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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?

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From the last equality you obtain that $F(g)\circ f_a=f_b\circ G(g)$, because $\lambda_b$ is a limiting cone.

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  • $\begingroup$ 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? $\endgroup$ – Jxt921 Dec 24 '18 at 12:21
  • $\begingroup$ @Jxt921 A composition of a natural transformation and a cone is always a cone. $\endgroup$ – Oskar Dec 24 '18 at 12:40
  • $\begingroup$ 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. $\endgroup$ – Jxt921 Dec 24 '18 at 12:49

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