# Passing to the quotient (function)

(Redirected from Function factorisation)
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
This page is waiting for a final review, then this notice will be removed.
See Passing to the quotient for a disambiguation of this term.

## Statement

 [ilmath]f[/ilmath] passing to the quotient $\begin{xy} \xymatrix{ X \ar[r]^w \ar[dr]_f & W \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy}$
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 $f=\tilde{f}\circ w$ (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

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 $\forall a\in w^{-1}(v)$ that $w(a)=v$ by definition of $w^{-1}$
This means $w(a)=w(b)$
But by hypothesis $w(a)=w(b)\implies f(a)=f(b)$
So $f(a)=f(b)$
Thus given an [ilmath]a\in w^{-1}(v)[/ilmath], $\forall b\in w^{-1}[f(a)=f(b)]$
We now know (formally) that: (given a [ilmath]v[/ilmath]) $\exists y\in Y\forall a\in w^{-1}(v)[f(a)=y]$ - notice the $\exists y$ comes first. We can uniquely define $f(w^{-1}(v))$
Since [ilmath]v\in W[/ilmath] was arbitrary we know $\forall v\in W\exists y\in Y\forall a\in w^{-1}(v)[f(a)=y]$
We have now shown that $\tilde{f}$ can be well defined (as the function that maps a [ilmath]v\in W[/ilmath] to a [ilmath]y\in Y[/ilmath].
To calculate $\tilde{f}(v)$ we may choose any $a\in w^{-1}(v)$ and define $\tilde{f}(v)=f(a)$ - we know $f(a)$ is the same for whichever $a\in w^{-1}(v)$ we choose.
So we know the function $\tilde{f}:W\rightarrow Y$ given by $\tilde{f}:x\mapsto f(w^{-1}(x))$ 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, $\tilde{f}':W\rightarrow Y$ that isn't the same as $\tilde{f}:W\rightarrow Y$
That means $\exists u\in W:[\tilde{f}(u)\ne\tilde{f}'(u)]$
• 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 $f=\tilde{f}\circ w=\tilde{f}'\circ w$ 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.

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. $\forall x,y\in X[w(x)=w(y)\implies f(x)=f(y)]$
2. $\forall x,y\in X[f(x)\ne f(y)\implies w(x)\ne w(y)]$
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. This is my (Alec's) own work