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) )
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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
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.