# set theory – Coding a function \$g:kappato V_{zeta+1}\$ by an element of \$V_{zeta+1}\$

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 $$zetageqkappa$$ is any ordinal.

Any function $$f:kappatozeta^+$$ can be coded by a wellordering $$R$$ of $$zeta$$ of ordertype sup(range($$f$$)) and a $$tilde g:kappato zeta$$. The pair $$langle R, tilde g rangle$$ can be coded by a $$g:kappato V_{zeta+1}$$. 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 $$g:kappato V_{zeta+1}$$ 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 $$f:kappato zeta^+$$ 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 $$Rsubseteq(zetatimeszeta)$$, and $$alpha$$ has cardinality $$zeta$$. Furthermore, if $$g’$$ is a bijection between $$alpha$$ and $$zeta$$, then we can define $$tilde g:kappatozeta$$ 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 $$g’$$.