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I don't understand the proof of Brouwer fixed-point theorem:

Every continuous function $f:D^2\rightarrow D^2$ has a fixed point, i.e. $\exists x, f(x)=x$

Here is our professor's proof:

Assume not then exists a retraction $r$ with

$r:D^2\rightarrow S^1$, $\quad$$r(x)=(1-t)f(x)+tx$ and $||r(x)||=1$

then; $r|_{S^1}=id_{S^1}$

$i:S^1\rightarrow D^2$, $\quad$$x\mapsto x$$\quad$($i$ is the inclusion i think)

$r_*:\pi_1(D^2,x_0)\rightarrow \pi_1(S^1,x_0)$

$i_*:\pi_1(S^1,x_0)\rightarrow \pi_1(D^2,x_0)$

So,

$id=r\circ i\Longrightarrow id_*=(r\circ i)_*=r_*\circ i_*$

$\pi_1(S^1,x_0)\overset{i_*}\longrightarrow \pi_1(D^2,x_0)\overset{r_*}\longrightarrow\pi_1(S^1,x_0)$

$\mathbb Z\rightarrow 0\rightarrow\mathbb Z$

$\textbf{What have we reached now ?}$

-if we compare the domains, $id_*=(r\circ i)_*$ can't be true or what ?

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    $\begingroup$ The domain of $r_\ast$ is trivial, hence the image is trivial too. But $\pi_1(S^1,x_0)$ isn't trivial. $\endgroup$ Dec 23, 2013 at 14:41

2 Answers 2

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Because the final map goes through 0, its image must be 0, but the final map is supposed to be the identity on $\mathbb{Z}$, which can't be the zero map.

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Notice that the map induced by the identity should be the identity itself (in the fundamental group), and you concluded that the identity is the composition of a $0$ map with the map induced by the retraction. There is no homomorphism $r_*$ with that property (consider the homomorphisms of the $0$ group).

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