# Surjective Proof

1. Sep 6, 2008

### nicksauce

1. The problem statement, all variables and given/known data
Let $f:X\rightarrow~Y$ and $g:Y\rightarrow~Z$ be surjections. Show that $g\circ~f$ is surjective.

2. Relevant equations

3. The attempt at a solution
Proof:
Suppose f and g are surjections.
Then (1)$\forall~y\in~Y \exists~x\in~X\textnormal{ st. }f(x)=y$
And (2) $\forall~z\in~Z \exists~y\in~Y\textnormal{ st. }g(y)=z$

(1) guarantees that we can write any y as f(x) for some x, so placing this into (2) gives:
(3)$\forall~z\in~Z \exists~x\in~X\textnormal{ st. }g(f(x))=g\circ~f=z$

And (3) shows that $g\circ~f$ is surjective.

Is my logic correct?

2. Sep 6, 2008

### tiny-tim

Hi nicksauce!

(have an exists: ∃ and an in/episilon: ε )

but it would be quicker and neater to start with (2) …

given z ε Z, ∃ y ε Y st g(y) = z, so ∃ x ε X st …

3. Sep 6, 2008

### HallsofIvy

Staff Emeritus
The proof is correct but, rather than saying "for all", better wording would be "if z is in Z, then, because g is surjective, there exist y in Y such that g(y)= z. Now, since f is surjective, there exist x in X such that f(x)= y (that specific y you got before). Then $g\circ f(x)= g(f(x))= g(y)= z$. That is, you have proved "if z is in Z, then there exist x in X such that $g\circ f(x)= z$.