Retract Subspaces of a Topological Space

## Retract Subspaces of a Topological Space

 Definition: Let $X$ be a topological space and let $A \subset X$ be a topological subspace. Then $A$ is said to be a Retract of $X$ if there exists a continuous function $r : X \to A$ called a Retraction Map such that $r \circ \mathrm{in} = \mathrm{id}_A$.

Here, $\mathrm{in} : A \to X$ is the Inclusion Map, which sends each point $a \in A$ to $a \in X$. Then $r \circ \mathrm{in} : A \to A$.

Intuitively, a subspace $A$ is a retract of $X$ if $X$ can be continuously changed to become $A$ by fixing all of the points in $A$.

For example, let $D^2$ be the closed unit disk in $\mathbb{R}^2$. That is:

(1)
\begin{align} \quad D^2 = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \} \end{align}

And let $\displaystyle{\frac{1}{2}D^2}$ be the closed disk centered at the origin with radius $\frac{1}{2}$ in $\mathbb{R}^2$, that is:

(2)
\begin{align} \quad \frac{1}{2}D^2 = \left \{ (x, y) \in \mathbb{R}^2 : x^2 + y^1 \leq \frac{1}{4} \right \} \end{align}

We claim that $\displaystyle{\frac{1}{2}D^2}$ is a retraction of $D^2$. Define a retraction map $r : X \to A$ by:

(3)
$$r(x, y) = (a, b)$$

Where $(a, b)$ is such that $d((x, y), (a, b)) = \mathrm{inf}_{(c, d) \in \frac{1}{2}D^2} d((x, y), (c, d))$. In other words, each point $(x, y) \in D^2$ is mapped to the point $(a, b)$ in $\displaystyle{\frac{1}{2}D^2}$ whose distance from $(x, y)$ is minimized. Clearly $r$ is a continuous map. Furthermore, $r \circ \mathrm{in} = \mathrm{id}_A$. So indeed, $\displaystyle{\frac{1}{2}D^2}$ is a retract of $D^2$.

Below is a visual representation of the retract:

 Theorem 1: Let $X$ be a topological space and let $a \in X$. Then $\{ a \}$ is a retract of $X$.
• Proof: Consider the function $r : X \to \{ a \}$ defined for all $x \in X$ by $r(x) = b$. Then $r$ is trivially a continuous function. Furthermore, for the inclusion map $in : \{ a \} \to X$ defined by $in(a) = a$, we have that:
(4)
\begin{align} \quad r \circ in (a) = r(a) = a \end{align}
• Hence [[$r \circ \in = \mathrm{id}_{\{ a \}} • So$\{ a \}$is a retract of$X$.$\blacksquare\$