Let $f\colon X\to Y$ and $g\colon Y\to X$ be functions and assume $g\circ f=I_X$. Prove if $g$ is injective then $f\circ g=I_Y$

Question

If $f\circ g = I_{Y}$, then NTS $f\circ g$ is a bijection by the definition of Identity function. Since $g\circ f = I_{X}$, then $f$ is injective because otherwise $g$ wouldn't be injective. But since $g$ and $f$ are injective, then $f\circ g$ is injective.

Is this correct and would this approach work the same way to show that $f\circ g$ is surjective?

Answer 1

$g \circ (f\circ g)=(g\circ f)\circ g =I_X \circ g =g\circ I_Y$. Since $g$ is injective this implies $f \circ g =I_Y$.