Passing to the quotient (function)

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See Passing to the quotient for a disambiguation of this term.

Statement

$f$ 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,

$f:X\rightarrow Y$ and another function,

$w:X\rightarrow W$ ^{[Note 1]} then " $f$ may be factored through $w$ " if^[1]:

$f$ is constant on the fibres of $w$ ^{[Note 2]}

If this condition is met then $f$ induces a mapping, $\tilde{f}:W\rightarrow Y$ , such that $f=\tilde{f}\circ w$ (equivalently, the diagram on the right commutes).

˜f:W→X may be given explicitly as: ˜f:v↦f(w−1(v))^{[Note 3]}
- We may also write $\tilde{f}=f\circ w^{-1}$ 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 $\tilde{f}$

We may then say:

" $f$ may be factored through $w$ to $\tilde{f}$ " or " $f$ descends to the quotient via $w$ to give $\tilde{f}$ "

Claims:

$\tilde{f}:W\rightarrow Y$ is given unambiguously by $\tilde{f}:v\mapsto f(w^{-1}(v))$
If $w:X\rightarrow W$ is surjective then $\tilde{f}:W\rightarrow Y$ is unique - the only function $(:W\rightarrow Y)$ such that the diagram commutes
If $f:X\rightarrow Y$ is surjective then $\tilde{f}:W\rightarrow Y$ 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 ˜f:W→Y such that all the information of f is "distilled" into w, notice that:
- if $w(x)=w(y)$ then $\tilde{f}(w(x))=\tilde{f}(w(y))$ just by composition of functions, regardless of $\tilde{f}$ !
- so if $f(x)\ne f(y)$ but $w(x)=w(y)$ then we're screwed and cannot use this.
So it is easy to see that we require $[w(x)=w(y)]\implies[f(x)=f(y)]$ in order to proceed.

Proof of claims

To see that if $f$ is surjective so is $\tilde{f}$ see my notes here:
- Talk:Passing_to_the_quotient_(function)/Induced_is_surjective_iff_function_is_surjective

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!

[Expand]
Claim: the induced function, $\tilde{f}$ exists and is given unambiguously by $\tilde{f}:v\mapsto f(w^{-1}(v))$

Existence

Let

$\tilde{f}:W\rightarrow Y$ be given by:

$f:v\mapsto f(w^{-1}(v))$ - 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

$v\in W$ be given

Let

$a\in w^{-1}(v)$ be given

Let

$b\in w^{-1}(v)$ 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

$a\in w^{-1}(v)$ ,

$\forall b\in w^{-1}[f(a)=f(b)]$

We now know (formally) that: (given a

$v$ )

$\exists y\in Y\forall a\in w^{-1}(v)[f(a)=y]$ - notice the

$\exists y$ comes first. We can uniquely define

$f(w^{-1}(v))$

Since

$v\in W$ 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

$v\in W$ to a

$y\in Y$ .

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]

[Expand]
Claim: if $w$ is surjective then the induced $\tilde{f}$ 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 $w:X\rightarrow W$ is surjective, that $\exists x'\in X[w(x')=u]$

However for both

$\tilde{f}$ and

$\tilde{f}'$ we have the property of

$f=\tilde{f}\circ w=\tilde{f}'\circ w$ so:

By hypothesis we have:

$\forall x\in X[f(x)=\tilde{f}(w(x))=\tilde{f}'(w(x))]$ however we know:

∃x′∈X[w(x′)=u] and ˜f(u)≠˜f′(u), this means:
- $f(x')=\tilde{f}(w(x'))\ne\tilde{f}'(w(x'))$ - which contradicts the hypothesis.

However if

$w$ is not surjective, then the parts of the domain on which

$\tilde{f}$ and

$\tilde{f}'$ disagree on may never actually come up; that is to say:

$\forall x\in X[\tilde{f}(w(x))=\tilde{f}'(w(x))]$ as $w:X\rightarrow W$ may never take an $x\in X$ to a $z\in W$ where $\tilde{f}(z)$ and $\tilde{f}'(z)$ differ; but they could still be different functions.

This completes the proof^[2]

Notes:

Notice that if $w$ is not surjective, the point(s) on which $\tilde{f}$ and $\tilde{f}'$ disagree on may never actually come up, so it is indeed not-unique if $w$ isn't surjective.

