# Relative Homotopy is Equivalence Relation

## Theorem

Let $X$ and $Y$ be topological spaces.

Let $K \subseteq X$ be a (possibly empty) subset of $X$.

Let $\map \CC {X, Y}$ be the set of all continuous mappings from $X$ to $Y$.

Define a relation $\sim$ on $\map \CC {X, Y}$ as:

- $f \sim g$ if and only if $f$ and $g$ are homotopic relative to $K$.

Then $\sim$ is an equivalence relation.

## Proof

We examine each condition for equivalence.

### Reflexivity

For any function $f: X \to Y$, define $H: X \times \closedint 0 1 \to Y$ by $\map H {x, t} := \map f x$.

This yields a homotopy between $f$ and itself.

Also, trivially, if $x \in K$ and $t \in \closedint 0 1$, then:

- $\map f x = H \left({x, t}\right)$

so that $H$ is a homotopy relative to $K$.

Thus $\sim$ is a reflexive relation.

$\Box$

### Symmetry

Given a $K$-relative homotopy:

- $H: X \times \closedint 0 1 \to Y$

from $\map f x = \map H {x, 0}$ to $\map g x = \map H {x, 1}$, the function:

- $\map G {x, t} = \map H {x, 1 - t}$

is a $K$-relative homotopy from $g$ to $f$.

Thus $\sim$ is a symmetric relation.

$\Box$

### Transitivity

Suppose that $f \sim g$ and $g \sim h$.

Let $F, G: X \times \closedint 0 1 \to Y$ be $K$-relative homotopies between $f$ and $g$, $g$ and $h$, respectively.

Define $H: X \times \closedint 0 1 \to Y$ by:

- $\map H {x, t} := \begin {cases} \map F {x, 2 t} & : 0 \le t \le \dfrac 1 2 \\ \map G {x, 2 t - 1} & : \dfrac 1 2 \le t \le 1 \end{cases}$

By Continuous Mapping on Finite Union of Closed Sets, $H$ is a $K$-relative homotopy between $f$ and $h$.

Thus $\sim$ is a transitive relation.

$\Box$

Having verified all three conditions, it follows that $\sim$ is an equivalence relation.

$\blacksquare$