If $M$ is elementarily $L$ -embedded into $M'$ then $\mathcal{M}$ is $L$ -isomorphic to an elementary $L$ -substructure of $\mathcal{M'}$

70 Views Asked by At

Suppose $M$ is the universe of the $L$-structure $\mathcal{M}$, and $M'$ is the universe of the $L$-structure $\mathcal{M'}$

Show if $M$ is elementarily $L$ -embedded into $M'$ then $\mathcal{M}$ is $L$ -isomorphic to an elementary $L$ -substructure of $\mathcal{M'}$.

Attempt:

Suppose $M$ is elementarily $L$ -embedded into $M'$. Then there exists a function $f: M \rightarrow M'$ such that for all formula $\phi(x)$, $M \models \phi(a)$ iff $M' \models \phi(a)$ $\forall a \in M$.

However, I'm not sure if I'm interpreting this correctly, as I'm not sure what the elementary $L$-embedding between two universes looks like. Here I'm just assuming it's the same as the elementary embedding between two $L$-structures.

Also, I'm confused because we know that an surjective $L$-embedding is an isomorphism.

So can we just let the elementary substructure of $\mathcal{M'}$ be built by the universe $f(M)$. Then we will be done?

Thanks in advance!