0,1]\rightarrow X$ that fulfill $\bar\alpha(0)=\bar\alpha(1)$. – Symplectic Witch Sep 05 '22 at 00:17