# Homotopy classes relative endpoints of the circle

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 $$q$$.

• 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? Mar 1, 2020 at 9:25
• @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 $\gamma^k: t \mapsto (\gamma(t))^k$.
– John
Mar 1, 2020 at 9:26
• But $p^k\neq p$ and $q^k\neq q$ in general, so the powers are not even paths from $p$ to $q$. Mar 1, 2020 at 9:28
• Oh yes, you are right, I made a mistake.
– John
Mar 1, 2020 at 9:29

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.