It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.2K
- Programming with Categories Course 21
- Exercises - Programming with Categories Course 15
- Applied Category Theory Course 341
- Applied Category Theory Seminar 4
- Exercises - Applied Category Theory Course 149
- Discussion Groups 50
- How to Use MathJax 15
- Chat 487
- Azimuth Code Project 108
- News and Information 147
- Azimuth Blog 149
- Azimuth Forum 29
- Azimuth Project 189
- - Strategy 108
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 711
- - Latest Changes 701
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 21
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 41

Options

## Comments

Definition 3.23 defines

Setto 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 inSet. 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 from1toSet, so there are exactly as many functors from1toSetas 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 \(F_S(\mathrm{id}_1)=\mathrm{id}_S\) (\(\forall s\in S\,\,\mathrm{id}_S(s)=s\)).`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\\)).`