Difference between revisions of "Passing to the quotient (function)"

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(Reclassified as theorem, cleaned up formatting, cleaned up writing style. Moved diagram into subpage, added explanation, corrected notation, basically redone it all!)
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*: So it is easy to see that we require {{M|1=[w(x)=w(y)]\implies[f(x)=f(y)]}} in order to proceed.
 
*: So it is easy to see that we require {{M|1=[w(x)=w(y)]\implies[f(x)=f(y)]}} in order to proceed.
 
==Proof of claims==
 
==Proof of claims==
 +
* To see that '''if {{M|f}} is surjective so is {{M|\tilde{f} }} see my notes here:
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** [[Talk:Passing_to_the_quotient_(function)/Induced_is_surjective_iff_function_is_surjective]]
 
{{Requires proof|grade=A|msg=Most of the proofs are done, I've done the surjective one like 3 times (CHECK THE TALK PAGE! SO YOU DON'T DO IT A FOURTH!) Also:
 
{{Requires proof|grade=A|msg=Most of the proofs are done, I've done the surjective one like 3 times (CHECK THE TALK PAGE! SO YOU DON'T DO IT A FOURTH!) Also:
 
* Move the proofs into sub-pages. It is just so much neater!}}
 
* Move the proofs into sub-pages. It is just so much neater!}}
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{{End Proof}}
 
{{End Proof}}
 
{{End Theorem}}
 
{{End Theorem}}
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==See also==
 
==See also==
 
* [[Passing to the quotient]] - disambiguation page
 
* [[Passing to the quotient]] - disambiguation page

Latest revision as of 20:49, 11 October 2016

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See Passing to the quotient for a disambiguation of this term.

Statement

[math] \begin{xy} \xymatrix{ X \ar[r]^w \ar[dr]_f & W \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy} [/math]
[ilmath]f[/ilmath] passing to the quotient
Given a function, [ilmath]f:X\rightarrow Y[/ilmath] and another function, [ilmath]w:X\rightarrow W[/ilmath][Note 1] then "[ilmath]f[/ilmath] may be factored through [ilmath]w[/ilmath]" if[1]:
  • [ilmath]f[/ilmath] is constant on the fibres of [ilmath]w[/ilmath][Note 2]

If this condition is met then [ilmath]f[/ilmath] induces a mapping, [ilmath]\tilde{f}:W\rightarrow Y [/ilmath], such that [math]f=\tilde{f}\circ w[/math] (equivalently, the diagram on the right commutes).

  • [ilmath]\tilde{f}:W\rightarrow X[/ilmath] may be given explicitly as: [ilmath]\tilde{f}:v\mapsto f(w^{-1}(v))[/ilmath][Note 3]
    • We may also write [ilmath]\tilde{f}=f\circ w^{-1}[/ilmath] but this is a significant abuse of notation and should be avoided! It is safe to use here because of the "well-defined"-ness of [ilmath]\tilde{f} [/ilmath]

We may then say:

  • "[ilmath]f[/ilmath] may be factored through [ilmath]w[/ilmath] to [ilmath]\tilde{f} [/ilmath]" or "[ilmath]f[/ilmath] descends to the quotient via [ilmath]w[/ilmath] to give [ilmath]\tilde{f} [/ilmath]"

Claims:

  1. [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] is given unambiguously by [ilmath]\tilde{f}:v\mapsto f(w^{-1}(v))[/ilmath]
  2. If [ilmath]w:X\rightarrow W[/ilmath] is surjective then [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] is unique - the only function [ilmath](:W\rightarrow Y)[/ilmath] such that the diagram commutes
  3. If [ilmath]f:X\rightarrow Y[/ilmath] is surjective then [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] is surjective also

Caveats

The following are good points to keep in mind when dealing with situations like this:

  • Remembering the requirements:
    We want to induce a function [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] such that all the information of [ilmath]f[/ilmath] is "distilled" into [ilmath]w[/ilmath], notice that:
    • if [ilmath]w(x)=w(y)[/ilmath] then [ilmath]\tilde{f}(w(x))=\tilde{f}(w(y))[/ilmath] just by composition of functions, regardless of [ilmath]\tilde{f} [/ilmath]!
    • so if [ilmath]f(x)\ne f(y)[/ilmath] but [ilmath]w(x)=w(y)[/ilmath] then we're screwed and cannot use this.
    So it is easy to see that we require [ilmath][w(x)=w(y)]\implies[f(x)=f(y)][/ilmath] in order to proceed.

