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Nov 23, 2018 at 13:53 history edited Gabe Goldberg CC BY-SA 4.0
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Nov 23, 2018 at 13:43 history edited Gabe Goldberg CC BY-SA 4.0
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Nov 22, 2018 at 21:29 vote accept James E Hanson
Nov 22, 2018 at 17:25 comment added Gabe Goldberg $V_{\kappa+2}^M = V_{\kappa+2}^{M_U}$ since the natural map $M_U \to M_Z \to M$ has critical point $j_U(\kappa)$. Since $V_\kappa^M = V_\kappa$, $V_{j_W(\kappa)}^{j_W(M)} = j_W(V^M_\kappa) = j_W(V_\kappa) = V_{j_W(\kappa)}^{M_W},$ which implies the second equality.
Nov 22, 2018 at 15:03 comment added Danielle Ulrich Why is $(V_{\kappa+2})^M = (V_{\kappa+2})^{M_U}$ and why is $(V_{\kappa+2})^{M_W} = (V_{\kappa+2})^{j_W(M)}$?
Nov 21, 2018 at 17:19 comment added Andrés E. Caicedo @Noah I suppose you would need an appropriate directed system of embeddings and then you would use directed limits.
Nov 21, 2018 at 16:40 comment added Noah Schweber If you don't require $S$ to be a well-ordering, how do you iterate along it? (I vaguely recall hearing that one could do this, but I don't recall the details.)
Nov 21, 2018 at 16:33 history edited Gabe Goldberg CC BY-SA 4.0
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Nov 21, 2018 at 15:54 history answered Gabe Goldberg CC BY-SA 4.0