Note: this is cross-posted from MSE.
This question is about the following remark (modified to be self-contained), found in Donald Martin's book on determinacy, page 340. The context is proving agreement between $V$ and some inner model $M$ containing a same extender (although this context doesn't seem to matter). Here, $\kappa$ is a cardinal and $\zeta\geq\kappa$ is any ordinal.

Any function $f:\kappa\to\zeta^+$ can be coded by a wellordering $R$ of $\zeta$ of ordertype sup(range($f$)) and a $\tilde g:\kappa\to \zeta$. The pair $\langle R, \tilde g \rangle$ can be coded by a $g:\kappa\to V_{\zeta+1}$. Such a $g$ can in turn be coded by an element of $V_{\zeta+1}$.

I don't see why the sentences in bold are true. Or more specifically, I don't know how to construct such a coding (Gödel pairing might not work, because we don't know if $\zeta$ is closed under the Gödel pairing function). I'm particularly puzzled by the last statement, that we can code $g:\kappa\to V_{\zeta+1}$ by an element of $V_{\zeta+1}$. The flat pairing construction in the style of Quine seems to only produce a subset of $V_{\zeta+1}$, hence an element of $V_{\zeta+2}$.

The first sentence is something like this: assume without loss of generality that $f:\kappa\to \zeta^+$ has range bounded by an ordinal $\alpha$ between $\zeta$ and $\zeta^+$. This is okay, because $\zeta^+$ is regular and we can always modify $f$ so that it sends something above $\zeta$. So $\alpha$ is the ordertype of some wellordering $R\subseteq(\zeta\times\zeta)$, and $\alpha$ has cardinality $\zeta$. Furthermore, if $g'$ is a bijection between $\alpha$ and $\zeta$, then we can define $\tilde g:\kappa\to\zeta$ by setting $\tilde g(x)=g'(f(x))$. So given such a wellordering $R$ and function $\tilde g$, we can recover $f$ by asking for the ordertype of $R$, and then seeing the pointwise preimage of $g'$.

Note: this is cross-posted from MSE.
This question is about the following remark (modified to be self-contained), found in Donald Martin's book on determinacy, page 340. The context is proving agreement between $V$ and some inner model $M$ containing a same extender (although this context doesn't seem to matter). Here, $\kappa$ is a cardinal and $\zeta\geq\kappa$ is any ordinal.

Any function $f:\kappa\to\zeta^+$ can be coded by a wellordering $R$ of $\zeta$ of ordertype sup(range($f$)) and a $\tilde g:\kappa\to \zeta$. The pair $\langle R, \tilde g \rangle$ can be coded by a $g:\kappa\to V_{\zeta+1}$. Such a $g$ can in turn be coded by an element of $V_{\zeta+1}$.

I don't see why the sentences in bold are true. Or more specifically, I don't know how to construct such a coding (Gödel pairing might not work, because we don't know if $\zeta$ is closed under the Gödel pairing function). I'm particularly puzzled by the last statement, that we can code $g:\kappa\to V_{\zeta+1}$ by an element of $V_{\zeta+1}$. The flat pairing construction in the style of Quine seems to only produce a subset of $V_{\zeta+1}$, hence an element of $V_{\zeta+2}$.

The coding referred to in the first sentence is something like this: assume without loss of generality that $f:\kappa\to \zeta^+$ has range bounded by an ordinal $\alpha$ between $\zeta$ and $\zeta^+$. This is okay, because $\zeta^+$ is regular and we can always modify $f$ so that it sends something above $\zeta$. So $\alpha$ is the ordertype of some wellordering $R\subseteq(\zeta\times\zeta)$, and $\alpha$ has cardinality $\zeta$. Furthermore, if $g'$ is a bijection between $\alpha$ and $\zeta$, then we can define $\tilde g:\kappa\to\zeta$ by setting $\tilde g(x)=g'(f(x))$. So given such a wellordering $R$ and function $\tilde g$, we can recover $f$ by asking for the ordertype of $R$, and then seeing the pointwise preimage of $g'$.