1
$\begingroup$

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 ?

$\endgroup$
  • 3
    $\begingroup$ The domain of $r_\ast$ is trivial, hence the image is trivial too. But $\pi_1(S^1,x_0)$ isn't trivial. $\endgroup$ – Daniel Fischer Dec 23 '13 at 14:41
2
$\begingroup$

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).

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.