Prove that if $g \circ f$ is injective, then $g$ is injective

Question

Consider two functions $f: S \to T$ and $g: T \to U$ for non-empty sets $S,T,U.$

1. prove or disprove if $g \circ f$ is injective then $f$ is injective.

2. prove or disprove if $g \circ f$ is injective then $g$ is injective

for question 1) the proof here makes sense but how would I prove that g is injective? This is what I have so far but I feel as though my argument is flawed.

Proof: Let x, y ∈ S, and suppose that g ◦ f is injective, then we have; (1) (g ◦ f)(x) = (g ◦ f)(y) and x = y. Also suppose that g is not injective, then we have (2) (g ◦ f)(x) = (g ◦ f)(y) where f(x) does not equal to f(y).

so g(f(x)) = g(f(y)). Since we know that x = y from (1), then f(x) = f(y), which contradicts the assumption in (2). Therefore, by contradiction, g must also be injective.

EDIT: counterexample for question 2)

Suppose that f and g go from reals to reals and let f(x) = e^(x) and g(x) = x^2, where x is a real number and g(x) is not injective. We have, g(f(x)) = e^(2x), which is injective. Therefore, if g o f is injective, then g is not necessarily injective.

Would the above be sufficient for a counter example or would I have to prove/disprove injectivity for the functions I defined?