Set Theory: Proving h is Surjective Implies f & g

[BubblesAreUs]

Homework Statement

Let

f: X ----> Y and g: Y ----> Z

be functions and let

h = g o f: X ----> Z

Homework Equations

a. If h is surjective then g is surjective

b. If h is surjective then f is surjective.

The Attempt at a Solution

Here

h: X ----> Z

a.
Suppose h: x ---> z is surjective for ∈ Z. Since h is surjective ∃a ∈ X such that
h(a) = g(f(a)) = k

Now let y = f(a) ∈ Y so...
g(y) = g(f(a)) = k; as declared QED.

b.
Suppose h: x ---> z is surjective for y...I'm not even sure how to start.

PS: To be honest, I really need to find a good textbook on proofs because my lecturer is outright atrocious. If anyone knows of any texts, do post me some recommendations as well.

 

[PeroK]

Homework Statement

Let

f: X ----> Y and g: Y ----> Z

be functions and let

h = g o f: X ----> Z

Homework Equations

a. If h is surjective then g is surjective

b. If h is surjective then f is surjective.

The Attempt at a Solution

Here

h: X ----> Z

a.
Suppose h: x ---> z is surjective for ∈ Z. Since h is surjective ∃a ∈ X such that
h(a) = g(f(a)) = k

Now let y = f(a) ∈ Y so...
g(y) = g(f(a)) = k; as declared QED.

b.
Suppose h: x ---> z is surjective for y...I'm not even sure how to start.

PS: To be honest, I really need to find a good textbook on proofs because my lecturer is outright atrocious. If anyone knows of any texts, do post me some recommendations as well.

Your proof of a) looks quite good. You didn't say what k is, but it's fairly obvious.

Why do you think b) is true?

 

[BubblesAreUs]

k is just an integer that belongs to set Z.

As for b, I think f is surjective because h is. Since f is an input of g, I'm not exactly sure how I can re-utilise my proof from part a.

 

[PeroK]

k is just an integer that belongs to set Z.

As for b, I think f is surjective because h is. Since f is an input of g, I'm not exactly sure how I can re-utilise my proof from part a.

If I can't see how to prove something, I usually try to disprove it and see what happens.