 Solution

Suppose $A=\mathrm{ran}f$ for a partial computable function $f$. Since $A\ne \varnothing$, fix $a_0\in A$. Decode each input $w$ as a pair $((w{)}_0,(w{)}_1)=(u,t)$. Using the truncated computation function, defineThis is a total computable function, and every output lies in $A$. Conversely, if $a=f(u)$, choosing $t$ at least the halting time gives an input $w$ with $h(w)=a$. Thus $\mathrm{ran}h=A$, proving (iii)$\Rightarrow$(iv).