$ g\circ f$ injective $\implies$ $f$ injective or $g$ injective

Question

First I have to prove that:

If $g\circ f$ injective $\implies$ $f$ injective or $g$ injective

And real functions that improve:

If $g\circ f$ injective $\implies$ $f$ injective and $g$ injective

I was thinking about $\sqrt x$ and $x^2$, so that $g\circ f$ will be bijective, but I don't really sure about that example

Answer 1

Proof for: If $g\circ f$ injective $\implies$ $f$ injective

Logically this proves that "If $g\circ f$ injective $\implies$ $f$ injective or $g$ injective" but no need to $g$-injective part, as your example says.

If not, there is x and y such that $x\ne y$ but $f(x)= f(y).$ So $g \circ f(x)= g \circ f(y).$

Answer 2

If $f(x) = f(y) \implies (g\circ f)(x) = (g\circ f)(y) \implies x = y$. Thus $f$ is injective. The condition $g$ is injective is not necessary. Example: $f(x) = \sqrt{x}, g(x) = x^2$ as you mentioned.