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

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this is rather messy, but it is better than nothing. Currently I'm not sure on the phrasing (eg: "[ilmath]f[/ilmath] is homotopic to [ilmath]g[/ilmath] is an equivalence relation" or "[ilmath]f[/ilmath] and [ilmath]g[/ilmath] being homotopic defines an equivalence relation"). It needs to be cleaned up, but it'll do for now

## Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces. Let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath]. Then:

• the relation that relates a continuous map, [ilmath]f:X\rightarrow Y[/ilmath] to another continuous map, [ilmath]g:X\rightarrow Y[/ilmath], if [ilmath]f[/ilmath] and [ilmath]g[/ilmath] are homotopic is an equivalence relation. Symbolically
• Let [ilmath]C^0(X,Y)[/ilmath] denote the set of all continuous maps of the form [ilmath](:X\rightarrow Y)[/ilmath]
• we define a relation, [ilmath]\mathcal{R}\subseteq C^0(X,Y)\times C^0(X,Y)[/ilmath] given by
• [ilmath]\forall f,g\in C^0(X,Y)[(f,g)\in\mathcal{R}\iff[f\simeq g\ (\text{rel }A)]][/ilmath] - recall: [ilmath]f\simeq g\ (\text{rel }A)[/ilmath] denotes that the maps [ilmath]f[/ilmath] and [ilmath]g[/ilmath] are homotopic relative to [ilmath]A[/ilmath]
• Then we claim [ilmath]\mathcal{R} [/ilmath] is an equivalence relation

We will denote this not by [ilmath]f\mathcal{R}g[/ilmath] (as is usual with relations) but by:

1. [ilmath]f\simeq g\ (\text{rel }A)[/ilmath] or
2. [ilmath]f\simeq g[/ilmath] if [ilmath]A=\emptyset[/ilmath]

## Proof

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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Check the proof!

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

1. 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]
2. 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]
3. 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].

### 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]