how to prove the fundamental group isomorphic?

Question

Suppose $f:S^1 \rightarrow S^1$ is continuous and has no fixed point.

How can we prove that $f_*:\pi_1(X,x_0)\rightarrow\pi_1(X,f(x_0))$ is a group isomorphism where $f_*([\sigma]) = [f \circ \sigma]$ for all $[\sigma] \in \pi_1(X,x_0)$?

Answer 1

I presume that $X$ is $S^1$.

Model $S^1$ by $\Bbb R/\Bbb Z$ and take $x_0$ to be the image of $0$. The map $f:\Bbb R/\Bbb Z\to\Bbb R/\Bbb Z$ induces $F:\Bbb R\to\Bbb R/\Bbb Z$. As $\Bbb R$ is simply connected, and also is the universal cover of $\Bbb R/\Bbb Z$, then $F$ lifts to $\tilde F:\Bbb R\to \Bbb R$. As $f$ has no fixed points, then $G(x)=\tilde F(x)-x$ is never an integer. But $\tilde F(1)-\tilde F(0)$ is an integer and so $G(1)-G(0)$ is an integer. By continuity $G(x)$ always lies in some interval $(m,m+1)$ ($m\in \Bbb Z$) and so $G(1)=G(0)$ that is $\tilde F(1)=\tilde F(0)+1$. Now $f$ is homotopic to the identity map by a homotopy $H:(x,t)\mapsto (1-t)\tilde F(x)-tx$. The crucial thing here is that $H(x+1,t)=H(x,t)$. As $f$ is homotopic to the identity map, it induces an isomorphism on the fundamental groups.