# Category theory: ex. 3.3.vi from Riehl's “Category Theory in Context”

Prove that for any small category $$\mathsf{A}$$, the functor category $$\mathsf{C^A}$$ again has any limits or colimits that $$\mathsf{C}$$ does, constructed objectwise. That is, given a diagram $$\mathcal{D}\colon\mathsf{J}\to\mathsf{C^A}$$, with $$\mathsf{J}$$ small, show that whenever limits of the diagrams $$\mathsf{ev}_a\mathcal{D}$$ exists for any $$a \in \mathsf{A}$$, then these values define the action on objects of $$\lim\mathcal{D} \in \mathsf{C^A}$$, a limit of the diagram $$F$$.

A functor $$\mathsf{ev}_a\colon\mathsf{C^A}\to\mathsf{C}$$ maps a functor $$F\colon\mathsf{A\to C}$$ to $$F(a)$$ and a natural transformation $$\alpha\colon F\Rightarrow G$$ to $$\alpha_a$$.

Here's what I have done:

I used the axiom of choice to obtain a family of limits $$\lim(\mathsf{ev}_a\mathcal{D})$$ together with limit cones $$\lambda_a\colon\lim(\mathsf{ev}_a\mathcal{D})\Rightarrow\mathsf{ev}_a\mathcal{D}$$. I have defined a functor $$F\colon\mathsf{A\to C}$$ which maps each $$a \in \mathsf{A}$$ to $$F(a)$$ and which maps each morphism $$f\colon a\to b$$ to the unique morphism $$F(f)$$ for which we have $$(\lambda_b)_i\circ F(f) = \mathcal{D}(i)(f) \circ (\lambda_a)_i$$ for any $$i \in \mathsf{J}$$ (this construction uses the fact that $$\lambda_b$$ is a limit cone and defines a cone $$\kappa\colon\lim(\mathsf{ev}_a\mathcal{D})\to\mathsf{ev}_b\mathcal{D}$$ by setting $$\kappa_i = \mathcal{D}(i)(f)\circ(\lambda_a)_i)$$.

Next, I have defined a cone $$\Lambda\colon F\Rightarrow\mathcal{D}$$ so that for any $$i \in \mathsf{J}$$, $$\Lambda_i$$ is such a natural transformation $$F\Rightarrow\mathcal{D}(i)$$ for which we have the following: for any $$a \in \mathsf{A}$$, $$(\Lambda_i)_a = (\lambda_a)_i$$.

What I need is to prove that $$\Lambda$$ is a limit cone of $$\mathcal{D}$$. Let $$M\colon G\Rightarrow\mathcal{D}$$ be a cone. From it we can obtain a family of cones $$\mu_a\colon G(a)\Rightarrow\mathsf{ev}_a\mathcal{D}$$ so that $$(\mu_a)_i = (M_i)_a$$. As $$\lambda_a$$ is a limit cone, we have a family of morphism $$f_a\colon G(a)\to\lim(\mathsf{ev}_a\mathcal{D})$$ unique in the sense that for any $$i \in I$$ we have $$(\lambda_a)_i\circ f_a = (\mu_a)_i$$.

What I can't do is to prove that $$(f_a)_{a \in \mathsf{A}}$$ is a natural transformation $$G\Rightarrow F$$. Let $$g\colon a\to b$$ be a morphism of $$\mathsf{A}$$. If $$(f_a)$$ is a natural transformation, then we must have $$F(g)\circ f_a = f_b\circ G(g)$$. I can't prove this identity. The best I can do is this:

As $$M_i$$ is a natural transformation $$G\Rightarrow\mathcal{D}(i)$$, we must have $$\mathcal{D}(i)(g)\circ (M_i)_a = (M_i)_b \circ G(g)$$, that is, we have $$\mathcal{D}(i)(g)\circ(\mu_a)_i = (\mu_b)_i \circ G(g)$$. Thus, $$(\lambda_b)_i \circ F(g)\circ f_a = \mathcal{D}(i)(g) \circ (\lambda_a)_i \circ f_a = \mathcal{D}(i)(g)\circ(\mu_a)_i = (\mu_b)_i\circ G(g) = (\lambda_b)_i\circ f_b\circ G(g)$$.

But it doesn't help much. So, what exactly am I missing?

From the last equality you obtain that $$F(g)\circ f_a=f_b\circ G(g)$$, because $$\lambda_b$$ is a limiting cone.
• But for that to be of use we have to prove that the family $(\mathcal{D}(i)(g)\circ(\mu_a)_i)_{i \in \mathsf{I}}$ defines a cone from $G(a)$ to $\mathsf{ev}_b\mathcal{D}$, right? – Jxt921 Dec 24 '18 at 12:21
• Yeah, I know this. What I meant is one needs to prove that $(\mathcal{D}(i)(g))_{i \in \mathsf{J}}$ indeed defines a natural transformation. But, turns out, I have already proven a similar result earlier (it is needed to defined the functor $F$ on morphisms). Thanks. – Jxt921 Dec 24 '18 at 12:49