One to one and onto in composite function

1. Feb 4, 2013

BlackDeath

1. The problem statement, all variables and given/known data

I just want to make sure that I am correct. if we have a composite function f(g(x)).

2. Relevant equations
f(g(x)) is onto if and only if both f(x) and g(x) are onto
f(g(x)) is one to one if and only if or both f(x) and g(x) are one to one

3. The attempt at a solution

when I try to make f(x) is onto, but not one to one. And g(x) is one to one but not onto, f(g(x)) is not onto

2. Feb 5, 2013

Kreizhn

This is false: The fact that f(g(x)) is one to one only guarantees that f(x) is one to one.

For example, let $f: \mathbb R \to \mathbb R^2$ by $f(x) = (x,0)$ and $g: \mathbb R^2 \to \mathbb R$ be $g(x,y) = x+y$. f is one-to-one but g is not one-to-one. However, the function g(f(x)) = x is just the identity function and is injective.

Precisely the same example shows that this does not work for onto functions. g(x,y) is onto but f(x) is not. However, g(f(x)) = x is onto, so again it breaks.