This is my main question:
Use the function $f$ to show that $S^1$ is a retract of $\mathbb{R}^2.$ Deduce that $\partial U_{i} = J.$ Where $J$ is a Jordan curve.
Here is the function $f$:
Let $U_{i}$ be a bounded path component of $X = \mathbb{R}^2 \setminus J,$ and assume $\partial U_{i} \neq J.$ Choose a point $c \in U_{i}.$ Use the retraction $r,$ we define the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 - \{c\} $ by the rule $$ f(\mathbf{x}) = \begin{cases} r(\mathbf{x}) & \text{if } \mathbf{x} \in \overline{U_{i}} \\ \mathbf{x} & \text{if } \mathbf{x} \in \mathbb{R}^2 \setminus U_{i} \end{cases} $$
where $r$ is a retraction from $\mathbb{R}^2$ to any arc $A$ of the normal space $\mathbb{R}^2.$
Could anyone give me a hint for the solution please? I see that this function will let me reach a contradiction.
Also, I was given an alternate choice either to solve the above question or to solve the following question: Show that the composition $S^1 \xrightarrow {j_{R}} \mathbb{R}^2 \xrightarrow {f} \mathbb{R}^2 \setminus \{c\}$ is a homotopy equivalence. Deduce that $\partial U_{i} = J.$
where $j_{R} : S^1 \rightarrow \mathbb{R}^2$ given by $f(\mathbf{x}) = R. \mathbf{x},$
Where $R$ is the radius of the ball centered at 0 that contained the Jordan curve $i.e. J \subseteq B_{R}(\mathbf{0})$ (I proved this before (with the help of many generous people on this site))
I just mentioned this second question to explain what I maybe doing in the first one. I am allowed to solve anyone of them. But because I do not have a clear vision of the solution of either I posted them both.
Also,my first question is related to this one Show that the function $f$ is well-defined and continuous.
My ultimate goal is to prove Jordan Curve Theorem the following statement of it:
If $J \subseteq \mathbb{R}^2$ is a jordan curve, then
$(a) \mathbb{R}^2 \setminus J$ has two path components.
$(b)$ one of them is bounded (the inside of the curve) and the other one is unbounded (the outside of the curve), and
$(c)$ the boundary of each component is precisely $J.$