[step:Use preservation of negations to recover reflection]
Assume conversely that, for every $k\in\mathbb N$, every $L$-formula $\psi(x_1,\dots,x_k)$, and every tuple $b\in M^k$,
\begin{align*}
M\models \psi(b)\implies N\models \psi(f(b)).
\end{align*}
Fix $k\in\mathbb N$, an $L$-formula $\varphi(x_1,\dots,x_k)$, and a tuple $a\in M^k$. The assumed implication applied to $\psi=\varphi$ gives
\begin{align*}
M\models \varphi(a)\implies N\models \varphi(f(a)).
\end{align*}
It remains to prove the reverse implication. Suppose that $N\models \varphi(f(a))$. If $M\not\models \varphi(a)$, then the satisfaction clause for negation gives
\begin{align*}
M\models \neg\varphi(a).
\end{align*}
Since $\neg\varphi(x_1,\dots,x_k)$ is also an $L$-formula, the assumed one-way preservation applied to $\psi=\neg\varphi$ gives
\begin{align*}
N\models \neg\varphi(f(a)).
\end{align*}
Again by the satisfaction clause for negation, this means $N\not\models \varphi(f(a))$, contradicting the choice of $a$ with $N\models \varphi(f(a))$. Hence $M\models \varphi(a)$.
Therefore, for every $k\in\mathbb N$, every $L$-formula $\varphi(x_1,\dots,x_k)$, and every tuple $a\in M^k$,
\begin{align*}
M\models \varphi(a)\iff N\models \varphi(f(a)).
\end{align*}
[/step]