Homotopically Equivalent Topological Spaces

Definition: Let $X$ and $Y$ be topological spaces. Then $X$ and $Y$ are said to be Homotopically Equivalent if there exists continuous functions $f : X \to Y$ and $g : Y \to X$ such that $g \circ f = \mathrm{id}_X$ and $f \circ g = \mathrm{id}_Y$.

For example, let $D^2$ denote 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 $p$ be any point in $\mathbb{R}^2$.

Let $X = D^2$ and $Y = \{ p \}$. We claim that $X$ homotopically equivalent to $Y$, i.e., that the closed unit disk is homotopically equivalent to a single point in $\mathbb{R}^2$. Define $f : X \to Y$ for all $(x, y) \in X = D^2$ by:

(2)

\begin{align} \quad f(x, y) = p \end{align}

And define $g : Y \to X$ by:

(3)

\begin{align} \quad g(p) = (0, 0) \end{align}

Then $f$ is continuous as it is a constant function, and $g$ is trivially continuous. Furthermore:

(4)

\begin{align} \quad (g \circ f)(x, y) = g(f(x, y)) = g(p) = (0, 0) \end{align}

And:

(5)

\begin{align} \quad (f \circ g)(p) = f(g(p)) = f(0,0) = p \end{align}

So $g \circ f = (0, 0)$ and $f \circ g = \mathrm{id}_Y$.

We have almost shown that $X$ is homotopically equivalent to $Y$. The only problem is that $g \circ f \neq \mathrm{id}_X$. We instead show that $g \circ f$ is homotopic to $\mathrm{id}_X$.

Let $H : X \times I \to Y$ be defined by:

(6)

\begin{align} \quad H(x, t) = (1 - t)(x, y) \end{align}

Then $H$ is clearly a continuous function, $H_0 = H(x, 0) = 1(x, y) = \mathrm{id}_X$, and $H_1 = H(x, 1) = 0(x, y) = (0, 0) = g \circ f$. So indeed, $g \circ f$ is homotopic to $\mathrm{id}_X$. So $D^2$ is homotopically equivalent to $\{ p \}$.