Prove that if $f \circ g$ is one-to-one, then $g$ if one-to-one

Question

Let $f$ and $g$ be functions and have domain $\mathbb{R}$. I have to prove or disprove two claims:

(1) if $f \circ g$ is one-to-one, then $f$ is one-to-one

(2) if $f \circ g$ is one-to-one, then $g$ is one-to-one

For (1), could you give me an intuition as to why the statement is wrong? What counter-example can be used?

For (2), how should I prove or approach the question? I have attempted direct proof, but I couldn't link the "IF" and "THEN" part.

Answer 1

The first statement is false: take $g(x)=e^x$ and $f(x)=x^2$. The $f\circ g$ is injective, but $f$ isn't.

But the second statement is true. If $f\circ g$ is injective and $x,y\in\Bbb R$ are such that $g(x)=g(y)$, then $f\bigl(g(x)\bigr)=f\bigl(g(y)\bigr)$, and therefore $x=y$. So, $g$ is indeed injective.