# Right adjoints preserve limits

In Awodey's book I read a slick proof that right adjoints preserve limits. If $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ is a pair of functors such that $(F,G)$ is an adjunction, then if $D:I\to \mathcal{D}$ is a diagram that has a limit, we have, for every $A\in \mathcal{C}$,

\begin{align*} \hom_\mathcal{C} (A, G(\varprojlim D)) &\simeq \hom_{\mathcal{D}} (F(A),\varprojlim D)\\ & \simeq \varprojlim \hom_{\mathcal{D}}(F(A),D)\\& \simeq \varprojlim \hom_{\mathcal{C}}(A,GD) \\& \simeq \hom_{\mathcal{C}}(A,\varprojlim GD)\end{align*}

because representables preserve limits. Whence, by Yoneda lemma, $G(\varprojlim D)\simeq \varprojlim GD$.

This is very slick, but I can't really see why the proof is finished. Yes, we proved that the two objects are isomorphic, but a limit is not just an object... Don't we need to prove that the isomorphism also respects the natural maps? That is,

if $\varphi:G(\varprojlim D)\to \varprojlim GD$ is the isomorphism, and $\alpha_i: \varprojlim D \to D_i$, $\beta_i:\varprojlim GD \to GD_i$ are the canonical maps for all $i\in I$, do we have that $\beta_i\varphi=G(\alpha_i)$?

I don't see how this follows from Awodey's proof. How can we deduce it?

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I've just seen this is also the proof in ncatlab: ncatlab.org/nlab/show/adjoint+functor – Bruno Stonek Jan 21 '12 at 14:41
Dear Bruno: Here the symbol $\simeq$ means "is canonically isomorphic to", not just "is isomorphic to". – Pierre-Yves Gaillard Jan 21 '12 at 14:47
@Pierre: Yes, I'm aware of this, and in fact it is necessary for Yoneda lemma to apply. I thought it was only necessary for this... Hmm... – Bruno Stonek Jan 21 '12 at 14:53
Detail: Here is a more direct link to the relevant part of the nLab entry: ncatlab.org/nlab/show/adjoint+functor#general_18. – Pierre-Yves Gaillard Jan 21 '12 at 15:00
Dear Bruno: I'll try to write a proof using the notation of these notes by Pierre Schapira. --- See more precisely Proposition 2.4.5 p. 36. – Pierre-Yves Gaillard Jan 21 '12 at 15:37

## 4 Answers

The proof is very nice, but one needs to be absolutely clear about what needs to be proven in order to understand it. The claim is, if $\lambda_i : \varprojlim D \to D_i$ is a limiting cone in $\mathcal{D}$, then $G \lambda_i : G(\varprojlim D) \to G D_i$ is a limiting cone in $\mathcal{C}$. We do not postulate the existence of $\varprojlim G D$; this is what we are going to prove.

So suppose we are given a cone $\mu_i : X \to G D_i$ in $\mathcal{C}$. This yields a cone of hom-sets $$(\mu_i)_* : \mathcal{C}(A, X) \to \mathcal{C}(A, G D_i)$$ and since $\varprojlim \mathcal{C}(A, G D_i) \cong \mathcal{C}(A, G(\varprojlim D))$ naturally in $A$ (by the argument you cited), by the Yoneda lemma it follows that there is a unique natural transformation $\varphi_* : \mathcal{C}(-, X) \Rightarrow \mathcal{C}(-, G(\varprojlim D))$ such that $$(\mu_i)_* = (G \lambda_i)_* \circ \varphi_*$$ where $\varphi_*$ comes from a morphism $\varphi : X \to G(\varprojlim D)$. Thus $G \lambda_i : G(\varprojlim D) \to G D_i$ is indeed a limiting cone.

