Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $g$.

share|cite|improve this question
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
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 $s\neq 0,1$.

share|cite|improve this answer
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
No, it also should be a loop. – Dirk Jul 13 '12 at 9:58
$F(t,s)$ 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: – 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.