2
$\begingroup$

Let be $ K:= \overline{K_1(0)} $ a closed unit disk in $ \mathbb{R^2} $.

I want to show that:

(i) There does not exist a continuous function $ f: K \rightarrow \partial K$, so that $f(x)=x ,\forall x \in \partial K $

(ii) Brouwer's fixed point theorem: Any continuous function $f: K \rightarrow K$ has a fixed point (by using (i) )

__

for (i) I need to to set a homotopy $H(t,s):= f(s \cos t, s \sin t )$ and $\omega = (x^2+y^2)^{-1} (-y dx+xdy))$ to prove contradiction. Do you have any idea for this?

for (ii) I have setted up a proof

$\endgroup$
  • 1
    $\begingroup$ This is the standard approach to proving the BFPT, and can be found in nearly any topology book or webpage about the theorem. $\endgroup$ – Randall Jan 22 at 12:33
  • $\begingroup$ $K$ is the closed unit disk. The closed unit circle is $S^1 = \partial K$. (i) is wrong, (ii) is only a rudiment of a statement. $\endgroup$ – Paul Frost Jan 22 at 12:52
  • $\begingroup$ @PaulFrost yes, sorry, I edited my question. $\endgroup$ – constant94 Jan 22 at 14:08
  • $\begingroup$ @Randall unfortunatly, I just found different approaches to the proof $\endgroup$ – constant94 Jan 22 at 14:09
  • 1
    $\begingroup$ Do you know what $\pi_1(\partial K)$ is ? Or maybe $H_1(\partial K; \mathbb{Z})$ ? $\endgroup$ – Max Jan 24 at 11:14
0
$\begingroup$

The map $f$, if it exists, will retract $K$ to $\partial K$.

Recall the following little lemma (Lemma 55.1, Munkres).

Lemma: If A is a retract of X, then the homomorphism of fundamental groups induced by inclusion $i:A\to X$ is injective.

Now it is clear what one should do. Observe that fundamental group of $\partial K$ is $\mathbb{Z}$, while that of $K$ is trivial. There is no way to give an injective homomorphism from $\mathbb{Z}$ to trivial group. Which proves that such a map $f$ is not possible.

For the second problem, assume that $f$ does not have a fixed point. This means for every $x$ and $f(x)$ are two distinct points. Great! If you have two distinct points in the disk, you can draw a line through them and hit the boundary. That’s exactly what you do, start from $f(x)$ and go to the boundary passing through $x$. Check that this gives you a continuous function from $K$ to its boundary which fixes the boundary. But, we just proved such a map is not possible.

$\endgroup$

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.