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Is there a way to determine explicitly the homotopy classes relative to their endpoints for the circle $S^1$?

We say that two paths $\gamma_1, \gamma_2 :[0,1] \to S^1$ are homotopic relative to their endpoints if there is a homotopy $F:[0,1] \times [0,1] \to S^1$ between $\gamma_1$ and $\gamma_2$ such that $F(0,t) = \gamma_1(0) = \gamma_2(0), \forall t \in [0,1]$ and $F(1,t) = \gamma_1(1) = \gamma_2(1), \forall t \in [0,1].$

Can we determine the general classes of such homotopies for the circle? That is, if we fix $p,q \in S^1$, how could we determine $$\{\gamma:[0,1] \to S^1 \ \mid \ \gamma(0) = p, \gamma(1) = q \} / \{\text{homotopic relative to their endpoints} \}?$$

Of course, if $p = q$, then we already know that every such path is homotopic to a path of the form $z \mapsto z^k$ for some $k \in \mathbb{Z}$.

When $p \neq q$, I believe we can do the same thing (i.e. classify paths by the number of "turns" they do around the circle): take the "usual" path between $p$ and $q$ on $S^1$, that is: $$\gamma:[0,1] \to S^1, \gamma(t) = \frac{tq+(1-t)p}{|tq+(1-t)p|}. $$ This is a path that does no "turns" around $S^1$, but how would a path that does $k$ "turns" around $S^1$ look and like and how would we prove that any other path between $p$ and $q$ must be homotopic to $\gamma$ raised to some integer?

Edit: the above $\gamma$ doesn't work if $p$ and $q$ are antipodal points, i.e. $p = -q$, since $tq + (1-t)p = 0$ when $t = 1/2$ in that case. But we can consider the paths between $p$ and another point $s \in S^1$ and then the path between $s$ and $q$.

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  • $\begingroup$ Your "usual path" from $p$ to $q$ is not well-defined since the denominator becomes zero for antipodal points. Furthermore, what does "raising a path to a power" mean? $\endgroup$
    – Christoph
    Mar 1, 2020 at 9:25
  • $\begingroup$ @Christoph Yes you are right, I noticed this and edited the question. By raising to a power $k \in \mathbb{Z}$ I mean the function $\gamma^k: t \mapsto (\gamma(t))^k$. $\endgroup$
    – John
    Mar 1, 2020 at 9:26
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    $\begingroup$ But $p^k\neq p$ and $q^k\neq q$ in general, so the powers are not even paths from $p$ to $q$. $\endgroup$
    – Christoph
    Mar 1, 2020 at 9:28
  • $\begingroup$ Oh yes, you are right, I made a mistake. $\endgroup$
    – John
    Mar 1, 2020 at 9:29

1 Answer 1

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Try to prove the following:

Two paths $\gamma_1,\gamma_2\colon I\to X$ from $p$ to $q$ are homotopic relative the endpoints if and only if the loop $\gamma_1*\overline{\gamma_2}$ at $p$ is null-homotopic (relative the basepoint).

Here $\overline{\gamma_2}$ denotes the reversed path of $\gamma_2$ and $*$ denotes concatenation of paths.

From this it then follows that the homotopy class of a path $\gamma\colon I\to S^1$ relative the endpoints consists of all paths $\gamma'$ such that the loop $\gamma*\overline{\gamma'}$ has winding number $0$.


The idea of going from a homotopy $H\colon I\times I\to X$ with $H(0,-)=\gamma_1*\overline{\gamma_2}$, $H(1,-)= p$ and $H(-,0)=H(-,1)=p$ to a homotopy from $\gamma_1$ to $\gamma_2$ is shown in the following picture. You can transform $\gamma_1$ into $\gamma_2$ like the blue lines show, keeping $\gamma(0)=p$ and $\gamma(1)=q$ at all times.

homotopy picture

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