Proof of claims

Grade: A
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Most of the proofs are done, I've done the surjective one like 3 times (CHECK THE TALK PAGE! SO YOU DON'T DO IT A FOURTH!) Also:
  • Move the proofs into sub-pages. It is just so much neater!

Claim: the induced function, [ilmath]\tilde{f} [/ilmath] exists and is given unambiguously by [ilmath]\tilde{f}:v\mapsto f(w^{-1}(v))[/ilmath]

Existence

Let [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] be given by: [ilmath]f:v\mapsto f(w^{-1}(v))[/ilmath] - I need to prove this is a Function
This means I must check it is well defined, a function must associate each point in its domain with exactly 1 element of its codomain
Let [ilmath]v\in W[/ilmath] be given
Let [ilmath]a\in w^{-1}(v)[/ilmath] be given
Let [ilmath]b\in w^{-1}(v)[/ilmath] be given
We know [math]\forall a\in w^{-1}(v)[/math] that [math]w(a)=v[/math] by definition of [math]w^{-1}[/math]
This means [math]w(a)=w(b)[/math]
But by hypothesis [math]w(a)=w(b)\implies f(a)=f(b)[/math]
So [math]f(a)=f(b)[/math]
Thus given an [ilmath]a\in w^{-1}(v)[/ilmath], [math]\forall b\in w^{-1}[f(a)=f(b)][/math]
We now know (formally) that: (given a [ilmath]v[/ilmath]) [math]\exists y\in Y\forall a\in w^{-1}(v)[f(a)=y][/math] - notice the [math]\exists y[/math] comes first. We can uniquely define [math]f(w^{-1}(v))[/math]
Since [ilmath]v\in W[/ilmath] was arbitrary we know [math]\forall v\in W\exists y\in Y\forall a\in w^{-1}(v)[f(a)=y][/math]
We have now shown that [math]\tilde{f}[/math] can be well defined (as the function that maps a [ilmath]v\in W[/ilmath] to a [ilmath]y\in Y[/ilmath].
To calculate [math]\tilde{f}(v)[/math] we may choose any [math]a\in w^{-1}(v)[/math] and define [math]\tilde{f}(v)=f(a)[/math] - we know [math]f(a)[/math] is the same for whichever [math]a\in w^{-1}(v)[/math] we choose.
So we know the function [math]\tilde{f}:W\rightarrow Y[/math] given by [math]\tilde{f}:x\mapsto f(w^{-1}(x))[/math] exists


This completes the proof[2]

Claim: if [ilmath]w[/ilmath] is surjective then the induced [ilmath]\tilde{f} [/ilmath] is unique

Uniqueness

Suppose another function exists, [math]\tilde{f}':W\rightarrow Y[/math] that isn't the same as [math]\tilde{f}:W\rightarrow Y[/math]
That means [math]\exists u\in W:[\tilde{f}(u)\ne\tilde{f}'(u)][/math]
  • Note, as [ilmath]w:X\rightarrow W[/ilmath] is surjective, that [ilmath]\exists x'\in X[w(x')=u][/ilmath]
However for both [ilmath]\tilde{f} [/ilmath] and [ilmath]\tilde{f}'[/ilmath] we have the property of [math]f=\tilde{f}\circ w=\tilde{f}'\circ w[/math] so:
By hypothesis we have: [ilmath]\forall x\in X[f(x)=\tilde{f}(w(x))=\tilde{f}'(w(x))][/ilmath] however we know:
  • [ilmath]\exists x'\in X[w(x')=u][/ilmath] and [ilmath]\tilde{f}(u)\ne \tilde{f}'(u)[/ilmath], this means:
    • [ilmath]f(x')=\tilde{f}(w(x'))\ne\tilde{f}'(w(x'))[/ilmath] - which contradicts the hypothesis.
However if [ilmath]w[/ilmath] is not surjective, then the parts of the domain on which [ilmath]\tilde{f} [/ilmath] and [ilmath]\tilde{f}'[/ilmath] disagree on may never actually come up; that is to say:
  • [ilmath]\forall x\in X[\tilde{f}(w(x))=\tilde{f}'(w(x))][/ilmath] as [ilmath]w:X\rightarrow W[/ilmath] may never take an [ilmath]x\in X[/ilmath] to a [ilmath]z\in W[/ilmath] where [ilmath]\tilde{f}(z)[/ilmath] and [ilmath]\tilde{f}'(z)[/ilmath] differ; but they could still be different functions.


