 Solution

For each $y\in Y$, surjectivity makes the fiber of a function $f^{-1}(\{y\})$ nonempty. Choose one element $x_y$ of each such fiber of a function and define $g(y)=x_y$. Then $f(g(y))=y$ for every $y$, proving $f\circ g=I_Y$. For arbitrary sets, the simultaneous selection invokes the usual axiom of choice; this is the right-inverse characterization of the axiom of choice. For finite $Y$ only finitely many choices are needed. If $Y$ is empty, the unique empty function gives the same conclusion.