Prove Surjectivity of g∘f: Homework Solution

[nicksauce]

Homework Statement

Let $f:X\rightarrow~Y$ and $g:Y\rightarrow~Z$ be surjections. Show that $g\circ~f$ is surjective.

Homework Equations

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?

 

[tiny-tim]

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?

Hi nicksauce!

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

Your logic is fine …

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 …

 

[HallsofIvy]

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 $g\circ f(x)= z$.