Math Online's 2500th Page - 9:31 PM CST on March 12th, 2017

Basic Theorems Regarding Deformation Retracts

Recall from the Deformation Retract Subspaces of a Topological Space page that if $X$ is a topological space and $A \subseteq X$ is a topological subspace then $A$ is said to be a deformation 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 \quad \mathrm{and} \quad \mathrm{in} \circ r = \mathrm{id}_X \end{align}

We will now look at some basic results regarding deformation retracts.

Theorem 1: Let $A$, $B$, and $X$ be topological spaces where $A \subseteq B \subseteq X$. If $A$ is a deformation retract of $B$ and $B$ is a deformation retract of $X$ then $A$ is a deformation retract of $X$.

• Proof: Since $A$ is a deformation retract of $B$ there exists a continuous function $r_1 : B \to A$ such that:

(2)

\begin{align} \quad r_1 \circ \mathrm{in} = \mathrm{id}_A \quad \mathrm{and} \quad \mathrm{in} \circ r_1 = \mathrm{id}_B \end{align}

• Similarly, since $B$ is a deformation retract of $X$ there exists a continuous function $r_2 : X \to B$ such that:

(3)

\begin{align} \quad r_2 \circ \mathrm{in} = \mathrm{id}_B \quad \mathrm{and} \quad \mathrm{in} \circ r_2 = \mathrm{id}_X \end{align}

• Let $r = r_1 \circ r_2 : X \to A$. Then $r$ is continuous since it is a composition of two continuous functions. Furthermore, we have that:

(4)

\begin{align} \quad r \circ \mathrm{in} = (r_1 \circ r_2) \circ \mathrm{in} = r_1 \circ \mathrm{id}_B |_A = r_1 \circ \mathrm{id}_A = \mathrm{id}_A \end{align}

(5)

\begin{align} \quad \mathrm{in} \circ r = \mathrm{in} \circ (r_1 \circ r_2) = \mathrm{id}_B \circ r_2 = \mathrm{id}_X \end{align}

• Therefore $A$ is a deformation retract of $X$. $\blacksquare$