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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?

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    $\begingroup$ Just to make sure, you do know how to prove (2), right? If not, see the proof of Theorem 4.4.2 in Emily Riehl, Category theory in context (version 14 Jul 2015), math.harvard.edu/~eriehl/161/context.pdf . When reading Riehl's notes (book, let's be honest), I have asked myself the same question -- whether the "slick" proof says anything of value about what happens to the limit cone. I found no reason to believe that it does, but someone more expert should have more to say. $\endgroup$ Jul 14 '15 at 21:18
  • $\begingroup$ Yes, thanks. And I know Emily Riehl's terrific notes. As you say, she uses the same proof (in fact "slick" is her word), which she probably picked up as a Part III student in Cambridge! $\endgroup$ Jul 14 '15 at 21:32
  • $\begingroup$ Sorry, @darijgrinberg -- I realised there was a garble in the way I put things before, so I've slightly edited. And points numbered (1) and (2) have gone. So, for others, (2) is the claim that G preserves limits in the full sense as now defined in the fourth para. $\endgroup$ Jul 14 '15 at 22:38
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    $\begingroup$ There is something to be extracted here. First of all, the limit cone is encoded into the counit of the adjunction $\Delta \dashv \varprojlim$. Secondly, the proof that adjoints compose also tells you something about the counits. So you just need to unfold these proofs a bit. $\endgroup$
    – Zhen Lin
    Jul 15 '15 at 8:06
  • $\begingroup$ By the way, am I seeing it right that the assumption that "categories $\mathscr{B}$ and $\mathscr{C}$ have all limits of shape $\mathbf{J}$" is a WLOG assumption, because one can embed any two categories $\mathscr{B}$ and $\mathscr{C}$ into $\mathsf{Set}^{\mathscr{B}^{\operatorname{op}}}$ and $\mathsf{Set}^{\mathscr{C}^{\operatorname{op}}}$ via the Yoneda embeddings (which reflect limits)? $\endgroup$ Jul 16 '15 at 11:50
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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.

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  • $\begingroup$ Nice! Essentially your refined limit functor keeps the limit and its limit cone together, while the standard limit functor throws the cone away. This trick should probably help out in various other places. $\endgroup$ Jul 16 '15 at 11:42
  • $\begingroup$ @darijgrinberg Thanks, glad you like it! $\endgroup$ Jul 16 '15 at 20:43
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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.

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  • $\begingroup$ In short: the set of limiting cones is closed under composing with an isomorphism. $\endgroup$
    – Berci
    Jul 14 '15 at 21:47
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    $\begingroup$ I don't get the last part. $G(\operatorname{Lim}D) ≅ \operatorname{Lim}(GD)$ guarantees that $G(\operatorname{Lim} D)$ is a limit of $GD$, with the limiting cone you get by dragging the $\operatorname{Lim}(GD)$'s limiting cone along the isomorphism. How do you know that this equals $Gπ$, where $π : Δ(\operatorname{Lim} D) → D$ is the limiting cone for $D$? $\endgroup$
    – user54748
    Jul 14 '15 at 22:28
  • $\begingroup$ I added. Assumed that $Lim(\Delta x)=x$ for all objects $x$. $\endgroup$
    – Berci
    Jul 14 '15 at 23:25
  • $\begingroup$ But in general this doesn't hold, and we have $\operatorname{Lim}(Δx) ≅ x^{cc(J)}$, where $cc(J)$ are connected components of $J$, no? And even if $J$ is connected, there's a lot of things involved in passing from the naturality square for $φ$ to the one in your post. The one I'm actually not sure about is can you assume that $φ_{Δ(\operatorname{Lim}D)} = \mathrm{id}$? $\endgroup$
    – user54748
    Jul 15 '15 at 0:16

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