Monoid homomorphisms to the additive naturals?

Question

Part of something I'm doing research in requires monoid homomorphisms $(\omega\times\omega,+,\bar{0})\to(\omega,+,0)$. (I'm just using component-wise addition for the product monoid). Are there any homomorphisms that aren't of the form $(n,m)\mapsto kn+lm$ (or $kn-lm$, or $lm-kn$) for $k,l\in\omega$? Google isn't being very friendly with this question. I'll obviously take any duplicate questions too, if there are any. Thank you!

Answer 1

Even in the special case of abelian groups this is not true. Take for example $\mathbb{Z}^2$ with the usual addition and $$ \mathbb{Z}^2 \times \mathbb{Z}^2 \to \mathbb{Z}^2, \quad ((k,l),(n,m)) \mapsto (k+l,n+m). $$

PS: In case you don’t mean an arbitrary abelian monoid $(\omega,+,0)$ but the natural numbers with addition, then the statement holds: If $f \colon \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ is an monoid homomorphism with $f((1,0)) = k$ and $f((0,1)) = l$ then $$ f((n,m)) = f(n \cdot (1,0)+ m \cdot (0,1)) = n \cdot f((1,0)) + m \cdot f((0,1)) = kn + lm. $$ (We just use that $\mathbb{N}^2$ is the free abelian monoid on $2$ generators.) Here we use the notation that for an abelian monoid $(M,+,0)$, an element $x \in M$ and a natural number $n \in \mathbb{N}$ we abreviate $n \cdot x = \sum_{i=1}^n x$.