Let $\sigma =\{R,f,c\}$ be first order logic where $R$ is a unary relation symbol, $f$ a binary function symbol and $c$ a constant symbol. Let $\mathcal{M}=(\mathbb{N}_0,\phi)$ and $\mathcal{N} = (\mathbb{R},\psi)$ be $\sigma$-structures, so that $\phi(R) = \{n \in \mathbb{N}_0 : (\exists k \in \mathbb{N}_0) \ n =8k +3\}$ is a unary relation on the set of nonnegative natural numbers, $\phi(f)(x,y) = x+2y$ is a binary function on the set of nonnegative natural numbers, $\phi(c) = 1$, whereas $\psi(R) =\{x \in \mathbb{R} : x > 9\}$ is a unary relation on the set of real numbers, $\psi(f)(x,y) = x + x·y$ is a binary function on the set of real numbers and $\psi(c) = 2$. Is there a homomorphism form the $\sigma$-structure $\mathcal{M}$ to the $\sigma$-structure $\mathcal{N}$?
Sorry I don't know how to write the set of natural or real numbers nicely. I've tried defining the rule for the homomorphism function but can't find it and don't really have a method to prove its existence other than finding the homomorphism.
As suggested by Andrés Caicedo, there is no homomorphism from the $\sigma$-structure $\mathcal{M}$ to the $\sigma$-structure $\mathcal{N}$.
The proof is by contradiction. Suppose there were such an homomorphism $h \colon \mathbb{N}_0 \to \mathbb{R}$ and consider $3 \in \mathbb{N}_0$. Then, $3 \in \phi(R)$ because $3 = 8 \cdot 0 + 3$. Note that $\phi(f)(1,1) = 1 + 2 \cdot 1 = 3$ and $\psi(f)(2,2) = 2 + 2 \cdot 2 = 6$. Since $h$ would preserve the interpretation of $f$ and $c$, we would have \begin{align} h(1) &= h(\phi(c)) = \psi(c) = 2 & h(3) &= h\big(\phi(f)(1,1)\big) = \psi(f)\big(h(1),h(1)\big) = \psi(f)(2,2) = 6. \end{align} Now, $h(3) \notin \psi(R)$ because $h(3) = 6 \not> 9$. Summing up, $3 \in \phi(R)$ but $h(3) \notin \psi(R)$, which is impossible because $h$ should preserve $R$.