[proofplan]
We prove the equivalence in both directions. If $C$ is creative, a productive function for $\mathbb{N} \setminus C$ is combined with the parameterized recursion theorem to build, uniformly in $x$, a self-referential c.e. set whose behaviour switches when $x$ enters a given c.e. set $A$. This yields a total computable many-one reduction from $A$ to $C$. Conversely, if $C$ is many-one complete, we transfer the elementary productivity of $\mathbb{N} \setminus K$ forward along a total computable many-one reduction from $K$ to $C$.
[/proofplan]
[step:Use the diagonal set to record a canonical productive complement]
Let
\begin{align*}
K := \{e \in \mathbb{N} : e \in W_e\}.
\end{align*}
The partial computable map
\begin{align*}
q: \mathbb{N} \to \mathbb{N}, \qquad e \mapsto e
\end{align*}
is productive for $\mathbb{N} \setminus K$. Indeed, suppose $e \in \mathbb{N}$ satisfies $W_e \subseteq \mathbb{N} \setminus K$. If $e \in K$, then by definition of $K$ we have $e \in W_e$, contradicting $W_e \subseteq \mathbb{N} \setminus K$. Hence $e \in \mathbb{N} \setminus K$. Since $W_e \subseteq \mathbb{N} \setminus K$, the membership $e \in W_e$ would again imply $e \in K$, which is false. Therefore
\begin{align*}
q(e)=e \in (\mathbb{N} \setminus K) \setminus W_e.
\end{align*}
Thus $\mathbb{N} \setminus K$ is productive.
[/step]
[step:Build the self-referential index used in the forward implication]
Assume $C$ is creative. Let
\begin{align*}
p: \mathbb{N} \rightharpoonup \mathbb{N}
\end{align*}
be a partial computable productive function for $\mathbb{N} \setminus C$.
Let $A \subseteq \mathbb{N}$ be computably enumerable. Choose $a \in \mathbb{N}$ such that $A = W_a$. For each pair $(x,e) \in \mathbb{N}^2$, define a c.e. set $E_{x,e} \subseteq \mathbb{N}$ by the following uniform enumeration procedure: wait until $x$ is enumerated into $W_a$; after that happens, simulate $p(e)$; if $p(e)$ converges to a value $y \in \mathbb{N}$, enumerate exactly $y$ into $E_{x,e}$.
By uniformity of this construction, there is a total computable map
\begin{align*}
h: \mathbb{N}^2 \to \mathbb{N}
\end{align*}
such that
\begin{align*}
W_{h(x,e)} = E_{x,e}
\end{align*}
for all $x,e \in \mathbb{N}$. By the parameterized recursion theorem (citing a result not yet in the wiki: Parameterized Recursion Theorem), there is a total computable map
\begin{align*}
n: \mathbb{N} \to \mathbb{N}
\end{align*}
such that, for every $x \in \mathbb{N}$,
\begin{align*}
W_{n(x)} = W_{h(x,n(x))} = E_{x,n(x)}.
\end{align*}
[guided]
The purpose of the recursion theorem is to make the enumeration procedure know its own index. We first define, uniformly in two parameters $x$ and $e$, the c.e. set $E_{x,e}$ as follows: the procedure watches the enumeration of $W_a$ until $x$ appears; once $x$ appears, it simulates the partial computable function $p$ at input $e$; if this simulation halts with value $y$, it enumerates $y$ and then enumerates nothing else.
Because the enumeration of $W_a$ and the computation of $p(e)$ can both be simulated effectively, there is a total computable indexing function
\begin{align*}
h: \mathbb{N}^2 \to \mathbb{N}
\end{align*}
with
\begin{align*}
W_{h(x,e)} = E_{x,e}.
\end{align*}
This still does not give self-reference: the procedure indexed by $h(x,e)$ uses the external number $e$, not necessarily its own index.
The parameterized recursion theorem supplies exactly this missing fixed point, uniformly in $x$. Applied to the total computable map $h$, it gives a total computable map
\begin{align*}
n: \mathbb{N} \to \mathbb{N}
\end{align*}
such that
\begin{align*}
W_{n(x)} = W_{h(x,n(x))} = E_{x,n(x)}
\end{align*}
for every $x \in \mathbb{N}$. Thus the c.e. set $W_{n(x)}$ behaves like the procedure that waits for $x \in W_a$ and then asks the productive function $p$ about the very index $n(x)$ of the set being enumerated.
[/guided]
[/step]
[step:Prove that the self-referential construction separates membership in $A$]
We claim that, for every $x \in \mathbb{N}$, the value $p(n(x))$ is defined and satisfies
\begin{align*}
x \in A \iff p(n(x)) \in C.
\end{align*}
Fix $x \in \mathbb{N}$. If $x \notin A$, then the enumeration procedure for $E_{x,n(x)}$ never passes its waiting stage, so
\begin{align*}
W_{n(x)} = E_{x,n(x)} = \varnothing.
\end{align*}
Since $\varnothing \subseteq \mathbb{N} \setminus C$, productivity gives that $p(n(x))$ is defined and
\begin{align*}
p(n(x)) \in (\mathbb{N} \setminus C) \setminus W_{n(x)}.
