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Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(V)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$. EDIT: this belief is correct, see Joel's comment below.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

• $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

• $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

• there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.

Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(j)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$. EDIT: this belief is correct, see Joel's comment below.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

• $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

• $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

• there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.

Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(V)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$. EDIT: this belief is correct, see Joel's comment below.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

• $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

• $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

• there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.

Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(V)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$. EDIT: this belief is correct, see Joel's comment below.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

• $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

• $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

• there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.

Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(V)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

• $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

• $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

• there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.

Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(V)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$. EDIT: this belief is correct, see Joel's comment below.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

• $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

• $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

• there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.