Notes:Quotient

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Terminology

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

This is best thought of as a map:

  • π:XX by π:x[x] where recall:
    • [a]={xX|xa}, the notation [a] makes sense, as by the reflexive property of we have a[a]

Quotient structure

Suppose that :X×XX is any map, and writing xy:=(x,y) when does induce an 'equivalent' mapping on X?

  • This is a mapping: :X×XX where [x][y]=[xy]
    • 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 on X, 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 xy
  • Then go forward again to [xy]

In functional terms we may say:

  • :X×XX given by:
    ([x],[y])π(π1([x])π1([y])if  makes sense)=[π1([x])π1([y])]

Here π1([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 AB (for AX and BX) if abab

As then

  • We can define π(A) (for AX) properly if xAyA[π(x)=π(y)]

This all seems very contrived

As a diagram

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

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)V×V and (u,u)V×V - from wanting the diagram to commute):
    • [(π×π)(v,v)=(π×π)(u,u)][π(+(v,v))=π(+(u,u))] then
      • there exists a unique function, :X×XX given by: :=(π+)(π×π)1

Note that while technically these are not functions (as they'd have to be bijective in this case) as FOR ANY x=(π×π)1(a,b) we have (x) being the same, it doesn't matter what element of (π×π)1 we take.