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# Exercise 54 - Chapter 3

edited June 2018

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

1. Figure out how to compose natural transformations. (Hint: an expert tells you “for each object $$c \in \mathcal{C}$$, compose the $$c$$-components”.)
2. 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 $$\textit{functor category}$$.

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Comment Source:1) ![Example](https://docs.google.com/drawings/d/e/2PACX-1vTthhQbJ3uZIjFaYDDIMLt6WPvNqAp77CzBQZByuZYHTMb1AN4MK_0bYDi7xSj5JZulb77cagXPFlCn/pub?w=607&h=383)
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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 $$\mathcal{D}$$.

Comment Source: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}\$$.