Suppose $\kappa$ is a measurable cardinal. Then if $U$ is the ultrafilter on $\kappa$, we can use this to generate an ultrapower of the entire universe. We can then embed $V$ into this ultrapower in the usual way, such that every set $S$ maps into the constant $\kappa$-sequence in which every entry is equal to $S$. (Noah Schweber has a good answer explaining this here).
What I don't get is: we can form the Mostowski collapse, giving us a transitive class $M$ which is isomorphic to the ultrapower. Then the embedding of $V$ into the ultrapower gives an nontrivial elementary embedding $j: V \to M$, of which the measurable cardinal $\kappa$ is the critical point.
But, shouldn't the ultrapower of $V$, and thus $M$, both be subclasses of all of $V$? So wouldn't this lead to an embedding of $V$ into itself? But Kunen's theorem says we can't have a nontrivial elementary embedding of $V$ into itself.
My reasoning: to form the ultrapower we first take the Cartesian power $V^\kappa$, which is the class of all $\kappa$-length tuples of sets — and since each of these tuples can be formalized as a set in the usual way, we clearly have a subclass of the entire universe $V$ of all possible sets. Then, when we map $V$ into these sequences, we're embedding $V$ into a proper subclass of itself. Then we quotient by the ultrafilter, and we can use Scott's trick to get a set-representative for each equivalence class, and again the result should be a proper subclass of $V$. Thus the embedding of $V$ into the ultrapower is again a map from $V$ into this proper subclass of itself.
We then use the Mostowski collapse to get a transitive class $M$ which is isomorphic to the ultrapower. This $M$ must somehow not be a subclass of $V$, I think, or else we would be embedding $V$ into itself, which is impossible via Kunen's theorem. But if there's a bijection with $M$ and the ultrapower, and if the ultrapower is a proper subclass of $V$, then we'd have an embedding of $M$ back into $V$, and thus a nontrivial elementary embedding of $V$ into itself.
Why isn't this a dealbreaker for measurable cardinals?
Best Answer
The ultrapower $(\mathrm{Ult}_U, E_U)$ is not a subclass of $(V, \in)$ since it uses a completely different membership relation. Its transitive collapse $(M,\in),$ on the other hand, is a subclass of $(V,\in).$
The map underlying the ultrapower embedding $(V,\in)\to (\mathrm{Ult}_U, E_U)$ is certainly a map $V\to V,$ but that's not the same thing as having an embedding $(V,\in)\to (V,\in).$ Similarly, the isomorphism $(M,\in)\to (\mathrm{Ult}_U, E_U)$ does not add up to an embedding $(M,\in)\to (V,\in).$
And the fact that $(M,\in)$ is a subclass of $(V,\in)$ doesn't run afoul of Kunen. While we do have an elementary embedding $j:(V,\in)\to (M,\in),$ $j$ is not elementary when viewed as an embedding $(V,\in)\to (V,\in).$