Provide a counter example to disprove that $ g\circ f$ surjective $ \implies $ $f$ surjective.

Question

Question:

Let $ \ f: A \to B$ and $ \ g : B \to C$ and $ g\circ f: A \to C. $

Provide a counter example to disprove that $ g\circ f$ surjective $ \implies $ $f$ surjective.

My attempt:

$ f : \mathbb{R} \to \mathbb{R}$ via $ \ f(x) = x^2$, not surjective

$ g : \mathbb{R} \to \ [0, \infty]$ via $ \ f(x) = x^2$

Then,

$ g\circ f: \mathbb{R} \to [0,\infty]$ via $ g(f(x)) = g(x^2) = x^4$, which is surjective

Is this counter example correct?

Answer 1

Is this counter example correct?

Yes, but it's more complicated than necessary. Don't bother with formulas. Can you build an example when $A$ and $C$ have one element each and $B$ has many? That's the essence of your argument.