# Homotopy is an equivalence relation on the set of all continuous maps between spaces/proof

## Proof

To be an equivalence relation we must show:[ilmath]\newcommand{\homo}[2]{#1\simeq #2\ (\text{rel }A)} [/ilmath]

- For all [ilmath]f\in C^0(X,Y)[/ilmath] that [ilmath]f\simeq f\ (\text{rel }A)[/ilmath], symbolically:
- Reflexive: [ilmath]\forall f\in C^0(X,Y)[\homo{f}{f}][/ilmath]

- if [ilmath]f\simeq g\ (\text{rel }A)[/ilmath] then [ilmath]g\simeq f\ (\text{rel }A)[/ilmath], symbolically:
- Symmetric: [ilmath]\forall f,g\in C^0(X,Y)[\homo{f}{g}\implies\homo{g}{f}][/ilmath]

- If [ilmath]\homo{f}{g} [/ilmath] and [ilmath]\homo{g}{h} [/ilmath] then [ilmath]\homo{f}{h} [/ilmath], symbolically:
- Transitive: [ilmath]\forall f,g,h\in C^0(X,Y)\left[\big(\homo{f}{g}\wedge\homo{g}{h}\big)\implies\homo{f}{h}\right][/ilmath]

Where we are given topological spaces, [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath], and also an arbitrary subset of [ilmath]X[/ilmath], [ilmath]A\in\mathcal{P}(X)[/ilmath]

### Reflexive property

Let [ilmath]f\in C^0(X,Y)[/ilmath] be given. We want to show that [ilmath]\homo{f}{f} [/ilmath].

- Define a map, [ilmath]H:X\times I\rightarrow Y[/ilmath] by [ilmath]H:(x,t)\mapsto f(x)[/ilmath].
- If we show [ilmath]H[/ilmath] is a homotopy [ilmath](\text{rel }A)[/ilmath] we have exhibited a homotopy between [ilmath]f[/ilmath] and itself, thus showing that [ilmath]f[/ilmath] is homotopic to [ilmath]f[/ilmath] [ilmath](\text{rel }A)[/ilmath]

### Symmetric property

Let [ilmath]f,g\in C^0(X,Y)[/ilmath] be given and suppose that [ilmath]H:\homo{f}{g} [/ilmath]^{[Note 1]} is also given. We want to show [ilmath]\homo{g}{f} [/ilmath]

- Define a map, [ilmath]H':X\times I\rightarrow Y[/ilmath] as [ilmath]H':(x,t)\mapsto H(x,1-t)[/ilmath].
- If this map is a homotopy [ilmath](\text{rel }A)[/ilmath] whose initial stage is [ilmath]g[/ilmath] and whose final stage is [ilmath]f[/ilmath] we are done.

### Transitive property

Let [ilmath]f,g,h\in C^0(X,Y)[/ilmath] be given and suppose that [ilmath]F:\homo{f}{g} [/ilmath] and [ilmath]G:\homo{g}{h} [/ilmath] are also given. We want to show that [ilmath]\homo{f}{h} [/ilmath]

- NOTE TO READERS: I've yet to do this page but this is the only non-trivial result, to answer the question:
- Define a map [ilmath]H:X\times I\rightarrow Y[/ilmath] as [ilmath]H:(x,t)\mapsto\left\{\begin{array}{lr}F(x,2t) & \text{if }t\in[0,\frac{1}{2}]\\ G(x,2t-1)&\text{if }t\in[\frac{1}{2},1]\end{array}\right.[/ilmath] (the [ilmath]t=\frac{1}{2}[/ilmath] case is a deliberate use of notation to draw attention to the fact that the maps are the same there, see pasting lemma) and showing that is a homotopy

## Notes

- ↑ Recall that if:
- [ilmath]\homo{f}{g} [/ilmath] denotes that [ilmath]f[/ilmath] and [ilmath]g[/ilmath] are homotopic [ilmath](\text{rel }A)[/ilmath]

- [ilmath]H:\homo{f}{g} [/ilmath] denotes that [ilmath]f[/ilmath] and [ilmath]g[/ilmath] are homotopic [ilmath](\text{rel }A)[/ilmath] and that we refer to any such homotopy between them by the letter [ilmath]H[/ilmath]. So [ilmath]H[/ilmath] is the map [ilmath]H:X\times I\rightarrow Y[/ilmath].
- [ilmath]H[/ilmath]'s initial stage is [ilmath]f[/ilmath] and [ilmath]H[/ilmath]'s final stage is [ilmath]g[/ilmath]