 Solution

True. The function composition $j=g\circ f$ is an injection: $j(a)=j(a^{\prime })$ first implies $f(a)=f(a^{\prime })$ by the injectivity of $g$, then $a=a^{\prime }$ by the injectivity of $f$. Fix one $a_0\in A$, using the given nonemptiness, and defineThe first case is well defined because $j$ is an injection. For every $a\in A$, $h(j(a))=a$. Thus this left inverse satisfies $\boxed{h\circ g\circ f={\mathrm{id}}_A}$, the identity function on $A$.