Proving adjoint functors preserve limits by lifting the adjunction to cone categories

Question

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?

Answer 1

It is indeed true that $F'\dashv G'$, and not too hard to show once you have the right picture in mind.

The thing to notice is that an object in $D'$ is a cone over $Q$, i.e. an object $y$ of $D$ together with a natural transformation $\beta:\Delta_D(y)\Rightarrow Q$, and an arrow $(y,\beta,Q)\to (y',\beta',Q)$ is just an arrow $g:y\to y'$ in $D$ that commutes with $\beta$ and $\beta'$, in the sense that $\beta_i'\circ g=\beta_i$ for all objects $i$ of $I$. $C'$ can be described similarly.

In particular, an arrow $(x,\alpha,GQ)\to (Gy,G\beta,GQ)$ is just an arrow $h:x\to Gy$ such that $G\beta_i\circ h=\alpha_i$, and an arrow $(Fx,\alpha^{ad},Q)\to (y,\beta,Q)$ is just an arrow $k:Fx\to y$ such that $\beta_i\circ k =\alpha_i^{ad}$; so the bijection induced by the adjunction $F\dashv G$ restricts (by naturality) to a bijection between arrows of one kind and arrows of the other kind, and it is still natural, since $F'$ and $G'$ acts on arrows like $F$ and $G$. Hence $F'\dashv G'$.

Answer 2

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: $$\begin{align} \mathsf{Hom}(X,G(\mathsf{Lim}(D))) & \cong \mathsf{Hom}(F(X),\mathsf{Lim}(D)) \\ & \cong \mathsf{Nat}(\Delta(F(X)),D) \\ & = \mathsf{Nat}(F\circ\Delta(X),D) \\ & \cong \mathsf{Nat}(\Delta(X),G\circ D) \\ & \cong \mathsf{Hom}(X,\mathsf{Lim}(G\circ D)) \end{align}$$ $\square$