 Solution

Can be false. Take $X=\{0,1,2,\dots \}$ and defineThen $(f\circ g)(n)=n$, so their function composition is the identity function and hence a bijection. However, $g$ is not surjective, since zero has no preimage, while $f$ is not injective, since $f(0)=f(1)=0$.

The valid implications are only that $g$ must be injective and $f$ must be surjective: equality of two $g$ values implies equality after composition, and every output of the composition is an output of $f$. On a finite $X$, either of these properties implies a bijection, so an infinite set is essential to this counterexample.