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Let $S^1 = \{z \in \mathbb{C} : |z| = 1\}$. Take the loops $f,g : [0,1] \rightarrow S^1$, $f(t) = 1$, $g(t) = e^{2\pi it}$. I know these represent different elements in $\pi_1(S^1, 1)$, but I don't see why $F(t,s) = e^{2\pi its}$ isn't a homotopy between $f$ and $g$.

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If you're allowed to move around endpoints then every map $[0,1] \to X$ is homotopic to a constant map. –  Justin Young Jul 13 '12 at 12:20

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up vote 2 down vote accepted

Loops are paths for which initial and end point coincide (in other words: closed paths), see here. The reason is simply that $F(\cdot,s)$ is not a closed path if $s\neq 0,1$.

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I thought the only condition on $F$ is that it's continuous, with $F(t,0) = f(t)$ and $F(t,1) = g(t)$.. –  Dog Jul 13 '12 at 9:54
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No, it also should be a loop. –  Dirk Jul 13 '12 at 9:58
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$F(t,s)$ should be a loop for each $t$. –  Host-website-on-iPage Jul 13 '12 at 10:04
    
Ok, I think I remember something that $F$ should leave the endpoints of $f$ and $g$ stable, so that explains why it needs to be a loop. I was confused since I couldn't find this in the definition on Wikipedia: en.wikipedia.org/wiki/Homotopy –  Dog Jul 13 '12 at 10:08
    
@Dog maybe it helps thinkink of the loops as maps with domain $S^1$ instead of $I$. Then a homotopy is a map with domain $S^1\times I$ and it is obvious that all $F(-,s)$ have to loops. –  Simon Markett Jul 13 '12 at 10:26

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