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

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^[1]

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^[1]

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.