Difference between revisions of "Notes:Quotient"
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| + | __TOC__ | ||
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| + | ==Terminology== | ||
Let {{M|X}} be a set and let {{M|\sim}} be an [[equivalence relation]] on the elements of {{M|X}}. | Let {{M|X}} be a set and let {{M|\sim}} be an [[equivalence relation]] on the elements of {{M|X}}. | ||
* Then {{M|\frac{X}{\sim} }} denotes the "[[equivalence class|equivalence classes]]" of {{M|~}} | * Then {{M|\frac{X}{\sim} }} denotes the "[[equivalence class|equivalence classes]]" of {{M|~}} | ||
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* We can say {{M|A\odot B}} (for {{M|A\subseteq X}} and {{M|B\subseteq X}}) if {{M|a\odot b\sim a'\odot b'}} | * We can say {{M|A\odot B}} (for {{M|A\subseteq X}} and {{M|B\subseteq X}}) if {{M|a\odot b\sim a'\odot b'}} | ||
As then | As then | ||
| − | * We can define {{M|\pi(A)}} (for {{M|A\subseteq X }}) properly if {{ | + | * We can define {{M|\pi(A)}} (for {{M|A\subseteq X }}) properly if {{MM|1=\forall x\in A\forall y\in A[\pi(x)=\pi(y)] }} |
This all seems very contrived | This all seems very contrived | ||
===As a diagram=== | ===As a diagram=== | ||
| − | I seem to be asking when a map ( | + | I seem to be asking when a map (dotted line) is induced such that the following diagram commutes: |
{| class="wikitable" border="1" | {| class="wikitable" border="1" | ||
|- | |- | ||
| style="font-size:1.4em;" | | | style="font-size:1.4em;" | | ||
| − | {{MM|1=\begin{xy}\xymatrix{X\times X \ar@{->}[r]^-\odot \ar@{->}[d]_-{\pi\times\pi}& X \ar@{->}[d]^-{\pi} \\ | + | {{MM|1=\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} }} | \frac{X}{\sim}\times\frac{X}{\sim} \ar@{.>}[r]_-{\odot} & \frac{X}{\sim} }\end{xy} }} | ||
|- | |- | ||
! Diagram | ! Diagram | ||
|} | |} | ||
| + | It is quite simple really: | ||
| + | # The dashed arrow exists by function composition. | ||
| + | # Using the [[Factor (function)]] idea, if we have (for {{M|(v,v')\in V\times V}} and {{M|(u,u')\in V\times V}} - from wanting the diagram to commute): | ||
| + | #* {{M|1=[(\pi\times\pi)(v,v')=(\pi\times\pi)(u,u')]\implies[\pi(+(v,v'))=\pi(+(u,u'))]}} then | ||
| + | #** there exists a unique function, {{M|\odot:\frac{X}{\sim}\times\frac{X}{\sim}\rightarrow\frac{X}{\sim} }} given by: {{M|1=\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 {{M|1=x=(\pi\times\pi)^{-1}(a,b)}} we have {{M|\odot(x)}} being the same, it doesn't matter what element of {{M|(\pi\times\pi)^{-1} }} we take. | ||
Latest revision as of 09:38, 24 November 2015
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 (dotted 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:
- The dashed arrow exists by function composition.
- 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]
- [ilmath][(\pi\times\pi)(v,v')=(\pi\times\pi)(u,u')]\implies[\pi(+(v,v'))=\pi(+(u,u'))][/ilmath] then
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.
