# Passing to the quotient (function)

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## 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:
• [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

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

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.