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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."

  1. Where is the "forgetful functor" the author referring to?
  2. 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.

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2 Answers 2

up vote 4 down vote accepted

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.

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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.

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