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# Exercise 44 - Chapter 3

edited June 2018

Let $$\textbf{1}$$ denote the category with one object, called 1, one identity morphism $$id_1$$, and no other morphisms. For any functor $$F : \textbf{1} \rightarrow \textbf{Set}$$ one can extract a set $$F(1)$$. Show that for any set $$S$$, there is a functor $$F_S : \textbf{1} \rightarrow \textbf{Set}$$ such that $$F_S (1) = S$$.

Since there is only one object in 1, each functor $$F:\textbf{1}\to\textbf{Set}$$ must select a single set out of $$\mathrm{Ob}(\textbf{Set})$$ to be the image of this object under $$F$$. As there is only one morphism, each functor must select a single function between sets in Set. Since functors must preserve identity morphisms, the image of the identity morphism must be the identity function on the set selected by $$F$$. These constitute the entirety of the constraints on functors from 1 to Set, so there are exactly as many functors from 1 to Set as there are sets. Thus, one may index these functors by the set they map to: for every set $$S\in\mathrm{Ob}(\textbf{Set})$$, there is a (unique) functor $$F_S:\textbf{1}\to\textbf{Set}$$ such that $$F_S(1)=S$$ and $$F_S(\mathrm{id}_1)=\mathrm{id}_S$$ ($$\forall s\in S\,\,\mathrm{id}_S(s)=s$$).
Comment Source:Definition 3.23 defines **Set** to be the category with sets as objects and functions as morphisms. Definition 3.34 defines functors as a function connecting the sets of objects and a set of functions connecting the sets of morphisms (one function for each pair of objects). Since there is only one object in **1**, each functor \$$F:\textbf{1}\to\textbf{Set}\$$ must select a single set out of \$$\mathrm{Ob}(\textbf{Set})\$$ to be the image of this object under \$$F\$$. As there is only one morphism, each functor must select a single function between sets in **Set**. Since functors must preserve identity morphisms, the image of the identity morphism must be the identity function on the set selected by \$$F\$$. These constitute the entirety of the constraints on functors from **1** to **Set**, so there are exactly as many functors from **1** to **Set** as there are sets. Thus, one may index these functors by the set they map to: for every set \$$S\in\mathrm{Ob}(\textbf{Set})\$$, there is a (unique) functor \$$F_S:\textbf{1}\to\textbf{Set}\$$ such that \$$F_S(1)=S\$$ and \$$F_S(\mathrm{id}_1)=\mathrm{id}_S\$$ (\$$\forall s\in S\,\,\mathrm{id}_S(s)=s\$$).