Passing to the quotient (function)

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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] then "[ilmath]f[/ilmath] may be factored through [ilmath]w[/ilmath]" if:

• [ilmath]f[/ilmath] is constant on the fibres of [ilmath]w[/ilmath]

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]
  • 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]

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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

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

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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

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

• Passing to the quotient - disambiguation page
• Equivalent conditions to being constant on the fibres of a map

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

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