 Solution

For a spatially flat model, ${\Omega }_{k,0}=0$, so ${\dot{a}}_{\min }^2/a_{\min }^2=3H_0^2{\Omega }_{\Lambda ,0}$. With the squared definition of $E$ used in the paper,Equivalently $1+z_{\min }=a_0/a_{\min }=(2{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}{)}^{1/3}$, and substitution into $H^2/H_0^2={\Omega }_{m,0}(1+z{)}^3+{\Omega }_{\Lambda ,0}$ gives the same answer. The minimum is in $\dot{a}$, not in the Hubble parameter $H$, which continues to decrease towards $H_0\sqrt{{\Omega }_{\Lambda ,0}}$.