Composition of two surjective functions

[cooldog182]

Hey, i'm trying to prove that the composition of two surjective functions is always surjective

however, the way I see it, something like this could happen:-

f: x^2
g: x^2

A = {1,2,3,4}
Z = {1,4,9,16)

f and g both map A - > Z happily

however, the composition " f o g " gives g(f( ))

hence (x^2)^2 = x^4

A does not map Z anymore. Not surjective.

Have I missed something?

 

[pka]

f and g both map A - > Z happily
the composition " g o f " gives g(f(x))

You have some very mistaken ideas about function composition!
First to have g o f the codomain of f must be the domain of g.
Thus in your example Z must be the domain of g.

Here is what must happen: \(\displaystyle f:A \mapsto B\quad \& \quad g:B \mapsto C\quad \Rightarrow \quad g \circ f:A \mapsto C\).

If both f & g are surjections then g o f is.
If \(\displaystyle t \in C\quad \Rightarrow \quad \exists s \in B\left[ {g(s) = t} \right]\).
But \(\displaystyle \exists r \in A\left[ {f(r) = s} \right]\).
Thus \(\displaystyle g \circ f(r) = g(f(r)) = g(s) = t\) or \(\displaystyle g \circ f\) is surjective.

 

[cooldog182]

Hahaha, oh thats so cool! Thanks for that!

Indeed, I did look this us on Wikipedia and what you said was there, it's just the book I have to support my lecture notes is a bit like........."This is surjection".........and that's about it!

Yes, Brilliant!