TITLE: Proving adjoint functors preserve limits by lifting the adjunction to cone categories
QUESTION [7 upvotes]: Let $C$ and $D$ be categories and let $I$ be any (maybe small?) category.
$\newcommand{\Hom}{\operatorname{Hom}}$
$\newcommand{\Ob}{\operatorname{Ob}}$
$\newcommand{\Ar}{\operatorname{Ar}}$
$\newcommand{\id}{\operatorname{id}}$
$\newcommand{\ad}{\mathrm{ad}}$
Say $F : C \leftrightarrows D : G$ is a pair of adjoint functors, $F \dashv G$.
Then there is some isomorphism pair
$$(~_\ad) : \Hom_D(FX,Y) \cong \Hom_C(X,GY) : (~^\ad)$$
natural in $X ∈ \Ob C$ and $Y ∈ \Ob D$.
I want to prove that $G$ respects limits like so:

Let $Q\colon I → D$ be any diagram in $D$. Define the categories of cones for $Q$ and $GQ$ as the comma categories $D' = (Δ_D,Q)$ and $C' = (Δ_C,GQ)$ where $Δ_C$ and $Δ_D$ are the diagonal functors $C → C^I$ and $D → D^I$ respectively (and $Q$ and $GQ$ are meant to be constant functors $\star → D^I$ and $\star → C^I$ here if one wants to keep things formal).
Then $Q$ and $GQ$ have limits in $D$ and $C$ if and only if $D'$ and $C'$ have terminal objects respectively.
Now, $F$ and $G$ lift to functors $F' \colon C' → D'$ and $G' \colon D' → C'$, using the adjunction. Formally,
  \begin{align}
F' \colon C' &→ D' & G' \colon D' &→ C' \\
(x,α,GQ) &↦ (Fx,α^\ad,Q) \quad\text{on $\Ob C'$}&
(y,β,Q) &↦ (Gy,Gβ,GQ)\quad\text{on $\Ob D'$}\\
(f,\id_\star) &↦ (Ff,\id_\star)\quad\text{on $\Ar C'$}&
(g,\id_\star) &↦ (Gg,\id_\star)\quad\text{on $\Ar D'$}.
\end{align}
  Less formally, while $G'$ is just the the application of $G$ on any cone over $Q$, the functor $F'$ takes the top of a cone over $GQ$ to $D$ by applying $F$ and transposes the actual cone using the adjunction to get a cone over $Q$. By the naturality of the adjunction, this is well-defined; as $F$ and $G$ are functors, so are $F'$ and $G'$.
So if now we could prove $F' \dashv G'$, then we would be done: As a right adjoint, $G'$ obviously preserves terminal objects, thus: $G$ preserves limits.

Is it true that $F' \dashv G'$? How can I go about proving it? Or is the situation too cluttered and another route might be preferable?

REPLY [4 votes]: This answer is meant to illustrate "another route" that "might be preferable".
First, we need a convenient characterizations of limits. The most flexible one for our purposes is via representability: $\mathsf{Hom}(X,\mathsf{Lim}_{I\in\mathcal{I}}D(I))\cong\mathsf{Lim}_{I\in\mathcal{I}}\mathsf{Hom}(X,D(I))$ natural in $X$. This can be taken as the definition of what a limit is assuming you know what limits are in $\mathbf{Set}$. Alternatively, you can easily prove this fact from other characterizations of limits.
The calculation that right adjoints preserves limits is then simple and direct: $$\begin{align}
\mathsf{Hom}(X,G(\mathsf{Lim}_{I\in\mathcal{I}}D(I)))
& \cong \mathsf{Hom}(F(X),\mathsf{Lim}_{I\in\mathcal{I}}D(I)) \\
& \cong \mathsf{Lim}_{I\in\mathcal{I}}\mathsf{Hom}(F(X),D(I)) \\
& \cong \mathsf{Lim}_{I\in\mathcal{I}}\mathsf{Hom}(X,G(D(I))) \\
& \cong \mathsf{Hom}(X,\mathsf{Lim}_{I\in\mathcal{I}}G(D(I)))
\end{align}$$
The desired result then follows by Yoneda. $\square$
Going one step further in our characterization of limits, we can calculate (using ends and the naturality formula or directly) that $\mathsf{Nat}(\Delta X,D)\cong\mathsf{Hom}(X,\mathsf{Lim}_{I\in\mathcal{I}}D(I))$ which is to say, if this is additionally natural in $D$, that $\Delta \dashv \mathsf{Lim}$. We can take this adjunction as our characterization of having limits of shape $\mathcal{I}$. (Note, before we were only talking about a specific limit of $D$, now we're talking about having the limit of any diagram of shape $\mathcal{I}$.) This characterization produces a basically identical derivation:
$$\begin{align}
\mathsf{Hom}(X,G(\mathsf{Lim}(D)))
& \cong \mathsf{Hom}(F(X),\mathsf{Lim}(D)) \\
& \cong \mathsf{Nat}(\Delta(F(X)),D) \\
& = \mathsf{Nat}(F\circ\Delta(X),D) \\
& \cong \mathsf{Nat}(\Delta(X),G\circ D) \\
& \cong \mathsf{Hom}(X,\mathsf{Lim}(G\circ D))
\end{align}$$
$\square$