Passing to the quotient (function)

Definition

Given a function, [ilmath]f:X\rightarrow Y[/ilmath] and another function, [ilmath]w:X\rightarrow W[/ilmath] (I have chosen [ilmath]W[/ilmath] to mean "whatever") we can say:

[ilmath]f[/ilmath] may be factored through [ilmath]w[/ilmath]

if [ilmath]f[/ilmath] and [ilmath]w[/ilmath] are such that:

• [math]\forall x,y\in X[w(x)=w(y)\implies f(x)=f(y)][/math]

  (this is the same as: [math]\forall x,y\in X[f(x)\ne f(y)\implies w(x)\ne w(y)][/math])

Then [ilmath]f[/ilmath] induces a function, [ilmath]\tilde{f} [/ilmath] such that [math]f=\tilde{f}\circ w[/math], or more simply that the following diagram commutes:

[math] \begin{xy} \xymatrix{ X \ar[r]^w \ar[dr]_f & W \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy} [/math]
Diagram

Note:

1. [ilmath]\tilde{f} [/ilmath] may be explicitly written as [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] by [ilmath]\tilde{f}:v\mapsto f(w^{-1}(v))[/ilmath]
   • Or indeed [ilmath]\tilde{f}:=f\circ w^{-1}[/ilmath]
   • This is actually an abuse of notation as [ilmath]w^{-1}(x\in W)[/ilmath] is a subset of [ilmath]X[/ilmath], however it is safe to use it because (as is proved below) [ilmath]f[/ilmath] of any element of [ilmath]w^{-1}(x\in W)[/ilmath] for a given [ilmath]x[/ilmath] is the same.
2. The function [ilmath]\tilde{f} [/ilmath] is unique if [ilmath]w[/ilmath] is surjective

Points to remember

• Remembering the requirements:

  We want to induce a function [ilmath]\tilde{f}:W\rightarrow Y[/ilmath] - if [ilmath]w(x)=w(y)[/ilmath] then [ilmath]\tilde{f}(w(x))=\tilde{f}(w(y))[/ilmath] just by composition.

  If [ilmath]f(x)\ne f(y)[/ilmath] we're screwed in this case. So it is easy to see that we must have [ilmath][w(x)=w(y)]\implies[f(x)=f(y)][/ilmath] otherwise we cannot proceed.

Proof of claims

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] (and as [ilmath]w[/ilmath] is surjective [ilmath]\exists x\in X[p(x)=u][/ilmath])

Both [ilmath]\tilde{f} [/ilmath] and [ilmath]\tilde{f}'[/ilmath] have the property of [math]f=\tilde{f}\circ w=\tilde{f}'\circ w[/math] so:

by hypothesis, for all [ilmath]x[/ilmath] however, we know [ilmath]\tilde{f} [/ilmath] and [ilmath]\tilde{f}'[/ilmath] don't agree over their entire domain, the [ilmath]p(x)[/ilmath] they do not agree on violate this property (as [ilmath]f[/ilmath] cannot be two things for a given [ilmath]x[/ilmath])

This contradicts that [ilmath]\tilde{f} [/ilmath] and [ilmath]\tilde{f}'[/ilmath] are different

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.