If I have two functions $f, g: R \to R$ is it possible for $g$ and $f$ to not be injective but $g(f(x))$ to be injective?

Question

My guess is that it's not possible since I couldn't find an example to fit the scenario. However, I can't seem to prove either, so I'm not sure if my guess was correct or not.

On a similar note, is it possible for one of them to not be injective but the composition still be injective?

You should read https://math.stackexchange.com/questions/1324627/composition-of-functions-injective-implies-one-of-them-is-injective — Mundron Schmidt, Nov 11 '17 at 23:42

score 2 · Accepted Answer · answered Nov 11 '17 at 23:38

2

If $f$ is not injective, then there exist $x$ and $y$ such that $f(x)=f(y)$. This necessarily implies $g(f(x))=g(f(y))$, so $g\circ f$ is not injective either. Note that no properties of $g$ (besides the fact that it is a function) were needed to arrive at this conclusion.

answered Nov 11 '17 at 23:38

angryavian

89,882

Is it possible for g to not be injective, but for the composition to still be injective then? – Nov 11 '17 at 23:42
@Manny See Lord Shark the Unknown's answer – angryavian Nov 11 '17 at 23:48

score 0 · Answer 2 · answered Nov 11 '17 at 23:42

0

You must have $f$ injective, but $g$ need not be, for instance $f(x)=e^x$ and $g(x)=x^2$.

answered Nov 11 '17 at 23:42

Angina Seng

158,341

If I have two functions $f, g: R \to R$ is it possible for $g$ and $f$ to not be injective but $g(f(x))$ to be injective?

2 Answers2