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Let’s look more deeply at how \(\mathcal{D}^\mathcal{C}\) is a category.

- Figure out how to compose natural transformations. (Hint: an expert tells you “for each object \(c \in \mathcal{C}\), compose the \(c\)-components”.)
- Propose an identity natural transformation on any object \(F \in \mathcal{D}^\mathcal{C}\) , and check that it is unital.

**Definition 3.44**
Let \(\mathcal{C}\) and \(\mathcal{D}\) be categories.
We denote by \(\mathcal{D}^\mathcal{C}\) the category whose objects are functors \(F : \mathcal{C} \rightarrow \mathcal{D}\) and whose morphisms \(\mathcal{D}^\mathcal{C}(F, G)\) are the natural transformations \( \alpha : F \rightarrow G\).
This category \(\mathcal{D}^\mathcal{C}\) is called the \(\textit{functor category}\).

## Comments

1)

`1) ![Example](https://docs.google.com/drawings/d/e/2PACX-1vTthhQbJ3uZIjFaYDDIMLt6WPvNqAp77CzBQZByuZYHTMb1AN4MK_0bYDi7xSj5JZulb77cagXPFlCn/pub?w=607&h=383)`

According to the Definition 3.48 (of a natural transformation), it is sufficient to specify each component of the natural transformation. As in Fredrick Eisele's figure (and the hint): compose the components, so \(\forall c\in\mathcal{C}\,\,(\beta\circ\alpha)_c:=\beta_c\circ\alpha_c\). Since \(\alpha_c\) and \(\beta_c\) are morphisms, their composite exists and is a morphism (since \(\mathcal{D}\) is a category), so \(\beta\circ\alpha\) is well-defined. The only thing to check is naturality. \((\beta\circ\alpha)_d\circ F(f)\) \(=\beta_d\circ\alpha_d\circ F(f)\) \(=\beta_d\circ G(f)\circ\alpha_c\) \(=H(f)\circ\beta_c\circ\alpha_c\) \(=H(f)\circ(\beta\circ\alpha)_c\).

It is sufficient to define it on objects. To act like the identity, it must map an object to itself. \(\forall F\in\mathcal{D}^\mathcal{C}\,\,(\mathrm{id}_F)_c(F(c))=F(c)\). This is a natural transformation since \((\mathrm{id}_F)_d\circ F(f)=\mathrm{id}_{F(d)}\circ F(f)=F(f)=F(f)\circ\mathrm{id}_{F(c)}=F(f)\circ(\mathrm{id}_F)_c\) (i.e., since each of its components is the identity morphism on the associated object, and unitality in \(\mathcal{D}\)). In the functor category \(\mathcal{D}^\mathcal{C}\), unitality says \(\forall\alpha\in\mathcal{D}^\mathcal{C}(F,G)\,\,\alpha\circ\mathrm{id}_F=\alpha=\mathrm{id}_G\circ\alpha\). The first equality follows from applying the above definition of equality on components in \(\mathcal{C}\), and similarly for the second (the only difference being the functor for which the natural transformation is the identity), i.e., \(\forall c\in\mathcal{C}\,\,(\alpha\circ\mathrm{id}_G)_c\) \(=\alpha_c\circ(\mathrm{id}_G)_c\) \(=\alpha_c\circ\mathrm{id}_{G(c)}\) \(=\alpha_c\) \(=\mathrm{id}_{F(c)}\circ\alpha_c\) \(=(\mathrm{id}_F)_c\circ\alpha_c\) \(=(\mathrm{id}_F\circ\alpha)_c\).

Note that associativity in \(\mathcal{D}^\mathcal{C}\) follows from associativity for morphisms in \(\mathcal{D}\).

`1. According to the Definition 3.48 (of a natural transformation), it is sufficient to specify each component of the natural transformation. As in Fredrick Eisele's figure (and the hint): compose the components, so \\(\forall c\in\mathcal{C}\,\,(\beta\circ\alpha)_c:=\beta_c\circ\alpha_c\\). Since \\(\alpha_c\\) and \\(\beta_c\\) are morphisms, their composite exists and is a morphism (since \\(\mathcal{D}\\) is a category), so \\(\beta\circ\alpha\\) is well-defined. The only thing to check is naturality. \\((\beta\circ\alpha)_d\circ F(f)\\) \\(=\beta_d\circ\alpha_d\circ F(f)\\) \\(=\beta_d\circ G(f)\circ\alpha_c\\) \\(=H(f)\circ\beta_c\circ\alpha_c\\) \\(=H(f)\circ(\beta\circ\alpha)_c\\). 2. It is sufficient to define it on objects. To act like the identity, it must map an object to itself. \\(\forall F\in\mathcal{D}^\mathcal{C}\,\,(\mathrm{id}\_F)\_c(F(c))=F(c)\\). This is a natural transformation since \\((\mathrm{id}\_F)\_d\circ F(f)=\mathrm{id}\_{F(d)}\circ F(f)=F(f)=F(f)\circ\mathrm{id}\_{F(c)}=F(f)\circ(\mathrm{id}\_F)\_c\\) (i.e., since each of its components is the identity morphism on the associated object, and unitality in \\(\mathcal{D}\\)). In the functor category \\(\mathcal{D}^\mathcal{C}\\), unitality says \\(\forall\alpha\in\mathcal{D}^\mathcal{C}(F,G)\,\,\alpha\circ\mathrm{id\}_F=\alpha=\mathrm{id}\_G\circ\alpha\\). The first equality follows from applying the above definition of equality on components in \\(\mathcal{C}\\), and similarly for the second (the only difference being the functor for which the natural transformation is the identity), i.e., \\(\forall c\in\mathcal{C}\,\,(\alpha\circ\mathrm{id}\_G)\_c\\) \\(=\alpha\_c\circ(\mathrm{id}\_G)\_c\\) \\(=\alpha\_c\circ\mathrm{id}\_{G(c)}\\) \\(=\alpha\_c\\) \\(=\mathrm{id}\_{F(c)}\circ\alpha\_c\\) \\(=(\mathrm{id}\_F)\_c\circ\alpha\_c\\) \\(=(\mathrm{id}\_F\circ\alpha)\_c\\). Note that associativity in \\(\mathcal{D}^\mathcal{C}\\) follows from associativity for morphisms in \\(\mathcal{D}\\).`