Notes:Quotient

Terminology

Let [ilmath]X[/ilmath] be a set and let [ilmath]\sim[/ilmath] be an equivalence relation on the elements of [ilmath]X[/ilmath].

• Then [ilmath]\frac{X}{\sim} [/ilmath] denotes the "equivalence classes" of [ilmath]~[/ilmath]

This is best thought of as a map:

• [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] by [ilmath]\pi:x\mapsto [x][/ilmath] where recall:
  • [ilmath][a]=\{x\in X\vert x\sim a\}[/ilmath], the notation [ilmath][a][/ilmath] makes sense, as by the reflexive property of [ilmath]\sim[/ilmath] we have [ilmath]a\in[a][/ilmath]

Quotient structure

Suppose that [ilmath]\odot:X\times X\rightarrow X[/ilmath] is any map, and writing [ilmath]x\odot y:=\odot(x,y)[/ilmath] when does [ilmath]\odot[/ilmath] induce an 'equivalent' mapping on [ilmath]\frac{X}{\sim} [/ilmath]?

• This is a mapping: [ilmath]\odot:\frac{X}{\sim}\times\frac{X}{\sim}\rightarrow\frac{X}{\sim} [/ilmath] where [ilmath][x]\odot[y]=[x\odot y][/ilmath]
  • should such an operation be 'well defined' (which means it doesn't matter what representatives we pick of [ilmath][x][/ilmath] and [ilmath][y][/ilmath] in the computation)

Alternatively

We have no concept of [ilmath]\odot[/ilmath] on [ilmath]\frac{X}{\sim} [/ilmath], but we do on [ilmath]X[/ilmath]. The idea is that:

• Given a [ilmath][x][/ilmath] and a [ilmath][y][/ilmath] we go back
• To an [ilmath]x[/ilmath] and a [ilmath]y[/ilmath] representing those classes.
• Compute [ilmath]x\odot y[/ilmath]
• Then go forward again to [ilmath][x\odot y][/ilmath]

In functional terms we may say:

• [ilmath]\odot:\frac{X}{\sim}\times\frac{X}{\sim}\rightarrow\frac{X}{\sim} [/ilmath] given by:

  [ilmath]\odot([x],[y])\mapsto\pi(\underbrace{\pi^{-1}([x])\odot\pi^{-1}([y])}_{\text{if }\odot\text{ makes sense} })=\Big[\pi^{-1}([x])\odot\pi^{-1}([y])\Big][/ilmath]

Here [ilmath]\pi^{-1}([x])[/ilmath] is a subset of [ilmath]X[/ilmath] containing exactly those things which are equivalent to [ilmath]x[/ilmath] (as these things all map to [ilmath][x][/ilmath]).

• We can say [ilmath]A\odot B[/ilmath] (for [ilmath]A\subseteq X[/ilmath] and [ilmath]B\subseteq X[/ilmath]) if [ilmath]a\odot b\sim a'\odot b'[/ilmath]

As then

• We can define [ilmath]\pi(A)[/ilmath] (for [ilmath]A\subseteq X [/ilmath]) properly if [math]\forall x\in A\forall y\in A[\pi(x)=\pi(y)][/math]

This all seems very contrived

As a diagram

I seem to be asking when a map (dashed line) is induced such that the following diagram commutes:

[math]\begin{xy}\xymatrix{X\times X \ar@{->}[r]^-\odot \ar@{->}[d]_-{\pi\times\pi} \ar@{-->}[dr]^{\pi\circ\odot} & X \ar@{->}[d]^-{\pi} \\ \frac{X}{\sim}\times\frac{X}{\sim} \ar@{.>}[r]_-{\odot} & \frac{X}{\sim} }\end{xy}[/math]
Diagram

It is quite simple really:

1. The dashed arrow exists by function composition.
2. Using the Factor (function) idea, if we have (for [ilmath](v,v')\in V\times V[/ilmath] and [ilmath](u,u')\in V\times V[/ilmath] - from wanting the diagram to commute):
   • [ilmath][(\pi\times\pi)(v,v')=(\pi\times\pi)(u,u')]\implies[\pi(+(v,v'))=\pi(+(u,u'))][/ilmath] then
     • there exists a unique function, [ilmath]\odot:\frac{X}{\sim}\times\frac{X}{\sim}\rightarrow\frac{X}{\sim} [/ilmath] given by: [ilmath]\odot:=(\pi\circ+)\circ(\pi\times\pi)^{-1}[/ilmath]

Note that while technically these are not functions (as they'd have to be bijective in this case) as FOR ANY [ilmath]x=(\pi\times\pi)^{-1}(a,b)[/ilmath] we have [ilmath]\odot(x)[/ilmath] being the same, it doesn't matter what element of [ilmath](\pi\times\pi)^{-1} [/ilmath] we take.