Inclusion of Natural Numbers in Integers is Epimorphism

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

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

Let $\struct {\Z, +}$ denote the 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 $\struct {M, \circ}$ 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:


 * $\forall n \in \N: \map f n = \map g n$

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

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


 * $\map f {- m} = \map f {-1} \circ \cdots \circ \map f {-1}$
 * $\map g {- m} = \map g {-1} \circ \cdots \circ \map g {-1}$

with on the $m$ copies of $\map f {-1}$ and $\map g {-1}$, respectively.

It thus is necessary and sufficient to show that $\map f {-1} = \map g {-1}$.

To this end, compute:

Hence the result.

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

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