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Exercise 44 - Chapter 3

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 \( F_S (1) = S \).

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  • 1.

    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 \(F_S(\mathrm{id}_1)=\mathrm{id}_S\) (\(\forall s\in S\,\,\mathrm{id}_S(s)=s\)).

    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\\)).
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