Inclusion of Natural Numbers in Integers is Epimorphism

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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 right hand side $m$ 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:

\(\ds \map f {-1}\) \(=\) \(\ds \map f {-1} \circ e\) Definition of Identity Element
\(\ds \) \(=\) \(\ds \map f {-1} \circ \map g 0\) Definition of Monoid Homomorphism
\(\ds \) \(=\) \(\ds \map f {-1} \circ \map g {1 + \paren {-1} }\)
\(\ds \) \(=\) \(\ds \map f {-1} \circ \map g 1 \circ \map g {-1}\) Definition of Monoid Homomorphism
\(\ds \) \(=\) \(\ds \map f {-1} \circ \map f 1 \circ \map g {-1}\) By the assumption on $f$ and $g$
\(\ds \) \(=\) \(\ds \map f {\paren {-1} + 1} \circ \map g {-1}\) Definition of Monoid Homomorphism
\(\ds \) \(=\) \(\ds \map f 0 \circ \map g {-1}\)
\(\ds \) \(=\) \(\ds e \circ \map g {-1}\) Definition of Monoid Homomorphism
\(\ds \) \(=\) \(\ds \map g {-1}\) Definition of Identity Element

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.


Sources