I am attaching the problem statement and proof of Brouwer Fixed Point Theorem (from Algebraic Topology by Hatcher). I am a newbie in Algebraic Topology.
And here is the proof.
I am confused on how $\pi_1(S^1)$ is zero, if $r$ is such a map as defined in the proof. Because clearly, $\pi_1(S^1)$ contains atleast one homotopy class which is defined as $f_t(s)$. Anyway retraction r is nothing but identity map, also $S_1 \subset D^2$. Hence all $rf_t(s)$ is nothing but $f_t(s)$. And this is a valid homotopy and should be a member of $\pi_1(S^1)$. But how the proof claims that $\pi_1(S^1)$ is zero if all the arguments in the proof are true (hence the contradiction). What I am getting wrong here.