# Inclusion of Natural Numbers in Integers is Epimorphism

## Theorem

Let $\mathbf{Mon}$ be the category of monoids.

Let $\left({\N, +}\right)$ denote the monoid of natural numbers as on Natural Numbers under Addition form Commutative Monoid.

Let $\left({\Z, +}\right)$ denote the monoid of integers as on additive group of integers.

Denote with $\iota: \N \to \Z$ the inclusion mapping.

Then $\iota: \N \to \Z$ is an epimorphism in $\mathbf{Mon}$.

## Proof

Let $\left({M, \circ}\right)$ be a monoid with identity $e$.

Let $f, g: \Z \to M$ be monoid homomorphisms such that:

$f \circ \iota = g \circ \iota$

that is, by definition of inclusion, such that:

$\forall n \in \N: f \left({n}\right) = g \left({n}\right)$

Now $\iota$ will be epic if we can show that $f = g$.

The morphism property of $f$ and $g$ yields that, for any $m > 0$:

$f \left({- m}\right) = f \left({-1}\right) \circ \cdots \circ f \left({-1}\right)$
$g \left({- m}\right) = g \left({-1}\right) \circ \cdots \circ g \left({-1}\right)$

with on the right-hand side $m$ copies of $f \left({-1}\right)$ and $g \left({-1}\right)$, respectively.

It thus is necessary and sufficient to show that $f \left({-1}\right) = g \left({-1}\right)$.

To this end, compute:

 $\displaystyle f \left({-1}\right)$ $=$ $\displaystyle f \left({-1}\right) \circ e$ $e$ is an identity $\displaystyle$ $=$ $\displaystyle f \left({-1}\right) \circ g \left({0}\right)$ $g$ is a monoid homomorphism $\displaystyle$ $=$ $\displaystyle f \left({-1}\right) \circ g \left({1 + \left({-1}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle f \left({-1}\right) \circ g \left({1}\right) \circ g \left({-1}\right)$ $g$ is a monoid homomorphism $\displaystyle$ $=$ $\displaystyle f \left({-1}\right) \circ f \left({1}\right) \circ g \left({-1}\right)$ By the assumption on $f$ and $g$ $\displaystyle$ $=$ $\displaystyle f \left({\left({-1}\right) + 1}\right) \circ g \left({-1}\right)$ $f$ is a monoid homomorphism $\displaystyle$ $=$ $\displaystyle f \left({0}\right) \circ g \left({-1}\right)$ $\displaystyle$ $=$ $\displaystyle e \circ g \left({-1}\right)$ $f$ is a monoid homomorphism $\displaystyle$ $=$ $\displaystyle g \left({-1}\right)$ $e$ is an identity

Hence the result.

$\blacksquare$

## Caution

The theorem statement does not assert that $\iota$ is an abstract-algebraic epimorphism.

This is plainly false, as $\iota$ is not a surjection.