# Talk:Passing to the quotient (function)/Induced is surjective iff function is surjective

From Maths

## Contents

## Reminder

Diagram |
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- [ilmath]\forall a,b\in X\left[\big(w(a)=w(b)\big)\implies\big(f(a)=f(b)\big)\right][/ilmath]

then we may "factor [ilmath]f[/ilmath] through [ilmath]w[/ilmath] to yield:

- A function, [ilmath]\bar{f}:W\rightarrow Y[/ilmath] well defined by [ilmath]\bar{f}:w\mapsto f(w^{-1}(w))[/ilmath] such that:
- [ilmath]f=\bar{f}\circ w[/ilmath]

- Furthermore, if [ilmath]w:X\rightarrow W[/ilmath] is surjective then the induced [ilmath]\bar{f}:W\rightarrow Y[/ilmath] is unique].

The [ilmath]w[/ilmath] and [ilmath]W[/ilmath] represent "whatever", anything.

## Claim

- If [ilmath]f[/ilmath] is surjective then [ilmath]\bar{f} [/ilmath] is surjective

## Utility

- Useful property to have when factoring, isn't difficult, can be applied to many things (especially anything involving isomorphisms)
- Motivated by the first group isomorphism theorem, in fact I'm writing this for use there.

## Proof

Suppose [ilmath]f[/ilmath] is surjective. We wish to show that [ilmath]\bar{f} [/ilmath] is too.

- Note: [ilmath]f[/ilmath] is surjective means:
- [ilmath]\forall y\in Y\exists x\in X[f(x)=y][/ilmath]

We may now begin the proof. We wish to show [ilmath]\forall y\in Y\exists a\in W[\bar{f}(a)=y][/ilmath]

- Let [ilmath]y\in Y[/ilmath] be given.
- By hypothesis: [ilmath]\exists x\in X[f(x)=y][/ilmath]
- Pick [ilmath]x\in X[/ilmath] such that [ilmath]f(x)=y[/ilmath].
- Note that by the other part of the hypothesis:
- [ilmath]f=\bar{f}\circ w[/ilmath] so

- [ilmath]y=f(x)=(\bar{f}\circ w)(x)=\bar{f}(w(x))[/ilmath]

- Note that by the other part of the hypothesis:
- Choose [ilmath]a\in W[/ilmath] by [ilmath]a:=w(x)[/ilmath]
- Now [ilmath]\bar{f}(a)=y[/ilmath]

- This completes the proof.

### Notes

On paper I've been rather unhappy with this proof, I actually tried to go by contradiction, starting with:

- [ilmath]\exists y\in Y\forall a\in W[\bar{f}(a)\ne y][/ilmath]

And hoping to reach a contradiction, but I didn't really "contradict". Obviously I got to "but we see [ilmath]\bar{f} [/ilmath] is the negation of (not surjective), but it didn't seem to have a contradiction in it as such.

I'm not sure what I was unhappy about though!