(Verification) If $g \circ f$ is injective, then $f$ must also be injective.

Question

Prove that if $g \circ f$ is injective, then $f$ must also be injective.

$g(f(x))$ is injective.

1. Then, $x\neq x’ \implies (g\circ f)(x) \neq (g\circ f)(x’)$.

2. Suppose for the sake of contradiction: $f(x)=f(x’) \nRightarrow x=x’$

3. Then this implies that it is logically possible that for some $x$ and $x’$:

$f(x)=f(x’) \implies (g\circ f)(x) \neq (g\circ f)(x’) \implies g(f(x))\neq g(f(x’))$

4. But if $f(x)=f(x’)$, then $f(x)=f(x’) \implies g(f(x))\neq g(f(x))$ which is a contradiction.

5. Therefore $f(x)=f(x’) \implies x=x’$, that is, $f$ is also injective.

I’ve numbered the lines to make it easier to reference a mistake if I’ve made one. My concern is (3.). I think it should be correct to say that if (2.) holds, then that suggests that it is logically possible that $x\neq x’$ could hold under $f(x)=f(x’)$.

Answer 1

The proof is, in principle, correct, but all the use of logic symbols are wrong. Proof: Suppose that $f$ is not injective. Then for some $x\ne x'$ we have $f(x)=f(x')$. But then $g(f(x))=g(f(x'))$, so $g\circ f$ is not injective, a contradiction.