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I am studying category theory and struggling to understand functors and the way they work with some elementary categories; in particular, I am interested in a functor from a single-object category made of a monoid to the $\mathbf{SET}$-category.

So I assume existence of any monoid $M$ with an underlying set $|M|$. As a consequence, there exists a single-object category $C$, $M \in \mathsf{Ob}(C)$. It seems to me that there must also exist a functor $F : C \mapsto \mathbf{SET}$, but I can't figure out how to map and preserve morphisms of the $C$, which are just elements of the $|M|$? As far as I understand, in the $\mathbf{SET}$ morphism are regular maps between regular sets. Does it mean that $F$ suppose to map elements of $|M|$ to a... maps $F(M) \in \mathsf{Ob}(\mathbf{SET}) \mapsto F(M)$?

Can someone kindly show me how such a functor should work and how can I verify that it does not break the functor laws?

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  • $\begingroup$ You can map the object of your one-object category to your favourite set, and all the morphisms of your one-object category to the identity map on your favourite set. $\endgroup$ – Lord Shark the Unknown Mar 10 '18 at 18:57
  • $\begingroup$ @LordSharktheUnknown, as far as I understand, identity map is indeed $F(M) \in \mathsf{Ob}(\mathbf{SET}) \mapsto F(M)$ - namely map which does not do nothing with it's input. Am I right here? $\endgroup$ – Sereja Bogolubov Mar 10 '18 at 19:00
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$F$ must map each element $m\in |M|$ to some function $F(m):F(M)\to F(M)$. To satisfy the functor axioms, this must preserve identities and composition: that is, $F(1)$ must be the identity map $F(M)\to F(M)$ and $F(mn)=F(m)\circ F(n)$ for each $m,n\in |M|$.

It may be more intuitive to think of this as an "action" of $M$ on the set $F(M)$. Given $x\in F(M)$ and $m\in |M|$, let us write $mx$ for $F(m)(x)$. Then $F(mn)(x)=(mn)x$ and $F(m)(F(n)(x))=F(m)(nx)=m(nx)$, so the condition $F(mn)=F(m)\circ F(n)$ just says that $(mn)x=m(nx)$ for all $x\in F(M)$ (that is, our "multiplication" is associative). Similarly, the requirement that $F(1)$ is the identity just means that $1x=x$.

If you want to describe concrete examples of such functors $F$, this will of course depend on what your monoid $M$ is! But there are some examples you can write down for any $M$. For instance, if $S$ is any set, then you could take $F(M)=S$ and let $F(m)$ be the identity map $S\to S$ for all $m$. In terms of actions, we have $mx=x$ for all $m$ and $x$. It is trivial to check that this is indeed a functor: $1x=x$ by definition and $(mn)x=x=nx=m(nx)$.

A more interesting example is to take $F(M)=|M|$ and define $F(m)(n)=mn$ for $m,n\in|M|$. That is, our action is just the usual multiplication operation on $|M|$ itself. The fact that this is indeed a functor follows immediately from the monoid axioms on $M$: $1x=x$ since $1$ is the unit of $M$, and $(mn)x=m(nx)$ since multiplication in $M$ is associative.

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