>**Puzzle 132.** For any category \\(\mathcal{C}\\), what's another name for a functor \\(F: \mathbf{1} \to \mathcal{C}\\)? There's a simple answer using concepts you've already learned in this course.

Since the category \\(\mathbf{1} \\) is just a one object category with a single identity morphism, \\(F: \mathbf{1} \to \mathcal{C}\\) must send this object to a single object in \\(\mathcal{C}\\). In other words, functors \\(F: \mathbf{1} \to \mathcal{C}\\) are objects of \\(\mathcal{C}\\).

>**Puzzle 133.** For any category \\(\mathcal{C}\\), what's another name for a functor \\(F: \mathbf{2} \to \mathcal{C}\\)? Again, there's a simple answer using concepts you've already learned here.

Since the category \\(\mathbf{2} \\) is a two object category with a single non-identity morphism, \\(F: \mathbf{2} \to \mathcal{C}\\) must send this arrow, and it's source and target objects, to a single arrow with mapped source anf target obects in \\(\mathcal{C}\\). In other words, functors \\(F: \mathbf{2} \to \mathcal{C}\\) are morphisms of \\(\mathcal{C}\\).

>**Puzzle 134.** For any category \\(\mathcal{C}\\), what's another name for a natural transformation \\(\alpha : F \Rightarrow G\\) between functors \\(F,G: \mathbf{1} \to \mathcal{C}\\)? Yet again there's a simple answer using concepts you've learned here.

Since both \\(F\\) and \\(G\\) map a single set to a chosen basepoints in \\(\mathcal{C}\\), so a natural transformation is a map that takes basepoints to basepoints, ie a based map, or point-preserving map.

>**Puzzle 135.** For any category \\(\mathcal{C}\\), what are functors \\(F : \mathcal{C} \to \mathbf{1} \\) like?

Such functors map all objects into a single object and maps all morphisms into the identity morphism.

>**Puzzle 138.** For any category, what are functors \\(F : \mathbf{0} \to \mathcal{C}\\) like?

Well, \\(\mathbf{0}\\) is empty, so a map \\(F : \mathbf{0} \to \mathcal{C}\\) just "pops" \\(\mathcal{C}\\) out of nothing. This mind as well be called *creatio ex nihilo,* since \\(\mathcal{C}\\) is created out of nothingness.

>**Puzzle 139.** For any category, what are functors \\(F : \mathcal{C} \to \mathbf{0} \\) like?

Such a map is as if \\(\mathcal{C}\\) just vanishes from existence as if it never existed. If my Latin isn't bad, this would be *destructio in nihilio*.

Also, what \\(\mathcal{C}\\)?

Since the category \\(\mathbf{1} \\) is just a one object category with a single identity morphism, \\(F: \mathbf{1} \to \mathcal{C}\\) must send this object to a single object in \\(\mathcal{C}\\). In other words, functors \\(F: \mathbf{1} \to \mathcal{C}\\) are objects of \\(\mathcal{C}\\).

>**Puzzle 133.** For any category \\(\mathcal{C}\\), what's another name for a functor \\(F: \mathbf{2} \to \mathcal{C}\\)? Again, there's a simple answer using concepts you've already learned here.

Since the category \\(\mathbf{2} \\) is a two object category with a single non-identity morphism, \\(F: \mathbf{2} \to \mathcal{C}\\) must send this arrow, and it's source and target objects, to a single arrow with mapped source anf target obects in \\(\mathcal{C}\\). In other words, functors \\(F: \mathbf{2} \to \mathcal{C}\\) are morphisms of \\(\mathcal{C}\\).

>**Puzzle 134.** For any category \\(\mathcal{C}\\), what's another name for a natural transformation \\(\alpha : F \Rightarrow G\\) between functors \\(F,G: \mathbf{1} \to \mathcal{C}\\)? Yet again there's a simple answer using concepts you've learned here.

Since both \\(F\\) and \\(G\\) map a single set to a chosen basepoints in \\(\mathcal{C}\\), so a natural transformation is a map that takes basepoints to basepoints, ie a based map, or point-preserving map.

>**Puzzle 135.** For any category \\(\mathcal{C}\\), what are functors \\(F : \mathcal{C} \to \mathbf{1} \\) like?

Such functors map all objects into a single object and maps all morphisms into the identity morphism.

>**Puzzle 138.** For any category, what are functors \\(F : \mathbf{0} \to \mathcal{C}\\) like?

Well, \\(\mathbf{0}\\) is empty, so a map \\(F : \mathbf{0} \to \mathcal{C}\\) just "pops" \\(\mathcal{C}\\) out of nothing. This mind as well be called *creatio ex nihilo,* since \\(\mathcal{C}\\) is created out of nothingness.

>**Puzzle 139.** For any category, what are functors \\(F : \mathcal{C} \to \mathbf{0} \\) like?

Such a map is as if \\(\mathcal{C}\\) just vanishes from existence as if it never existed. If my Latin isn't bad, this would be *destructio in nihilio*.

Also, what \\(\mathcal{C}\\)?