Let $\mathcal{A}, \mathcal{B}$ be $\mathcal{L}$-structures, and let $S\subseteq A$. I want to show that $f:S\to B$ is an elementary mapping if and only if $(\mathcal{A},a)_{a\in S}\equiv(\mathcal{B},f(a))_{a\in S}$.


($\Rightarrow$) First suppose $f$ is elementary. Then pick a sentence, $\phi$ in $\text{Th}((\mathcal{A},a)_{a\in S})$. This is a sentence in $\mathcal{L}$ as well with parameters $a_1,...,a_n$ for each constant symbol in $\phi$ as an $\mathcal{L}(S)$ formula. Also $\mathcal{A}\models\phi(a_1,...,a_n)$. So $\mathcal{B}\models\phi(fa_1,...,fa_n)$ since $f$ is elementary, and so extending to an $\mathcal{L}(f(S))$-structure, we have $\mathcal{B}\models\phi$, and so $\phi\in\text{Th}((\mathcal{B},f(a))_{a\in S})$

For the flipside, when we pick a sentence in $\text{Th}((\mathcal{B},f(a))_{a\in S})$, we can just use the fact that $f$ is elementary and hence injective (such that it is invertible) to do the same proof in reverse to show that the same sentence is also in $\text{Th}((\mathcal{A},a)_{a\in S})$.

($\Leftarrow$) Here is where I get a bit confused. I want to show that $\mathcal{A}\models\phi(a_1,...,a_n)$ if and only if $\mathcal{B}\models\phi(fa_1,...,fa_n)$, given that the two structures in their respective extended languages are elementarily equivalent. So, I take an arbitrary $\mathcal{L}$-formula such that $\mathcal{A}\models\phi(a_1,...,a_n)$. This is really just a sentence. Extending the language to $\mathcal{L}(S)$ we have that this is actually a sentence in $\text{Th}((\mathcal{A},a)_{a\in S})$ so it is a sentence in $\text{Th}((\mathcal{B},f(a))_{a\in S})$. So going back down to $\mathcal{L}$ we have $\mathcal{B}\models\phi(fa'_1,...,fa'_m)$.

How can we get that the set $a'_1,...,a'_m$ is the same as $a_1,...a_n$, because it doesn't seem obvious to me that this must be the case.

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    $\begingroup$ I don't see how these are different notions, even superficially. Doesn't $(\mathcal A,a)_{a\in S}\equiv (\mathcal B,f(a))_{a\in S}$ mean precisely that $\mathcal A\models \phi(a_1,\ldots, a_n)$ iff $\mathcal B\models \phi(f(a_1),\ldots f(a_n))$? In particular I don't understand this "going back down to $\mathcal L$" part and what $a_i'$ are. $\endgroup$ – spaceisdarkgreen Sep 22 '19 at 16:23
  • $\begingroup$ @spaceisdarkgreen I think $(\mathcal{A},a)_{a\in S}\equiv(\mathcal{B},f(a))_{a\in S}$ means that the theory of $\mathcal{A}$ as an $\mathcal{L}(S)$-structure is equivalent to the theory of $\mathcal{B}$ as an $\mathcal{L}(f(S))$-structure. All we can conclude from this (as far as I know) is that any sentence in the former is in the latter and vice versa. $\endgroup$ – quanticbolt Sep 22 '19 at 16:32
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    $\begingroup$ $\equiv$ usually means elementarily equivalent, and we can only talk about elementary equivalence between two theories if they have the same language. Here the language is $\mathcal L$ with a constant symbol $c_a$ added for each element of $S$ and $(\mathcal A,a)_{a\in S}$ interprets $c_a$ is $a$ and $(\mathcal B,f(a))_{a\in S}$ interprets $c_a$ as $f(a).$ At least that's how I interpret things. $\endgroup$ – spaceisdarkgreen Sep 22 '19 at 16:42
  • $\begingroup$ @spaceisdarkgreen ah. Makes sense. Thanks! I was getting a bit confused by the notation. $\endgroup$ – quanticbolt Sep 22 '19 at 16:48

When you think of $\phi(a_1,...,a_n)$ as a sentence $\psi$ in the language $\mathcal{L}(S)$, the $a_i$'s are not just arbitrary constant symbols--they are specifically the constant symbols corresponding to the elements $a_i$ of $S$. So, when you then interpret the same sentence in the structure $(\mathcal{B},f(a))_{a\in S}$, these constant symbols get interpreted as the elements $f(a_i)$. Thus to say $(\mathcal{B},f(a))_{a\in S}\models\psi$ means exactly that $\mathcal{B}\models\phi(f(a_1),...,f(a_n))$.


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