Solution
= Solution
\b[Always true.] If $(f\circ g)(x)=(f\circ g)(x')$, <injectivity> of $f$ gives $g(x)=g(x')$, and <injectivity> of $g$ gives $x=x'$. 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 $g^{-1}\circ f^{-1}$.