proving property ( and Brouwer's fixed point theorem)

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

• This is the standard approach to proving the BFPT, and can be found in nearly any topology book or webpage about the theorem. – Randall Jan 22 at 12:33
• $K$ 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. – Paul Frost Jan 22 at 12:52
• @PaulFrost yes, sorry, I edited my question. – constant94 Jan 22 at 14:08
• @Randall unfortunatly, I just found different approaches to the proof – constant94 Jan 22 at 14:09
• Do you know what $\pi_1(\partial K)$ is ? Or maybe $H_1(\partial K; \mathbb{Z})$ ? – Max Jan 24 at 11:14

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