The problem:
Let $N$ be an $L$-structure for a language $L$. The $diagram$ of $N$, $D(N)$ is the set of all quantifier-free $L_N$-sentences true in $N$. Suppose $M$ is a model of $D(N)$. Show that $M$ has a substructure isomorphic to $N$.
My approach
At first I wasn't sure how to approach this, but I've tried something and I'm not sure if it's right or not.
Define the isomorphism $\beta$ as $\beta$: $N \rightarrow im(\beta) \leq M$. This map has the properties:
- $\beta(c^N) = c^M$ if $c$ is a constant of the language $L$
- $\beta(c_n^N) = c_n^M$ where $c_n$ is the constant of $L_N$ 'corresponding' with $n \in N$
Now for every $n$-airy function $f$ of $L$ we can do the following:
If $f^N(c_1^N, c_2^N, \dots, c_n^N) = c_x^N$, then this is a quantifier-free $L_N$-sentence. So it is true in $M$, in particular this is: $f^M(c_1^M, c_2^M, \dots, c_n^M) = c_x^M$. This means that $\beta(f^N(c_1^N, c_2^N, \dots, c_n^N)) = \beta(c_x^N) = c_x^M = f^M(c_1^M, c_2^M, \dots, c_n^M)=f^M(\beta(c_1^N), \beta(c_2^N), \dots, \beta(c_n^N))$
So for every function we have the correct property for an isomorphism. For the relations, I'd take an analogous approach.
Is this correct? I feel like I'm doing it a little bit too simple.
It looks like you have the right idea, but you need to define $\beta$ more concretely than just telling what its domain is (and stating that its codomain is whatever its image is, which is just a roundabout way of saying it is shap'd like itself).
Did you mean to define $\beta(n)=(c_n)^M$ for $n\in N$, by any chance?
Don't worry about it being "too simple"; this is not a very deep property. But do make sure to get all the details there. In particular, you seem to have omitted an argument that $\beta$ is injective.