----- > [!proposition] Proposition. ([[homotopy class]]) > The [[homotopy]] and [[path homotopy]] relations $\simeq$ and $\simeq_{p}$ on a [[topological space]] $X$ are [[equivalence relation|equivalence relations]], the [[equivalence class|equivalence classes]] of which are called **homotopy classes**. > [!proof]- Proof. ([[homotopy class]]) > Let $f, f', f''$ be [[continuous]] maps between two [[topological space]]s $X,Y$. > > **Reflexivity.** We have $f \simeq f$, the [[homotopy]] being $\begin{align} > F: X \times [0,1] & \to Y \\ > (x,t) & \mapsto f(x). > \end{align}$ > Clearly $F(x,0)=f(x)$ and $F(x,1)=f(x)$ as required. If $f$ is a [[parameterized curve]], then $F$ is a [[path homotopy]] because we further have $F(0,t)=f(0)$ and $F(1,t)=f(1)$ for all $t$. > > **Symmetry.** Suppose $f \simeq f'$; we want to show $f' \simeq f$. Since $f \simeq f'$, there exists [[continuous]] $\begin{align}F: X \times [0,1] & \to Y \\\end{align}$ such that $F(x,0)=f(x)$ and $F(x,1)=f'(x)$. We now 'reversely' define $\begin{align} > F': X \times [0,1] & \to Y \\ > (x,t) & \mapsto F(x, 1-t). > \end{align}$ > Observe that $F'(x,0)=F(x, 1)=f'(x)$, while $F'(x,1)=F(x,0)=f(x)$. And $F'$ is [[continuous]], since it equals the composition $F \circ g$, where $g:X \times [0,1] \to X \times [0,1]$ is given by $g(x,t):=(x,1-t)$ ($g$ is [[continuous]] due to [[continuous on product space iff continuous on coordinates]]). > If we further have that $f$ and $f'$ are [[parameterized curve|paths]] with common endpoints $x_{0}$ and $x_{1}$, and $F$ is a [[path homotopy]] between them, then $F'$ is a [[path homotopy]] too: observe $F'(0, t)=F(0, 1-t)=x_{0}$ > and $F'(1, t)=F(1, 1-t)=x_{1}.$ > > **Transitivity.** Suppose $f \simeq f'$ and $f' \simeq f''$. We want to show that $f \simeq f''$. [[the pasting lemma|The pasting lemma]] is useful. Let $F$ be a [[homotopy]] between $f$ and $f'$, and let $F'$ be a [[homotopy]] between $f'$ and $f''$. Define $G:X \times [0,1] \to Y$ by $G(x,t)= \begin{cases} > F(x,2t) & t \in \left[ 0, \frac{1}{2} \right] \\ > F'(x, 2t-1) & t \in \left[ \frac{1}{2}, 1 \right]; > \end{cases}$ > then $G$ well-defined because $F\left( x,2 \cdot \frac{1}{2} \right)=f'(x)=F'(x, 0)=F'\left(x, 2 \cdot \frac{1}{2} -1 \right)$ > and since $G$ is [[continuous]] on the two [[closed set|closed subsets]] $X \times \left[ 0, \frac{1}{2} \right]$ and $X \times \left[ \frac{1}{2}, 1 \right]$ of $X \times I$, it is [[continuous]] on all of $X \times I$ by [[the pasting lemma]]. > > If we further have that if $F$ and $F'$ are [[path homotopy|path homotopies]] then so is $G$. See picture below (page 324 of munkres). ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```