Notes:Quotient
Terminology
Let X be a set and let ∼ be an equivalence relation on the elements of X.
- Then X∼ denotes the "equivalence classes" of
This is best thought of as a map:
- π:X→X∼ by π:x↦[x] where recall:
- [a]={x∈X|x∼a}, the notation [a] makes sense, as by the reflexive property of ∼ we have a∈[a]
Quotient structure
Suppose that ⊙:X×X→X is any map, and writing x⊙y:=⊙(x,y) when does ⊙ induce an 'equivalent' mapping on X∼?
- This is a mapping: ⊙:X∼×X∼→X∼ where [x]⊙[y]=[x⊙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 ⊙ 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 x⊙y
- Then go forward again to [x⊙y]
In functional terms we may say:
- ⊙:X∼×X∼→X∼ given by:
- ⊙([x],[y])↦π(π−1([x])⊙π−1([y])⏟if ⊙ makes sense)=[π−1([x])⊙π−1([y])]
Here π−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⊙B (for A⊆X and B⊆X) if a⊙b∼a′⊙b′
As then
- We can define π(A) (for A⊆X) properly if ∀x∈A∀y∈A[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:
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Diagram |
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It is quite simple really:
- The dashed arrow exists by function composition.
- 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∼×X∼→X∼ given by: ⊙:=(π∘+)∘(π×π)−1
- [(π×π)(v,v′)=(π×π)(u,u′)]⟹[π(+(v,v′))=π(+(u,u′))] then
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.