The Retract Theorem for Subspaces of a Topological Space

# The Retract Theorem for Subspaces of a Topological Space

Recall from the Retract Subspaces of a Topological Space page that if $X$ is a topological space and $A \subset X$ is a topological subspace of $X$ then $A$ is said to be a retract of $X$ if there exists a continuous function $r : X \to A$ such that:

(1)\begin{align} \quad r \circ \mathrm{in} = \mathrm{id}_A \end{align}

Where $\mathrm{in}_A : A \to X$ is the inclusion map.

Lemma 1: Let $X$ be a topological space and let $A \subset X$ be a topological subspace of $X$. If $A$ is a retract of $X$ with retraction mapping $r : X \to A$ then:a) $r_* : \pi_1(X, a) \to \pi_1(A, a)$ is surjective.b) $\mathrm{in}_* : \pi_1(A, a) \to \pi_1(X, a)$ is injective. |

**Proof:**Suppose that $r : X \to A$ is a retraction mapping. Then:

\begin{align} \quad r \circ \mathrm{in} = \mathrm{id}_A \end{align}

- Therefore:

\begin{align} \quad r_* \circ \mathrm{in}_* &= \mathrm{id}_{A*}\\ &= \mathrm{id}_{\pi_1(A, a)*} \end{align}

- So $r_* \circ \mathrm{in}_*$ is bijective which implies that $r_*$ is surjective and $\mathrm{in}_*$ is injective.

Theorem 2 (Retract Theorem for Subspaces of a Topological Space): Let $X$ be a topological space and let $A \subset X$ be a topological subspace of $X$. If $\pi_1(X, a)$ is simply connected and $\pi_1(A, a)$ is not simply connected then $A$ is NOT a retract of $X$. |

**Proof:**Suppose that $A$ is a retract of $X$. Let $r : X \to A$ be a retraction mapping. By Lemma 1 we have that $r_* : \pi_1(X, a) \to \pi_1(A, a)$ must be surjective. But $\pi_1(X, a)$ is the trivial group, and $\pi_1(A, a)$ is not the trivial group. There exists no surjection from a one element set to a set with more than one elements. Hence the assumption that $A$ was a retract of $X$ is false.

- So $A$ is not a retract of $X$. $\blacksquare$

For example, let $X = D^2$ be the closed unit disk and let $A = S^1$ be the circle. Then $A \subset X$. We have that $\pi_1(X, a)$ is the trivial group and $\pi_1(A, a) \cong \mathbb{Z}$. So the circle is not a retract of the closed unit disk.

Some caution must be taken when reading Theorem 2. The converse of Theorem 2 is not true in general.