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?

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

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


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