# Free and based homotopy of self-maps of $S^1$

I would like understand how this task is meant:

"Show that for self-maps of $S^1$ free and based homotopy are the same."

My interpretation would be that if we take arbitrary $f, g: S^1 \rightarrow S^1$ with $f(1) = g(1) = 1 \in S^1$, then free homotopy of $f,g$ would imply based homotopy relatively to $1 \in S^1$. But this can not be true, since $S^1$ is path connected (so that all such maps $f,g$ are homotopic, but not relatively homotopic, since $\Pi(S^1,1) = \mathbb{Z})$.

• The pathconnectedness of $S^1$ tells you no such thing. For instance, just compare the identity $S^1 \to S^1$ with any constant mapping $S^1 \to S^1$. What you are presumably thinking of is the fact that any two paths (not loops) in a pathconnected space are homotopic. – Mees de Vries Oct 24 '16 at 20:24
• Okay, thanks. But I still cannot solve the task, assuming my interpretation which I've given above is right. We get a homotopy F which transforms $f$ into $g$, and using the path-connectedness of $S^1$ we can obtain paths $a_t$ which connect $F(1, t)$ with $1 \in S^1$ for every $t$, but I do not see how to combine these two things to obtain a based homotopy. – CHwC Oct 24 '16 at 22:15
• My interpretation of the question would be: Given self maps $f,g:S^1\rightarrow S^1$, that are homotopic through a free homotopy $h$, then in fact they are homotopic through a based homotopy $\widetilde{h}$ (which you should be able to construct from $h$). Maybe you also need $f$ and $g$ to be based self-maps of $S^1$. – Jack Davies Oct 26 '16 at 8:36