Brouwer Fixed Point Theorem - Non Zero Homotopy Class

Question

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.

Answer 1

You have a composition of maps $$ S^1 \stackrel{\iota}{\longrightarrow} D^2 \stackrel{r}{\longrightarrow} S^1 $$ where $\iota$ is the inclusion of $S^1$ into $D^2$. This composition is the identity map on $S^1$. So, it should give the identity map on $\pi_1(S^1)$. Since $\pi_1(D^2) = 0$, this is impossible.

Answer 2

The idea here can perhaps be made clearer by seperating the proof in to three steps:

step 1: $\pi_1(S^1)=\mathbb Z$

I will omit the proof for this statement.

step 2: no retract theorem

There is no map $r:D^2 \to S^1$ so that $r \circ i=id$, where $i$ is inclusion, and $id$ is identity. Suppose that there were. Then the induced morphisms on the fundamental group would be $i_*:\mathbb Z \to \pi_1 (D^2)$ and $r_*:\pi_1(D^2) \to \mathbb Z$ with the additional property that $r_* \circ i_*: \mathbb Z \to \mathbb Z$ should be identity, but $\pi_1(D^2)=0$, so this composition is impossible. [For clarity's sake, I replaced $\pi_1(S^1)$ with $\mathbb Z$ in the previous statement]

step 3: fixed point theorem:

Using the map defined by hatcher, notice that if there were no fixed point, the map $r$ would be a retract, which is a contradiction.