Why are epimorphisms in the category of Monoids not necessarily surjections?

Question

The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?

Answer 1

Quoting Wikipedia, we see:

In the category of monoids, Mon, the inclusion map $\mathbb{N}\to\mathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $\mathbb{Z}$ to some monoid $M$. Then for some $n\in\mathbb{Z}$, $g_1(n)\neq g_2(n)$, so $g_1(-n)\neq g_2(-n)$. Either $n$ or $-n$ is in $\mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $\mathbb{N}$ are unequal.

Answer 2

Let $\Bbb Z$ be the additive group of integers, and $\Bbb N_0=\{0,1,2,\ldots\}$. Then the inclusion $\Bbb N_0\to\Bbb Z$ is an epimorphism in Mon.

Answer 3

Consider the inclusion $i: \mathbb{N}\hookrightarrow\mathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:\mathbb{Z} \to M$ such that $f \circ i = g \circ i$, we must have $f = g$, by a simple argument.

Answer 4

Actually, the situation in the other answers are all instances of a more general phenomenon: If $\mathbb M_A$ is the free monoid generated by a set $A$, and $\mathbb F_A$ is the free group generated by the same set, then the inclusion map $\mathbb M_A \hookrightarrow \mathbb F_A$ is epic! To see why, observe that the image of a monoid homomorphism $h : \mathbb F_A \to M$ is always a group. By the universal property of the free group construction, then, $h$ is determined by its values at the elements of $A$. Given any two homomorphisms $h, g : \mathbb F_A \to M$ that agree on $\mathbb M_A$, since they must also agree on $A$, $h = g$.