Solutions

By Exercise 23 a), $f$ must be injective. For there not to be any function $h$ such that $f\circ h=I$, there has to be an input $x$ whose output $h(x)$ the function $f$ cannot map back to $x$. Consider the function $f(x)=2^x$. Such a function is injective, as it is strictly increasing. But for $x=-1$, where the output of $h(-1)$ is, $f$ cannot map it back to $-1$ because that value is not in its range; there is no number $x$ such that $-1=2^x$.