An Example of a Retract that is NOT a Deformation Retract

Let $X$ be a topological space and let $A \subseteq X$ be a topological subspace. Recall the following definitions:

• On the Retract Subspaces of a Topological Space page we said that $A$ is a retract of $X$ if there exists a continuous function (called a retract) $r : X \to A$ such that:

(1)

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

• On the Deformation Retract Subspaces of a Topological Space page we said that $A$ is a deformation retract of $X$ if there exists a continuous function $r : X \to A$ such that:

(2)

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

We are now able to give an example of two topological spaces $A$ and $X$ where $A$ is a retract of $X$ but such that $A$ is not a deformation retract of $X$.

Let $X = S^1\times S^1$ be the torus and let $A \subseteq X$ be the equatorial circle of $X$ depicted below:

Showing that $A$ is a retract of $X$:

Define a function $r : X \to A$ as follows. Cut $X$ up into vertical circle sections. Then each circle intersects $A$ uniquely. Define $r(x)$ to be the unique point $a \in A$ for which the unique circle intersects both $x$ and $a$.

Alternatively, recall that the torus $X$ can be identified as a quotient topological space by taking $[0, 1] \times [0, 1]$ and identified opposite edges of the square in the same orientation. Then the circle $A$ can be viewed as a horizontal line segment in $[0, 1] \times [0, 1]$ and we map every point in $[0, 1] \times [0, 1]$ to its vertical projection onto this line segment:

It is easy to see that $r$ is a continuous map. This is because if $U$ is an open subset of $X$, then $r^{-1}(U)$ will be a union of open intervals in the equatorial circle. So $r^{-1}(U)$ is open. Furthermore, we have that $r(a) = a$ for all $a \in A$, and so:

(3)

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

So $A$ is a retract of $X$.

Showing that $A$ is NOT a deformation retract of $X$.

Observe that:

(4)

\begin{align} \quad \pi_1(X, x) = \mathbb{Z} \times \mathbb{Z} \end{align}

(5)

\begin{align} \quad \pi_1(A, x) = \mathbb{Z} \end{align}

These fundamental groups are not isomorphic and so $A$ cannot be a deformation retract of $X$.