Doubt in Brouwer fixed point theorem.

Question

Consider $S^1 =\{(x,y)\in \mathbb{R}^2 : x^2 + y^2 =1\}$ , $D=\{(x,y)\in \mathbb{R}^2: x^2 + y^2 \leq 1\}$

How to show that If $f : D \to S^1$ is a continuous mapping, then there exists $x \in S^1$ such that $f(x)=x$.

From here (Which of the following statements are correct? (topology)) I know that by Brouwer fixed point theorem, as $S^1 \subset D$ there exists a point $x \in D$ such that $f(x) = x$.

But how to ensure that $x \in S^1$?

Answer 1

Notice that $f(x) \in S^1$ by definition, so the $x$ from Brouwer's fixed point must satisfy $x \in S^1$.

Answer 2

In order to use Brouwer Fixed Point Theorem, you must have $f: X \rightarrow X$, $f$ continuous, $X$ a compact (as $D$). If you note that $S^1 \subset D$, you can apply the theorem.

As the point $x$ is fixed, it must be a point of $S^1$ (in fact, all the points of $D \backslash S^1$ are applied in a point of $S^1$ and, consequently, can't remain fixed by $f$).