# Show that if $f$ is an elementary map then structures are elementarily equivalent in extended language

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 $$(\mathcal{A},a)_{a\in S}\equiv(\mathcal{B},f(a))_{a\in S}$$.

Attempt:

($$\Rightarrow$$) 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.

• 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. – spaceisdarkgreen Sep 22 '19 at 16:23
• @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. – quanticbolt Sep 22 '19 at 16:32
• $\equiv$ 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. – spaceisdarkgreen Sep 22 '19 at 16:42
• @spaceisdarkgreen ah. Makes sense. Thanks! I was getting a bit confused by the notation. – 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 $$\mathcal{B}\models\phi(f(a_1),...,f(a_n))$$.