# Brouwer's fixed point theorem $n=1$ case

I have read a sketch of the proof for Brouwer's fixed-point theorem, i.e. that every continuous $f:\mathbb{D}^n\to\mathbb{D}^n$ has a fixed-point ($n\ge1$).

The idea is to construct a retract $r:\mathbb{D}^n\to\mathbb{S}^{n-1}$ such that $r\circ i=id_{\mathbb{S}^{n-1}}$. Taking $H_{n-1}$ yields the contradiction $[\mathbb{Z}\to 0\to\mathbb{Z}]=id_{\mathbb{Z}}$.

But as far as I see this only works for $n>1$, not for $n=1$, since $H_0(\mathbb{D}^1)=\mathbb{Z}\neq 0$.

So what can I do in the $n=1$ case?

• There still does not exist a retract $[-1,1]\to \{-1,1\}$. Or use the Intermediate Value Theorem. – Hagen von Eitzen May 16 '17 at 15:06
• Use reduced homology $\tilde H_0$ rather than $H_0$ itself. – anomaly May 16 '17 at 15:08
• Is $\tilde{H}_{n-1}$ still a covariant functor? – user369147 May 16 '17 at 15:11
• Are you asking how to show that a continuous function $f:[-1,1] \to [-1,1]$ has a fixed point? – copper.hat May 16 '17 at 15:27
• $\mathbb{S}^0$ consists of two points, so $\tilde{H}_0 (\mathbb{S}^0) \cong \mathbb{Z}$. And $\mathbb{D}^1$ is connected, thus $\tilde{H}_0 (\mathbb{D}^1) = 0$. – user144221 May 16 '17 at 15:39

Basically just expanding on von Eitzen's comment here if you don't see it immediately.

For $$n=1$$, $$f:[-1,1]\rightarrow[-1,1]$$ and $$f$$ is continuous.

Define $$g:[-1,1]\rightarrow[-2,2]$$ such that $$g=f(x)-x$$. Note that $$g$$ is also continuous.

Now if either $$g(1)$$ or $$g(-1)$$ is equal to zero, we are done.

If not, $$g(-1)<0$$ and $$g(1)>0$$.

By the intermediate value theorem, $$\exists c\in (-1,1)$$ such that $$g(c)=0$$.

$$g(c)=0\implies f(x)=x$$.