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Thank you for your answer; however, I'm confused by notation and nomenclature. I don't really understand the argument... – Bruno Stonek Jan 21 '12 at 15:28
Can you be more specific about what needs clarification? – Zhen Lin Jan 21 '12 at 15:30
I dont't see how you conclude there is a unique natural transformation such that [...] – Bruno Stonek Jan 21 '12 at 15:32
For each $A$ we get a unique map $(\phi_*)_A : \mathcal{C}(A, X) \to \mathcal{C}(A, G(\varprojlim D))$ such that various diagrams commute, and the collection of all of these is a natural transformation of functors. I would write out the whole argument, but it's just a matter of drawing the right diagrams (which is difficult to do here). – Zhen Lin Jan 21 '12 at 15:41

This is essentially taken from the proof of Proposition 2.4.5 p. 36 in these notes by Pierre Schapira.

Let me use the abbreviation $$L:=\lim_{\underset{i}{\longleftarrow}}\quad.$$

Let $C$ and $D$ be categories, let $b:I^{op}\to C$, $i\mapsto b_i$, be a projective system, let $Lb$ be its limit (we assume it exists), let $F:C\to D$ be a functor, let $G:D\to C$ be its left adjoint (we assume it exists), and let $y$ be an object of $D$.

We have the following commuting squares of morphisms and isomorphisms, where the vertical arrows are induced by the $i$ th canonical projections: $$\begin{matrix} D(y,FLb)&\simeq&C(Gy,Lb)\\ \downarrow&&\downarrow\\ D(y,Fb_i)&\simeq&C(Gy,b_i), \end{matrix}$$

$$\begin{matrix} C(Gy,Lb)&\simeq&LC(Gy,b)\\ \downarrow&&\downarrow\\ C(Gy,b_i)&=&C(Gy,b_i), \end{matrix}$$

$$\begin{matrix} LC(Gy,b)&\simeq&LD(y,Fb)\\ \downarrow&&\downarrow\\ C(Gy,b_i)&\simeq&D(y,Fb_i), \end{matrix}$$

$$\begin{matrix} LD(y,Fb)&\simeq&D(y,LFb)\\ \downarrow&&\downarrow\\ D(y,Fb_i)&=&D(y,Fb_i). \end{matrix}$$ By splicing these squares, we get the following commuting square of morphisms and isomorphisms: $$\begin{matrix} D(y,FLb)&\simeq&D(y,LFb)\\ \downarrow&&\downarrow\\ D(y,Fb_i)&=&D(y,Fb_i), \end{matrix}$$ which is what we wanted.

EDIT A. Let's go back to the first square: $$\begin{matrix} D(y,FLb)&\simeq&C(Gy,Lb)\\ \downarrow&&\downarrow\\ D(y,Fb_i)&\simeq&C(Gy,b_i). \end{matrix}$$ The isomorphisms are given by the adjunction. If $p_i:Lb\to b_i$ denotes the $i$ th canonical projection, then the first downward arrow is $D(y,Fp_i)$, and the second is $C(Gy,p_i)$. To show that the square commutes, we only need to invoke the fact that the adjunction is functorial in the second variable.

Now to the second square: $$\begin{matrix} C(Gy,Lb)&\simeq&LC(Gy,b)\\ \downarrow&&\downarrow\\ C(Gy,b_i)&=&C(Gy,b_i). \end{matrix}$$ By assumption, we have chosen a representing object $Lb$ and an isomorphism $$C(x,Lb)\simeq LC(x,b)$$ functorial in $x\in\text{Ob}(C)$. We have a natural map from $LC(x,b)$ to $C(x,b_i)$ --- because $LC(x,b)$ is a projective limit of sets. Then we define the map from $C(x,Lb)$ to $C(x,b_i)$ as the one which makes the above square commutative. All this being functorial in $x$, the Yoneda Lemma yields the morphism $p_i:Lb\to b_i$ used above.

EDIT B. The third and fourth squares are handled similarly. So we end up with the square $$\begin{matrix} D(y,FLb)&\simeq&D(y,LFb)\\ \downarrow&&\downarrow\\ D(y,Fb_i)&=&D(y,Fb_i), \end{matrix}$$ which commutes for all $i$. What we want to prove is the existence of an isomorphism $FLb\simeq LFb$ such that the square $$\begin{matrix} FLb&\simeq&LFb\\ \downarrow&&\downarrow\\ Fb_i&=&Fb_i \end{matrix}$$ commutes for all $i$. But, in view of Yoneda, the above square commutes because the previous one does.

