Domains and ranges of partial computable functions are computably enumerable

Let $E_D:=\{e_X(x):x\in D\}\subseteq \{0,1{\}}^{\ast }$. On an input word $z\in \{0,1{\}}^{\ast }$, run the computable decoder from the given data for the input encoding of $X$. If $z$ is malformed, diverge forever. If $z=e_X(x)$ is valid, simulate the machine from [L1] on the same word $z$ and accept exactly when that simulation halts. This recognizes exactly the binary language $E_D$, so [L2] makes $E_D$ computably enumerable. Since $E_D$ consists exactly of the codes of elements of $D$, the same enumeration viewed through the fixed encoding $e_X$ witnesses that $D$ is computably enumerable as a subset of $X$.

Let $E_R:=\{e_Y(y):y\in f[D]\}\subseteq \{0,1{\}}^{\ast }$. On an input word $z\in \{0,1{\}}^{\ast }$, dovetail the following branches over all binary words $w$ in shortlex order: first run the computable decoder for the input encoding of $X$ on $w$, and discard the branch if $w$ is malformed; otherwise simulate the machine from [L1] on $w$. Accept as soon as some valid input code halts with output $z$. If $z=e_Y(f(x))$ for some $x\in D$, then [L3] ensures that the halting computation on the valid code $e_X(x)$ is eventually reached, so the recognizer accepts. If $z\mathrm{\notin }{E}_R$, no valid branch can witness acceptance. Thus $E_R$ is recognizable, [L2] makes it computably enumerable as a binary language, and therefore $f[D]$ is computably enumerable as a subset of $Y$.

Therefore both the domain and the range of a partial computable function are computably enumerable.