Notes

Jump up ↑ I have chosen $W$ to mean "whatever"
Jump up ↑
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
Jump up ↑
Of course, only bijections have inverse functions, we indulge in the common practice of using w−1(v) to mean w−1({v}), in general for sets A and B and a mapping f:A→B we use f−1(C) to denote (for some C∈P(B) (a subset of X)) the pre-image of C under the function f, f−1(C):={a∈A | f(a)∈C}. Just as for D∈P(A) (a subset of A) we use f(D) to denote the image of D under f, namely: f(D):={f(d)∈B | d∈D}
[Expand]Caution: Writing ˜f:v↦f(w−1(v)) is dangerous as it may not be "well-defined"
A function (considered as a relation) of the form f:X→Y must associate every x∈X with exactly one y∈Y.
Suppose that $w^{-1}(v)$ is empty or contains 2 (or more!) elements, then what do we define $\tilde{f}$ 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 $f$ to be constant on the fibres of $w$ , as if we have $w(x)=w(y)$ but $f(x)\ne f(y)$ then no function composed with $w$ can ever be equal to $f$ !
- Suppose that $g:W\rightarrow Y$ is such that $f=g\circ w$ , then $f(x)=g(w(x))$ , and we have $f(x)\ne f(y)$ , then:
  $w(x)=w(y)$ so we must have $g(w(x))=g(w(y))$ , so we must have $f(x)=f(y)$ ! 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

Jump up ↑ 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
↑ ^{Jump up to: 2.0} ^2.1 This is my (Alec's) own work

[1] Jump up ↑ I have chosen $W$ to mean "whatever"

[3] Jump up ↑ We can state this in 2 other equivalent ways:
$\forall x,y\in X[w(x)=w(y)\implies f(x)=f(y)]$

$\forall x,y\in X[f(x)\ne f(y)\implies w(x)\ne w(y)]$
See equivalent conditions to being constant on the fibres of a map for proofs and more details

[3] $\forall x,y\in X[w(x)=w(y)\implies f(x)=f(y)]$

[4] $\forall x,y\in X[f(x)\ne f(y)\implies w(x)\ne w(y)]$

[4] Jump up ↑ Of course, only bijections have inverse functions, we indulge in the common practice of using $w^{-1}(v)$ to mean $w^{-1}(\{v\})$ , in general for sets $A$ and $B$ and a mapping $f:A\rightarrow B$ we use $f^{-1}(C)$ to denote (for some $C\in\mathcal{P}(B)$ (a subset of $X$ )) the pre-image of $C$ under the function $f$ , $f^{-1}(C):=\{a\in A\ \vert\ f(a)\in C\}$ . Just as for $D\in\mathcal{P}(A)$ (a subset of $A$ ) we use $f(D)$ to denote the image of $D$ under $f$ , namely: $f(D):=\{f(d)\in B\ \vert\ d\in D\}$
Caution: Writing $\tilde{f}:v\mapsto f(w^{-1}(v))$ is dangerous as it may not be "well-defined"
A function (considered as a relation) of the form $f:X\rightarrow Y$ must associate every $x\in X$ with exactly one $y\in Y$ .
Suppose that $w^{-1}(v)$ is empty or contains 2 (or more!) elements, then what do we define $\tilde{f}$ 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 $f$ to be constant on the fibres of $w$ , as if we have $w(x)=w(y)$ but $f(x)\ne f(y)$ then no function composed with $w$ can ever be equal to $f$ !

Suppose that $g:W\rightarrow Y$ is such that $f=g\circ w$ , then $f(x)=g(w(x))$ , and we have $f(x)\ne f(y)$ , then:
$w(x)=w(y)$ so we must have $g(w(x))=g(w(y))$ , so we must have $f(x)=f(y)$ ! 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

[6] Suppose that $g:W\rightarrow Y$ is such that $f=g\circ w$ , then $f(x)=g(w(x))$ , and we have $f(x)\ne f(y)$ , then:
$w(x)=w(y)$ so we must have $g(w(x))=g(w(y))$ , so we must have $f(x)=f(y)$ ! A contradiction!

[7] $w(x)=w(y)$ so we must have $g(w(x))=g(w(y))$ , so we must have $f(x)=f(y)$ ! A contradiction!

[2] Jump up ↑ 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

[Alec-5] {Jump up to: 2.0} ^2.1 This is my (Alec's) own work

[Note 1]

[1]

[Note 2]

[Note 3]

[2]

Passing to the quotient (function)

Contents

Statement

Caveats

Proof of claims

See also

Notes

References

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