Extending Semiring $\mathbb{N}$ to $\mathbb{Z}$ through exact sequence

Question

I am working on extensions in the form of $$A\hookrightarrow B\twoheadrightarrow C$$ in my thesis and I am just wanting to add as an extra note, IF POSSIBLE, this. We have that that $\mathbb{Z}$ is isomorphic to $$\frac{\mathbb{N}\oplus\mathbb{N}}{\Delta}$$ Where $\Delta=\{i\oplus i:i\in\mathbb{N}\}$ when it comes to semirings, for quotients of semirings we have that $a\cong b$ in $R/I$ if there exists $i,j\in I$ such that $a+i=b+j$ and addition is done component wise, e.i $(a,b)+(c,d)=(a+c,b+d)$ along with the multiplication $(a,b)\cdot(c,d)=(ac+bd,ac+bd)$. I have checked (partially from peano) that this works out and makes sense which it does. However my question is more along with this, we have the homomorphism $\imath:\mathbb{N}\to\mathbb{Z}$ with $\imath(x)=(x,0)$ and as such we have $$\mathbb{N}\xrightarrow{\imath}\mathbb{Z}$$ which is clearly monomorphic, however my question is this, does there exist a homomorphism $\pi$ such that the following sequence is exact for some $X$? $$\mathbb{N}\xrightarrow{\imath}\mathbb{Z}\xrightarrow{\pi}X$$ I know that we have that $\Delta$ must naturally be mapped to $0$ and $\mathbb{N}\oplus0$ must also be in the kernal, however I am having trouble to figure out if this homomorphism between semirings exists or not, does it?

Answer 1

As Captain Lama commented, the usual definition of "exact sequences" doesn't work well at all for semirings (or rings, for that matter). Specifically, if $f:R\to S$ is a homomorphism of semirings and $Q\subseteq R$ is a subring of $R$ such that $Q\subseteq\ker{f}$, then $S$ must be the zero semiring (i.e., the semiring with one element). Indeed, since $Q$ is a subring, it must contain $1$, so $f(1)=0$. But a homomorphism of semirings must send $1$ to $1$, and so this means $1=0$ in $S$. It follows that for any $s\in S$, $s=s\cdot 1=s\cdot 0=0$. Thus $S$ has only one element.

This means that if you were ever to have an "exact sequence of semirings" $Q\stackrel{i}\to R\stackrel{f}\to S$ in the sense that $\operatorname{im}(i)=\ker(f)$, then $i$ is surjective and $S=\{0\}$, since $\operatorname{im}(i)$ is always a subsemiring of $R$.