A sentence is existential if it is of the form $\exists x_1 ... \exists x_nR$ and $R$ has no further quantifiers. A sentence is preserved upwards if and only if whenever it is true in an interpretation $\mathcal{P}$, and there is an embedding of $\mathcal{P}$ in another interpretation $\mathcal{Q}$, then it is true in $\mathcal{Q}$. Show that every existential sentence is preserved upwards.
I'm not sure how proofs of this nature are ordinarily constructed. So lets assume the question, that every existential sentence is preserved upwards. This means that for a sentence $\exists x_1 ...\exists x_nR$ that is true in an interpretation $\mathcal{P}$, where $\mathcal{P}$ is embedded within another interpretation $\mathcal{Q}$, the sentence is also true in $\mathcal{Q}$. I figure that this is a direct result of isomorphism:
If there is an isomorphism between interpretations $\mathcal{P}$ and $\mathcal{Q}$ under language $L$, then for every sentence $A$ of $L$: $\mathcal{P} \models A$ if and only if $\mathcal{Q} \models A$.
But I'm not sure how to apply that result to existential quantifiers in a particular sentence $R$.
Let $\varphi:\mathcal{P}\to\mathcal{Q}$ be an embedding between two interpretation. Let $P(x_1,\cdots,x_n)$ be a quantifier-free formula which only have free variables $x_1$, $\cdots$, $x_n$. Then $\mathcal{P}\models P(x_1,\cdots,x_n)$ iff there is $d_1,\cdots,d_n\in|\mathcal{P}|$ such that $d_1$, $\cdots$, $d_n$ satisfies $P$ (in $\mathcal{P}$). Consider $\varphi(d_1)$, $\cdots$, $\varphi(d_n)$ in $\mathcal{Q}$. You can check that $\varphi(d_1)$, $\cdots$, $\varphi(d_n)$ satisfies $P(x_1,\cdots,x_n)$ in $\mathcal{Q}$.