Let $L=\{R\}$ where $R$ is a binary relation symbol, and let $\mathcal M=(\mathbb Z,\equiv_2)$ and $\mathcal N=(\mathbb Z,\equiv_3)$ be $L$-structures where $m\equiv_2 n \iff 2|m-n$ and $m\equiv_3 n \iff 3|m-n$.
Is there a homomorphism/embedding/isomorphism $F:\mathbb Z\to \mathbb Z$ from $\mathcal M$ into $\mathcal N$?
By definition for $F$ to be a homomorphism, it should satisfy:
- For every relation symbol $R$ and $\vec a\in M^n$, we should have $\vec a \in R^{\mathcal M}\implies F(\vec a)=(F(a_1),\cdots,F(a_n))\in R^{\mathcal N}$
So, let $(m,n)\in \mathbb Z^2$,
is there a map $F$ such that $(m,n)\in R^{\mathcal M}\implies F((m,n))=(F(m),F(n))\in R^{\mathcal N}$ ?
i.e is there a map $F$ such that $2|m-n \implies 3|F(m)-F(n)$ ?