This completes the proof[2]

Notes:
  1. Notice that if [ilmath]w[/ilmath] is not surjective, the point(s) on which [ilmath]\tilde{f} [/ilmath] and [ilmath]\tilde{f}'[/ilmath] disagree on may never actually come up, so it is indeed not-unique if [ilmath]w[/ilmath] isn't surjective.


See also

TODO: Factoring a map through the canonical projection of the equivalence relation it generates

Notes

  1. ↑ I have chosen [ilmath]W[/ilmath] to mean "whatever"
  2. ↑ We can state this in 2 other equivalent ways:
    1. [math]\forall x,y\in X[w(x)=w(y)\implies f(x)=f(y)][/math]
    2. [math]\forall x,y\in X[f(x)\ne f(y)\implies w(x)\ne w(y)][/math]
    See equivalent conditions to being constant on the fibres of a map for proofs and more details
  3. ↑ Of course, only bijections have inverse functions, we indulge in the common practice of using [ilmath]w^{-1}(v)[/ilmath] to mean [ilmath]w^{-1}(\{v\})[/ilmath], in general for sets [ilmath]A[/ilmath] and [ilmath]B[/ilmath] and a mapping [ilmath]f:A\rightarrow B[/ilmath] we use [ilmath]f^{-1}(C)[/ilmath] to denote (for some [ilmath]C\in\mathcal{P}(B)[/ilmath] (a subset of [ilmath]X[/ilmath])) the pre-image of [ilmath]C[/ilmath] under the function [ilmath]f[/ilmath], [ilmath]f^{-1}(C):=\{a\in A\ \vert\ f(a)\in C\}[/ilmath]. Just as for [ilmath]D\in\mathcal{P}(A)[/ilmath] (a subset of [ilmath]A[/ilmath]) we use [ilmath]f(D)[/ilmath] to denote the image of [ilmath]D[/ilmath] under [ilmath]f[/ilmath], namely: [ilmath]f(D):=\{f(d)\in B\ \vert\ d\in D\}[/ilmath]
    Caution: Writing [ilmath]\tilde{f}:v\mapsto f(w^{-1}(v))[/ilmath] is dangerous as it may not be "well-defined"
    A function (considered as a relation) of the form [ilmath]f:X\rightarrow Y[/ilmath] must associate every [ilmath]x\in X[/ilmath] with exactly one [ilmath]y\in Y[/ilmath].

    Suppose that [ilmath]w^{-1}(v)[/ilmath] is empty or contains 2 (or more!) elements, then what do we define [ilmath]\tilde{f} [/ilmath] as?

    As it turns out it doesn't matter, but is really important to see why we must be so careful! This is why we require [ilmath]f[/ilmath] to be constant on the fibres of [ilmath]w[/ilmath], as if we have [ilmath]w(x)=w(y)[/ilmath] but [ilmath]f(x)\ne f(y)[/ilmath] then no function composed with [ilmath]w[/ilmath] can ever be equal to [ilmath]f[/ilmath]!

    • Suppose that [ilmath]g:W\rightarrow Y[/ilmath] is such that [ilmath]f=g\circ w[/ilmath], then [ilmath]f(x)=g(w(x))[/ilmath], and we have [ilmath]f(x)\ne f(y)[/ilmath], then:
      • [ilmath]w(x)=w(y)[/ilmath] so we must have [ilmath]g(w(x))=g(w(y))[/ilmath], so we must have [ilmath]f(x)=f(y)[/ilmath]! A contradiction!

    Lastly note the alternate forms of the "constant on fibres" (in the note above) is very similar to the definition of a function being injective

    TODO: Develop that last thought

References

  1. ↑ Alec's own work, "distilled" from passing to the quotient (topology) which is defined by Mond (2013, Topology) and Lee (Intro to Top manifolds), by further abstracting the claim
  2. ↑ 2.0 2.1 This is my (Alec's) own work


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