= Solution
\b[Can be false.] Take $X=\{0,1,2,\ldots\}$ and define
$$
g(n)=n+1,\qquad f(0)=0,\qquad f(n)=n-1\quad(n\geq1).
$$
Then $(f\circ g)(n)=n$, 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.
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