# Brouwer Fixed Point Theorem - Non Zero Homotopy Class

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.

• "Anyway retraction r is nothing but identity map"... The retraction is $D^2 \mapsto S^1$ is not the identity map. The domain and range of an identity map must be equal. – Lee Mosher May 27 '17 at 14:44
• en.wikipedia.org/wiki/Retract this says it is identity @LeeMosher – user3001408 May 27 '17 at 14:49
• No, it says given a space $X$ and a subset $A$, a function $f : X \to A$ is a retraction if the restriction of $f$ to $A$ is the identify function on $A$, in other words the function $f \bigm| A$ is equal to the identify function $\text{Id}_A$ on the set $A$. The definition of a retraction applies to many important situations where $A$ is a proper subset of $X$, for example in your question where $X=D^2$ and $A=S^1$, and in such a situation the restricted function $f \bigm| A = \text{Id}_A$ is not the same as the given function $f : X \to A$. – Lee Mosher May 27 '17 at 15:24

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.

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.