 Solution

Always true. If $(f\circ g)(x)=(f\circ g)(x^{\prime })$, injectivity of $f$ gives $g(x)=g(x^{\prime })$, and injectivity of $g$ gives $x=x^{\prime }$. Thus the function composition is injective. For any $y\in X$, since $f$ is surjective, choose $z$ with $f(z)=y$. Since $g$ is surjective, choose $x$ with $g(x)=z$. Then $(f\circ g)(x)=y$, proving that the composition is surjective. Consequently $f\circ g$ is a bijection; its inverse function is $g^{-1}\circ f^{-1}$.