Let $f:S^1\rightarrow S^1$ be a map such that $f(-x)=-f(x)$ for all $x\in S^1$. Show that the induced homomorphism $f_\ast:\pi_1(S^1,x_0)\rightarrow\pi_1(S^1,f(x_0))$ is non-trivial.
What I know: 1. a trivial homomorphism maps everything to the identity; in this case $[c_{x_0}]$
- The induced homomorphism is defined by $f_\ast([\omega])=[f\circ\omega]$
- $f_\ast$ is trivial iff $f_\ast':\pi_1(X,-x_0)\rightarrow\pi_1(X,-f(x_0))$ is trivial
So what we have is a function which maps antipodal points to antipodal points, and we need to show that the induced homomorphism maps all loops around $x_0$ to the point $f(x_0)$. I started my attempt with a proof by contradiction, but I wasn't sure where the contradiction would come in.