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

Reminder

[ilmath]\xymatrix{ X \ar[r]^f \ar[d]_w & Y \\ W \ar@{.>}[ur]_{\bar{f} } }[/ilmath]
Diagram

If [ilmath]f:X\rightarrow Y[/ilmath] is a function and [ilmath]w:X\rightarrow W[/ilmath] another function such that:

• [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]
  • 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!