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

  • 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 '19 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 '19 at 12:52
  • $\begingroup$ @PaulFrost yes, sorry, I edited my question. $\endgroup$
    – constant94
    Jan 22 '19 at 14:08
  • $\begingroup$ @Randall unfortunatly, I just found different approaches to the proof $\endgroup$
    – constant94
    Jan 22 '19 at 14:09
  • 1
    $\begingroup$ Do you know what $\pi_1(\partial K)$ is ? Or maybe $H_1(\partial K; \mathbb{Z})$ ? $\endgroup$ Jan 24 '19 at 11:14

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.


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