If $f\circ g$ is injective then $g$ is injective

Question

If $f:S\to T$ and $g:R\to S$ are functions such that $f \circ g$ is injective, then prove that $g$ must be injective.

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I dont know how to prove it. I only know that composition of two injective functions is an injective function and composition of two surjective functions is surjective.

Answer 1

If there are some $x,y \in \Bbb{R}$ such that $x \neq y$ and $g(x) = g(y)$, then $f\circ g(x) = f\circ g(y)$; hence $f\circ g$ is not injective, a contradiction.

Answer 2

Alternatively, suppose that $g(x_1)=g(x_2)$. Since $f$ is a function, $f\big(g(x_1)\big)=f\big(g(x_2)\big)$, i.e., $(f\circ g)(x_1)=(f\circ g)(x_2)$. Since $f\circ g$ is injective, this implies that $x_1=x_2$, hence the injectivity of $g$ follows.