# Factoring a function through the projection of an equivalence relation induced by that function yields an injection

**Stub grade: A***

## Contents

## Statement

Let [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be sets, let [ilmath]f:X\rightarrow Y[/ilmath] be any function between them, and let [ilmath]\sim\subseteq X\times X[/ilmath] denote the*equivalence relation*induced by the function [ilmath]f[/ilmath], recall that means:

- [ilmath]\forall x,x'\in X[x\sim x'\iff f(x)=f(x')][/ilmath]

Then we claim we can factor^{[Note 1]} [ilmath]f:X\rightarrow Y[/ilmath] through [ilmath]\pi:X\rightarrow \frac{X}{\sim} [/ilmath]^{[Note 2]} to yield a unique injective map^{[1]}:

- [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath]

Furthermore, if [ilmath]f:X\rightarrow Y[/ilmath] is surjective then [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] is not only injective but surjective to, that is: [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] is a bijection^{[Note 3]}.

## Applications

Topology:

- If [ilmath]f:X\rightarrow Y[/ilmath] is a continuous map then factoring it through the projection of the equivalence relation it induces yields a continuous injection - topological version of this theorem almost exactly. We can extend this slightly in the case [ilmath]f:X\rightarrow Y[/ilmath] is surjective:
- If [ilmath]f:X\rightarrow Y[/ilmath] is continuous
*and*surjective then factoring it through the canonical projection of the equivalence relation it induces yields a continuous bijection - this is an extension of the previous statement, recall that when we take a surjection and apply passing-to-the-quotient we get surjection, we know from above it's injective. So now it's bijective

- If [ilmath]f:X\rightarrow Y[/ilmath] is continuous

## Proof

Outline of proof method

- We must check the set up satisfies the requirements of the passing-to-the-quotient theorem (to yield [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath]).
- We apply the theorem to get [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath].
- Next we must show [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] is injective

### Proof body

The message provided is:

- To do so requires that "[ilmath]f[/ilmath] be constant on the fibres of [ilmath]\pi[/ilmath]", ie:
- [ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath] (see Equivalent conditions to being constant on the fibres of a map)
- But if [ilmath]\pi(x)=\pi(y)[/ilmath] then [ilmath]x\sim y[/ilmath] and by definition [ilmath]x\sim y\iff f(x)=f(y)[/ilmath]! Trivial!

- [ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath] (see Equivalent conditions to being constant on the fibres of a map)

#### Step 3

We wish to show that [ilmath]\tilde{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] is injective, ie: [ilmath]\forall a,b\in\frac{X}{\sim}[\tilde{f}(a)=\tilde{f}(b)\implies a=b][/ilmath]

- Let [ilmath]a,b\in\frac{X}{\sim} [/ilmath] be given.
- Suppose [ilmath]\tilde{f}(a)\ne\tilde{f}(b)[/ilmath] then by the nature of logical implication we do not care whether or not [ilmath]a=b[/ilmath], either way it is true, we're done in this case.
- Suppose [ilmath]\tilde{f}(a)=\tilde{f}(b)[/ilmath], by the nature of logical implication we now require this leads to [ilmath]a=b[/ilmath]
- By the surjectivity of [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] we see:
- [ilmath]\exists x\in X[\pi(x)=a][/ilmath] and [ilmath]\exists y\in X[\pi(y)=b][/ilmath]
- Note that [ilmath]\tilde{f}(a)=\tilde{f}(\pi(x))[/ilmath] and [ilmath]\tilde{f}(b)=\tilde{f}(\pi(y))[/ilmath], also by assumption we have [ilmath]\tilde{f}(a)=\tilde{f}(\pi(x))= \tilde{f}(\pi(y))=\tilde{f}(b)[/ilmath]
- Notice also that [ilmath]f(x)=\tilde{f}(\pi(x))[/ilmath] and [ilmath]f(y)=\tilde{f}(\pi(y))[/ilmath] (this is the whole point of [ilmath]\tilde{f} [/ilmath])
- So we see [ilmath]f(x)=\tilde{f}(a)=\tilde{f}(\pi(x))= \tilde{f}(\pi(y))=\tilde{f}(b)=f(y)[/ilmath], or importantly:
- [ilmath]f(x)=f(y)[/ilmath]

- Recall that [ilmath]x\sim y\iff f(x)=f(y)[/ilmath], so now we know [ilmath]x\sim y[/ilmath]

- So we see [ilmath]f(x)=\tilde{f}(a)=\tilde{f}(\pi(x))= \tilde{f}(\pi(y))=\tilde{f}(b)=f(y)[/ilmath], or importantly:
- An elementary property of [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] is that if [ilmath]x\sim y[/ilmath] then [ilmath]\pi(x)=\pi(y)[/ilmath]
^{[Note 4]}^{[Note 5]}- Explicitly, notice we have: [ilmath]\pi(x)=\pi(y)[/ilmath]

- Recall that [ilmath]a=\pi(x)[/ilmath] and [ilmath]b=\pi(y)[/ilmath] by definition of [ilmath]x[/ilmath] and [ilmath]y[/ilmath], so:
- [ilmath]a=\pi(x)=\pi(y)=b[/ilmath] or more simply: [ilmath]a=b[/ilmath]

- [ilmath]\exists x\in X[\pi(x)=a][/ilmath] and [ilmath]\exists y\in X[\pi(y)=b][/ilmath]
- We have shown that supposing [ilmath]f(a)=f(b)[/ilmath] is given then [ilmath]a=b[/ilmath]

- By the surjectivity of [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] we see:
- Both cases have been evaluated and the implication holds for both

- Since [ilmath]a,b\in\frac{X}{\sim} [/ilmath] were arbitrary we have shown this for all [ilmath]a,b\in\frac{X}{\sim} [/ilmath]

## See also

- If a surjective function is factored through the canonical projection of the equivalence relation induced by that function then the yielded function is a bijection
- Factoring a continuous map through the projection of an equivalence relation induced by that map yields an injective continuous map

## Notes

- ↑ AKA: passing to the quotient
- ↑ the canonical projection of the equivalence relation, given by [ilmath]\pi:x\mapsto [x][/ilmath] where [ilmath][x][/ilmath] denotes the equivalence class containing [ilmath]x[/ilmath]
- ↑ See "
*If a surjective function is factored through the canonical projection of the equivalence relation induced by that function then the yielded function is a bijection*" for details - ↑

TODO: Link to statement, this ought to be mentioned somewhere explicit on this site!

- ↑

TODO: This can be strengthened to [ilmath]\iff[/ilmath] surely!

## References

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