Notes:Quotient

Terminology

Let $X$ be a set and let $\sim$ be an equivalence relation on the elements of $X$.

• Then $\frac{X}{\sim}$ denotes the "equivalence classes" of $~$

This is best thought of as a map:

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

Quotient structure

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

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

Alternatively

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

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

In functional terms we may say:

• $\odot:\frac{X}{\sim}\times\frac{X}{\sim}\rightarrow\frac{X}{\sim}$ given by:

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

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

• We can say $A\odot B$ (for $A\subseteq X$ and $B\subseteq X$) if $a\odot b\sim a'\odot b'$

As then

• We can define $\pi(A)$ (for $A\subseteq X$) properly if

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:

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

It is quite simple really:

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

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