Show that the function $f$ is well-defined and continuous.

Question

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 $$ f(\mathbf{x}) = \begin{cases} r(\mathbf{x}) & \text{if } x \in \overline{U_{i}} \\ \mathbf{x} & \text{if } x \in \mathbb{R}^2 \setminus U_{i} \end{cases} $$

Show that the function $f$ is well-defined and continuous.

My previous knowledge is:

I know that $\partial U_{i}$ is either all of $J$(where $J$ is a Jordan curve) or it is contained in an arc $A \subseteq J.$

And this question explains some prior knowledge:

Show that $\partial U_{i}$ is either all of $J$ or it is contained in an arc $A \subseteq J.$

My questions are:

1-Should not the $\mathbf{x}$ and the $x$ in the question be the same?

2- I know that the retraction is a continuous function by definition and I know that the identity function is also a continuous function. Does this implies that our $f$ is continuous? But it is a piecewise function. Could anyone help me with the proof of continuity please?

3-I know that to show that a function is well-defined you have to start with $x=y$ and ends up with $f(x) = f(y)$.... I am not sure how to implement this in our case. Could anyone help me with this proof also?

EDIT:

I was given this hint for the solution of this problem : $\overline{U_{i}}$ and $X \setminus U_{i}$ are both closed in $\mathbb{R}^2$

Answer 1

1. Yes, $\mathbf x$ and $x$ are the same. All we see is what you typeset, but if your source typeset them the same way, they differ only because someone goofed.
2. No, being pieced together from two continuous pieces does not by itself prove that $f$ is continuous. $$h(t) = \begin{cases}1 & t \le 0\\0 & t > 0\end{cases}$$ is pieced together from continuous pieces, but is not continuous. You need to address continuity separately on the interiors of the pieces (where it does follow from continuity of the piece functions) and on the boundary, where you need to show that the limit converges to the function value, no matter from which side the boundary point is approached.
3. "Well-defined" means that for each point $x$ in the domain, $f(x)$ has one and only one value assigned to it by the definition. You have definitions on $\overline{U_i}$ and on $\Bbb R^2 \setminus U_i$. What you need to show is that for each point in $\overline{U_i} \cap (\Bbb R^2 \setminus U_i)$, these two definitions agree.

Note that you need to show "well-defined" before you examine continuity, since you can hardly show that the limit as you approach the boundary is the function value at the boundary before you have established there is a function value at the boundary. (Besides which, in this case, knowing that the two pieces agree on the boundary is critical to proving continuity there.)