Why is the underlying homomorphism function a "forgetful functor"? Reading category theory book, Awodey states (during the introduction to Free monoids, p.18):

First, every monoid $N$ has an underlying set $|N|$, and every monoid homomorphism $f:N\to M$ has an underlying function $|f|:|N|\to |M|$. It is easy to see that this is a functor, called the "forgetful functor."



*

*Where is the "forgetful functor" the author referring to?

*Also, what does the notation with vertial lines around the function name $|f|$ means?


As I understand it, monoid homomorphism is simply a functor, but $|f|:|N|\to |M|$ is simply a function between sets, and not a functor... or it's a functor in Sets, but it's not forgetting anything.
Thank you for any clarification.
 A: The " forgetful functor" is the map 
$$N \mapsto |N| \text{ the underlying set of N }$$ 
$$f \mapsto |f| \text{ the morphism viewed as a map of sets }$$ 
Its a functor from the category of monoids to the category sets. And it is forgetting about the properties of being a monoid and just stripping down to the underlying set. 
A: Consider the category $\mathbf{Mon}$ of monoids.  Note that every object of $\mathbf{Mon}$ is essentially of the form $( A , e , \cdot )$ where $A$ is a set, $e \in A$ and $\cdot$ is a binary operation on $A$ with identity $e$.  A morphism in $\mathbf{Mon}$ between $(A , e , \cdot )$ and $(B , i , \odot )$ is a function $f : A \to B$ with $f(e) = i$ and $f ( a \cdot b ) = f(a) \odot f(b)$ for all $a,b \in A$.
Consider now the mapping $\mathbf{Mon} \to \mathbf{Sets}$ denoted by $| \cdot |$ given by


*

*$| ( A , e , \cdot ) | = A$;

*$| f | = $ the underlying mapping between sets.


Therefore $\mid \cdot \mid$ forgets about the additional structure that defined the category $\mathbf{Mon}$.  It is easy to show that this is a functor from $\mathbf{Mon}$ into $\mathbf{Sets}$, and such a functor is called forgetful.
It is not necessary to go to $\mathbf{Sets}$ to be a forgetful functor:  there is a perfectly good forgetful functor $\mathbf{Ring} \to \mathbf{Group}$ obtained by forgetting about the multiplication operation on rings.
