Embedding and existentially quantified formula I wanted to prove that given an existentially quantified formula $\varphi(\bar{y})=\exists\bar{x}\psi(\bar{x}, \bar{y})$ and an embedding $f : \mathfrak{M} \hookrightarrow \mathfrak{N}$ between two $\mathscr{L}$-structures $\mathfrak{M}$ and $\mathfrak{N}$, there is for any $\bar{a} \in M^k$,
$$\mathfrak{M} \models \varphi(\bar{a}) \Rightarrow \mathfrak{N} \models \varphi(f(\bar{a})). \tag{*}$$
($f(\bar{a})$ denotes $f(a_1), \ldots, f(a_n)$).
I already proven that given a quantifier-free formula $\psi$,  $$\mathfrak{M} \models \psi(\bar{a}) \iff \mathfrak{N} \models \varphi(f(\bar{a})) \tag{**}.$$ I have a wrong proof for (*) but wasn't able to figure out wherever was erroneous:
$$
\begin{align*}
\mathfrak{M} \models \exists \bar{x} \psi(\bar{x}, \bar{a}) &\iff \textrm{exists $\bar{b} \in M^k$, s.t. $\mathfrak{M} \models \psi(\bar{b}, \bar{a})$} \tag{Definition}\\
&\iff \textrm{exists $f(\bar{b}) \in N^k$, s.t. }\mathfrak{N} \models \psi(f(\bar{b}), f(\bar{a})) \tag{**}\\
& \iff \mathfrak{N} \models \exists\bar{x} \psi(\bar{x}, f(\bar{a})) \tag{Definition}
\end{align*}
$$
I'm not sure which logical equivalence is wrong.
 A: First let's show that the converse $$\mathfrak{N}\models\varphi(f(\overline{a}))\quad\implies\quad\mathfrak{M}\models\varphi(\overline{a})$$ is false in general; this will help uncover the issue with your proof.
Working in the empty language for simplicity, let $\mathfrak{M}$ be a one-element structure and let $\mathfrak{N}$ be a two-element structure. Let $a$ be the unique element of $\mathfrak{M}$ and fix some $f:\mathfrak{M}\rightarrow\mathfrak{N}$; note that any such $f$ is an embedding. Then we have $$\mathfrak{N}\models\exists x(x\not=f(a))\quad\mbox{but}\quad \mathfrak{M}\models\neg\exists x(x\not=a).$$ So $\exists$-sentences are only "upwards absolute" in general.

OK, so based on this where is the issue in your argument?
Well, the problem above was that the witness to the existential statement holding in $\mathfrak{N}$ was not itself in the image of $f$, so it couldn't be "pulled down" to a witness in $\mathfrak{M}$. This shows us the problem in reversing the final implication:

 We do have $$\mathfrak{N}\models\varphi(f(\overline{a}))\quad\implies\quad \exists \overline{c}\in\mathfrak{N}\mbox{ such that }\mathfrak{N}\models\psi(\overline{c},f(\overline{a})),$$ but we cannot assume that there is $\overline{b}\in\mathfrak{M}$ such that $f(\overline{b})=\overline{c}$. So we do not have $$\mathfrak{N}\models\varphi(f(\overline{a}))\quad\implies\quad \exists \overline{b}\in\mathfrak{M}\mbox{ such that }\mathfrak{N}\models\psi(f(\overline{b}),f(\overline{a}))$$ in general.

