Fixed point of interior of closed disk

Question

Let $D = \{(x,y)\in \mathbb{R}^2: x^2 + y^2 \leq 1 \}$. Let $A \subset \mathrm{int}D$. Let $A$ be connected and compact and let $D \setminus A$ be connected. Let $f:A \longrightarrow A$ be a continuous function and let $g:D \longrightarrow D$ be also continuous function such that $g_{|A} = f$.$\\$ Does it imply that there exists $x \in \mathrm{int}D$ such that $g(x)=x$?

Answer 1

Brouwer (https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem) states, that $g$ already has a fixed point in the convex closure of $A$ (or the smallest closed ball containing $A$). Since $A$ lies in the interior of $D$, also the convex closure lays in the interior (or you deform $A$ continuously into something convex and don't need the convex closure), hence the fix point lays in the interior of $D$.