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For $x, a$ sets, let the Mostowski rank of $x$ at set $a$, $MR_a(x)$, be the rank of $x$ in the Mostowski collapse applied to $R\cap a^2$. Let the Mostowski spectrum of a set $x$, $MS(x)$, be the set of possible values of the Mostwoski rank of $x$, as $a$ varies over all sets. On the face of it, $MS(x)$ is a class, rather than a set. However, by induction on $H(x)$, $MS(x)$ is bounded; specifically, for each ordinal $\alpha$ there is some ordinal $b(\alpha)$ such that $H(x)\le \alpha$ implies $MS(x)\subseteq \beta$. For example, if $\alpha=0$, then $b(\alpha)=1$, since every $x$ with $H(x)=0$ can only ever be collapsed to $0$. (Indeed, regardless of $R$ we'll always have $b(\alpha)\le\alpha+1$, but we don't need that here.)

In fact, $MS(x)$ has a maximum! Why? Well, for every $\alpha\in MS(x)$, let $S_\alpha$ be a set containing $x$, such that $MR_{S_\alpha}(x)=\alpha$. Now consider the set $T=\bigcup_{\alpha\in MS(\alpha)} S_\alpha$. Mostowski rank is monotonic: $a\subseteq b$ implies $MR_a(x)\le MR_b(x)$. So $MR_T(x)\ge \alpha$ for every $\alpha\in MS(\alpha)$. But also $MR_T(x)\in MS(x)$, by definition. So it is the maximum of this set.

Now consider the map $\mu(x)=\max(MS(x))$.

For $x, a$ sets, let the Mostowski rank of $x$ at set $a$, $MR_a(x)$, be the rank of $x$ in the Mostowski collapse applied to $R\cap a^2$. Let the Mostowski spectrum of a set $x$, $MS(x)$, be the set of possible values of the Mostwoski rank of $x$, as $a$ varies over all sets. On the face of it, $MS(x)$ is a class, rather than a set. However, by induction on $H(x)$, $MS(x)$ is bounded; specifically, for each ordinal $\alpha$ there is some ordinal $b(\alpha)$ such that $H(x)\le \alpha$ implies $MS(x)\subseteq \beta$. For example, if $\alpha=0$, then $b(\alpha)=1$, since every $x$ with $H(x)=0$ can only ever be collapsed to $0$. (Indeed, regardless of $R$ we'll always have $b(\alpha)\le\alpha+1$, but we don't need that here.)

In fact, $MS(x)$ has a maximum! Why? Well, for every $\alpha\in MS(x)$, let $S_\alpha$ be a set containing $x$, such that $MR_{S_\alpha}(x)=\alpha$. Now consider the set $T=\bigcup_{\alpha\in MS(\alpha)} S_\alpha$. Mostowski rank is monotonic: $a\subseteq b$ implies $MR_a(x)\le MR_b(x)$. So $MR_T(x)\ge \alpha$ for every $\alpha\in MS(\alpha)$. But also $MR_T(x)\in MS(x)$, by definition. So it is the maximum of this set.

Now consider the map $\mu(x)=\max(MS(x))$.

Let the Mostowski spectrum of a set $x$, $MS(x)$, be the set of ordinals $\alpha$ which $x$ can be mapped to under the Mostowski collapse applied to some restriction of $R$ to a set - that is, some $R\cap a^2$ for $a\in V$. (Note that such are automatically set-like, so the Mostowski collapse works as usual.)

By induction on $H(x)$, $MS(x)$ is bounded; specifically, for each ordinal $\alpha$ there is some ordinal $b(\alpha)$ such that $H(x)\le \alpha$ implies $MS(x)\subseteq \beta$. For example, if $\alpha=0$, then $b(\alpha)=1$, since every $x$ with $H(x)=0$ can only ever be collapsed to $0$. (Indeed, regardless of $R$ we'll always have $b(\alpha)\le\alpha+1$, but we don't need that here.)

Let $\mu(x)=\sup MS(x)$; this map, then, has the desired property.

For $x, a$ sets, let the Mostowski rank of $x$ at set $a$, $MR_a(x)$, be the rank of $x$ in the Mostowski collapse applied to $R\cap a^2$. Let the Mostowski spectrum of a set $x$, $MS(x)$, be the set of possible values of the Mostwoski rank of $x$, as $a$ varies over all sets. On the face of it, $MS(x)$ is a class, rather than a set. However, by induction on $H(x)$, $MS(x)$ is bounded; specifically, for each ordinal $\alpha$ there is some ordinal $b(\alpha)$ such that $H(x)\le \alpha$ implies $MS(x)\subseteq \beta$. For example, if $\alpha=0$, then $b(\alpha)=1$, since every $x$ with $H(x)=0$ can only ever be collapsed to $0$. (Indeed, regardless of $R$ we'll always have $b(\alpha)\le\alpha+1$, but we don't need that here.)

In fact, $MS(x)$ has a maximum! Why? Well, for every $\alpha\in MS(x)$, let $S_\alpha$ be a set containing $x$, such that $MR_{S_\alpha}(x)=\alpha$. Now consider the set $T=\bigcup_{\alpha\in MS(\alpha)} S_\alpha$. Mostowski rank is monotonic: $a\subseteq b$ implies $MR_a(x)\le MR_b(x)$. So $MR_T(x)\ge \alpha$ for every $\alpha\in MS(\alpha)$. But also $MR_T(x)\in MS(x)$, by definition. So it is the maximum of this set.

Now consider the map $\mu(x)=\max(MS(x))$.

Yes. Let $J$ be the range of $H$ - that is, the class $\{\alpha\in ON: \exists x(H(x)=\alpha)\}$. We can take the Mostowski collapse of $J$, ordered as usual; that is, there is a map $F: J\rightarrow ON$ such that $f(\alpha)=\{f(\beta): \beta\in J\cap \alpha\}$ for all $\alpha\in J$. (This class function $f$ is defined by transfinite recursion.)

Then the map $G=F\circ H$ has the desired property.

The point is that while $R$ may not be set-like, $H$ maps $R$ onto a relation which is set-like in an appropriate way.

Note that the usual caveats about reasoning about classes inside of ZFC apply: really, we're arguing externally that in any model of ZFC, given any definable $R, H$ with the desired properties, there is a desired $G$ with the desired properties.

Let the Mostowski spectrum of a set $x$, $MS(x)$, be the set of ordinals $\alpha$ which $x$ can be mapped to under the Mostowski collapse applied to some restriction of $R$ to a set - that is, some $R\cap a^2$ for $a\in V$. (Note that such are automatically set-like, so the Mostowski collapse works as usual.)

By induction on $H(x)$, $MS(x)$ is bounded; specifically, for each ordinal $\alpha$ there is some ordinal $b(\alpha)$ such that $H(x)\le \alpha$ implies $MS(x)\subseteq \beta$. For example, if $\alpha=0$, then $b(\alpha)=1$, since every $x$ with $H(x)=0$ can only ever be collapsed to $0$. (Indeed, regardless of $R$ we'll always have $b(\alpha)\le\alpha+1$, but we don't need that here.)

Let $\mu(x)=\sup MS(x)$; this map, then, has the desired property.