Csdrhrt

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 $mathcal{D}colonmathsf{J}tomathsf{C^A}$, with $mathsf{J}$ small, show that whenever limits of the diagrams $mathsf{ev}_amathcal{D}$ exists for any $a in mathsf{A}$, then these values define the action on objects of $limmathcal{D} in mathsf{C^A}$, a limit of the diagram $F$.

A functor $mathsf{ev}_acolonmathsf{C^A}tomathsf{C}$ maps a functor $Fcolonmathsf{Ato C}$ to $F(a)$ and a natural transformation $alphacolon FRightarrow G$ to $alpha_a$.

Here's what I have done:

I used the axiom of choice to obtain a family of limits $lim(mathsf{ev}_amathcal{D})$ together with limit cones $lambda_acolonlim(mathsf{ev}_amathcal{D})Rightarrowmathsf{ev}_amathcal{D}$. I have defined a functor $Fcolonmathsf{Ato C}$ which maps each $a in mathsf{A}$ to $F(a)$ and which maps each morphism $fcolon ato b$ to the unique morphism $F(f)$ for which we have $(lambda_b)_icirc F(f) = mathcal{D}(i)(f) circ (lambda_a)_i$ for any $i in mathsf{J}$ (this construction uses the fact that $lambda_b$ is a limit cone and defines a cone $kappacolonlim(mathsf{ev}_amathcal{D})tomathsf{ev}_bmathcal{D}$ by setting $kappa_i = mathcal{D}(i)(f)circ(lambda_a)_i)$.

Next, I have defined a cone $Lambdacolon FRightarrowmathcal{D}$ so that for any $i in mathsf{J}$, $Lambda_i$ is such a natural transformation $FRightarrowmathcal{D}(i)$ 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 $Mcolon GRightarrowmathcal{D}$ be a cone. From it we can obtain a family of cones $mu_acolon G(a)Rightarrowmathsf{ev}_amathcal{D}$ so that $(mu_a)_i = (M_i)_a$. As $lambda_a$ is a limit cone, we have a family of morphism $f_acolon G(a)tolim(mathsf{ev}_amathcal{D})$ unique in the sense that for any $i in I$ we have $(lambda_a)_icirc f_a = (mu_a)_i$.

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

As $M_i$ is a natural transformation $GRightarrowmathcal{D}(i)$, 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, $(lambda_b)_i circ F(g)circ f_a = mathcal{D}(i)(g) circ (lambda_a)_i circ f_a = mathcal{D}(i)(g)circ(mu_a)_i = (mu_b)_icirc G(g) = (lambda_b)_icirc f_bcirc G(g)$.

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

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