\end{align*}
In particular, $p(n(x)) \notin C$.
Now suppose $x \in A$. If $p(n(x))$ were undefined, then the procedure $E_{x,n(x)}$ would enumerate nothing even after $x$ appears in $W_a$, so
\begin{align*}
W_{n(x)} = \varnothing \subseteq \mathbb{N} \setminus C.
\end{align*}
Productivity would then force $p(n(x))$ to be defined, a contradiction. Hence $p(n(x))$ is defined. Let
\begin{align*}
y := p(n(x)).
\end{align*}
By construction of $E_{x,n(x)}$, once $x$ appears in $W_a$ and $p(n(x))$ converges to $y$, the set $W_{n(x)}$ enumerates exactly $y$. Thus
\begin{align*}
W_{n(x)} = \{y\}.
\end{align*}
If $y \in \mathbb{N} \setminus C$, then $W_{n(x)} \subseteq \mathbb{N} \setminus C$, so productivity applied to the index $n(x)$ gives
\begin{align*}
p(n(x)) \in (\mathbb{N} \setminus C) \setminus W_{n(x)}.
\end{align*}
This says $y \notin W_{n(x)}$, contradicting $W_{n(x)} = \{y\}$. Therefore $y \in C$, meaning $p(n(x)) \in C$.
Combining the two cases gives, for every $x \in \mathbb{N}$,
\begin{align*}
x \in A \iff p(n(x)) \in C.
\end{align*}
[/step]
[step:Turn the pointwise construction into a many-one reduction]
Define the partial computable map
\begin{align*}
g: \mathbb{N} \rightharpoonup \mathbb{N}, \qquad x \mapsto p(n(x)).
\end{align*}
The map $n: \mathbb{N} \to \mathbb{N}$ is total computable, and the preceding step proves that $p(n(x))$ is defined for every $x \in \mathbb{N}$. Therefore $g$ is a total computable map. Moreover, for every $x \in \mathbb{N}$,
\begin{align*}
x \in A \iff g(x) \in C.
\end{align*}
Thus $A$ many-one reduces to $C$. Since $A$ was an arbitrary computably enumerable set, $C$ is many-one complete for the computably enumerable sets.
[/step]
[step:Transfer productivity through a many-one reduction from $K$ to $C$]
Conversely, assume $C$ is many-one complete for the computably enumerable sets. Since $K$ is computably enumerable, there is a total computable map
\begin{align*}
f: \mathbb{N} \to \mathbb{N}
\end{align*}
such that, for every $u \in \mathbb{N}$,
\begin{align*}
u \in K \iff f(u) \in C.
\end{align*}
We construct a productive function for $\mathbb{N} \setminus C$. Given $e \in \mathbb{N}$, let $V_e \subseteq \mathbb{N}$ be the c.e. set
\begin{align*}
V_e := \{u \in \mathbb{N} : f(u) \in W_e\}.
\end{align*}
Because $f$ is total computable and $W_e$ is c.e. uniformly in $e$, the family $(V_e)_{e \in \mathbb{N}}$ is uniformly c.e. Hence there is a total computable map
\begin{align*}
r: \mathbb{N} \to \mathbb{N}
\end{align*}
such that
\begin{align*}
W_{r(e)} = V_e
\end{align*}
for every $e \in \mathbb{N}$.
Define the total computable map
\begin{align*}
P: \mathbb{N} \to \mathbb{N}, \qquad e \mapsto f(r(e)).
\end{align*}
We verify that $P$ is productive for $\mathbb{N} \setminus C$. Suppose $W_e \subseteq \mathbb{N} \setminus C$. If $u \in V_e$, then $f(u) \in W_e \subseteq \mathbb{N} \setminus C$, so by the reduction equivalence $u \notin K$. Therefore
\begin{align*}
V_e \subseteq \mathbb{N} \setminus K.
\end{align*}
By the productivity of $\mathbb{N} \setminus K$ from the first step, applied to the index $r(e)$ of $V_e$, we have
\begin{align*}
r(e) \in (\mathbb{N} \setminus K) \setminus V_e.
\end{align*}
Since $r(e) \notin K$, the reduction equivalence gives
\begin{align*}
f(r(e)) \notin C.
\end{align*}
Since $r(e) \notin V_e$, the definition of $V_e$ gives
\begin{align*}
f(r(e)) \notin W_e.
\end{align*}
Thus
\begin{align*}
P(e)=f(r(e)) \in (\mathbb{N} \setminus C) \setminus W_e.
\end{align*}
So $\mathbb{N} \setminus C$ is productive, and therefore $C$ is creative.
[/step]
[step:Conclude the equivalence]
The forward implication shows that every creative computably enumerable set is many-one complete for the computably enumerable sets. The reverse implication shows that every many-one complete computably enumerable set is creative. Hence, for every computably enumerable set $C \subseteq \mathbb{N}$,
\begin{align*}
C \text{ is creative} \iff C \text{ is many-one complete for the computably enumerable sets}.
\end{align*}
[/step]
