A proof that right adjoints preserve limits?

Question

Assume that categories $\mathscr{B}$ and $\mathscr{C}$ have all limits of shape $\mathbf{J}$. Then there's a slick proof that if $G\colon \mathscr{C} \to \mathscr{B}$ is a right adjoint, $G \circ Lim_\mathscr{C} \cong Lim_\mathscr{B} \circ [\mathbf{J}, G]$.

[$Lim_\mathscr{C} \colon [\mathbf{J}, \mathscr{C}] \to \mathscr{C}$ sends a diagram $D \colon \mathbf{J} \to \mathscr{C}$ to the vertex of a limit cone over $D$. $[\mathbf{J}, G]$ is a functor which, acting on objects in $[\mathbf{J}, \mathscr{C}]$ sends $D$ to $GD$. And the slick proof is to take left adjoints, noting that the left adjoint of $Lim$ is a constant functor, show very easily that the composite left adjoints are equal, so their right adjoints must be equal up to natural isomorphism. But proof details do not matter here.]

Now, I have seen it very briskly said -- e.g. in notes of a course by Peter Johnstone -- that this result shows that, at least when dealing with categories that have all limits of shape $\mathbf{J}$, right adjoints preserve those limits.

The usual definition of "$G$ preserves limits" is that if $(L, \pi_J)$ is a limit cone over $D$, then $(GL, G\pi_j)$ is a limit cone over $GD$.

But does the stated result establish that? One issue is that not every limit cone over $D$ has a vertex $Lim_\mathscr{C}D$ -- for the functor $Lim_\mathscr{C}$ has to be defined in terms of a choice from limit cones over $D$. So strictly speaking, the slick proof doesn't (as it stands, without augmentation) tell us what $G$ does to those other limit cones. [Added: OK: We can prove that if $G$ preserves one limit cone over $D$ it preserves all limit cones over $D$, and then make use of this fact here.]

But more basically, even when we have a limit cone with a vertex $Lim_\mathscr{C}D$, what the result shows is that applying $G$ gives a result $X$ which a certain natural bijection maps to the vertex of a limit cone over $GD$. But it doesn't actually tell us that $X$ is the vertex of a limit cone over $GD$ as we require for full preservation. Or so it seems.

Given Peter Johnstone can do no wrong(!) where am I missing the point?

Answer 1

Well, indeed there is a choice when defining either $Lim_{\mathscr B}$ or $Lim_{\mathscr C}$.
However, this choice picks one object only among isomorphic objects, since any two limiting cones are naturally isomorphic (hence their summit vertices are isomorphic).
Note on the other hand, if $X$ is a limit vertex for a diagram $D$, and $f:Y\to X$ is an isomorphism, then composing every arrow by $f$, we get a limiting cone with vertex $Y$.

Moreover, assume that we fixed a limit vertex $L_D:=Lim_{\mathscr C}D$ (with limiting cone!) for each and every diagram $D:{\bf J}\to\mathscr C$, and we also fix arbitrary isomorphisms $f_D:L'_D\to L_D$. Then, $Lim'_{\mathscr C}(D):=L'_D$ (with the limiting cone composed by $f_D$) will be an equally legal limit functor.

So, knowing $G\circ Lim_{\mathscr C}\simeq Lim_{\mathscr B}\circ[{\bf J},G]$ is sufficient, and it does express that $G$ preserves limits of shape $\bf J$, as $G(Lim_{\mathscr C}(D))\cong Lim_{\mathscr B}(GD)$ expresses nothing but that $G(Lim_{\mathscr C}(D))$ is a limit of diagram $GD$ (with corresponding limiting cone).

Letting $L:=Lim D$, and using naturality of these isomorphisms $\phi_D$ at the limiting cone $\pi:\Delta L\to D$ as a morphism of category $[{\bf J},\mathscr C]$, we get the commutative square $$\matrix{G(L) & = & G(L) \\ \Vert && \downarrow & \hspace{-2.4pc} \scriptstyle{Lim(G\pi)} \\ G(L) &\underset{\phi_D}\to& Lim(GD) }$$ where $Lim(G\pi)$ is the arrow induced by factoring $G\pi$ through the $Lim_{\mathscr B}$-cone.

Answer 2

I don't think that this proof shows the fully explicit mapping of any limit cone over $D$ to a limit cone over $GD$ that you want. That's a hard claim to justify precisely, so instead perhaps I can suggest how the slick proof could be refined to get exactly the claim you want. Let $J^l$ be the general cone over $J$, that is, $J$ with a disjoint initial object attached. Then we can refine the limit functor to a functor $\lim:[J,\mathcal{C}]\to [J^l,\mathcal{C}]$, with $\lim D$ simply a choice of limit cone over $D$. This limit functor is right adjoint to the restricton to $J$, and the same argument as that you alluded to shows that $G$ send the limit cone on $D$ specified by $\lim$ to something naturally isomorphic to the specified limit cone on $GD$, that is, to a limit cone. Then this composes with the natural isomorphism of any limit cone on $D$ to the specifed one to get the full result desired.