EDIT C. Alternative wording of the poof that $$\begin{matrix} C(x,Lb)&\simeq&LC(x,b)\\ \downarrow&&\downarrow\\ C(x,b_i)&=&C(x,b_i) \end{matrix}$$ commutes:

A morphism $f\in C(x,Lb)$ is given by a family $f_\bullet=(f_j)_{j\in I}\in LC(x,b)$ satisfying the obvious compatibility conditions, and we have $f_j=p_j\circ f$ for all $j$. So, $f$ and $f_\bullet$ correspond under the isomorphism in the above square. Moreover, the first vertical arrow maps $f$ to $f_i$, and the second vertical arrow maps $f_\bullet$ to $f_i$.

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Dear Pierre: thank you very much for your answer. I understand this a bit more, I think, but I still can't quite grasp it. For example, why is the fourth line an equality? Yes, I know the objects are the same, but why do you know the isomorphism that makes the square commute is the equality? (same for the last square). Also, why does the last commuting square give us what we want? My guess is that the left vertical arrow is $(F(\lambda_i))_*$ and the right vertical arrow is $(\beta_i)_*$, where the $\lambda_i$ and $\beta_i$ are the canonical morphisms; if we view the horizontal isomorphism.. – Bruno Stonek Jan 21 '12 at 20:21
...as $\varphi_*$, then since the Yoneda embedding is faithful, it must be that $F(\lambda_i)=\beta_i\varphi$ which is what we wanted. But I don't know if this is fine. – Bruno Stonek Jan 21 '12 at 20:22
Surely it is proof that I still don't quite grasp these things, but I find it bewildering that the proof given above skips these steps. Are they supposed to be obvious? Are they really encoded in the natural isomorphisms, and I'm failing to see them? – Bruno Stonek Jan 21 '12 at 20:25
In the second comment: we can write the isomorphism as $\varphi_*$ because the Yoneda embedding is full. (I dislike not being able to edit the comments after five minutes...) – Bruno Stonek Jan 21 '12 at 21:05
Dear @Bruno: I tried to make things as clear as I could. It seems to me Schapira's handout is a nice reference. He knows all this much better than I, and he explains it very well. I find your questions and observations highly interesting. – Pierre-Yves Gaillard Jan 21 '12 at 21:47

I would say you're overcomplicating the issue. You've postulated these maps $\beta_i$...where are they coming from? Better is to show simply that if $(X, \mu_i)$ is a limit of a diagram $H:I \to C$, and $\phi: Y \xrightarrow{\sim} X$, then $\phi^{-1}: (X, \mu_i) \xrightarrow{\sim} (Y, \mu_i \circ \phi)$ is an isomorphism of cones (further, one can show that it is the unique isomorphism $(X, \mu_i) \to (Y, \mu_i \circ \phi)$, since an isomorphism where one of the objects is final is a unique isomorphism, and $(X, \mu_i)$ is final in the category of cones on $H$). Now morphisms of cones $(Z, \nu_i) \to (X, \mu_i)$ are in bijection with morphisms of cones $(Z, \nu_i) \to (Y, \mu_i \circ \phi)$. So $Y$ is a limit.

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I don't think this will work. You need to show the cone <GlimD(i), Gλ(i)> is a limit cone. – Chilango Feb 20 '15 at 15:31

Awodey sometimes hand waves his proofs, especially the ones in which you have to get your hands dirty--for instance, his proof that every presheaf is a colimit of representables.

Anyway, in this case, application of Yoneda is overkill because it is no harder to prove RAPL just by using adjointness:

Suppose $\left \langle L,\lambda _{i} \right \rangle$ is a limit cone to $D$. Then $\left \langle GL,G\lambda _{i} \right \rangle$ is a cone to $GD$. Suppose $\left \langle X,\mu _{i} \right \rangle$ is a cone to $GD$. Then taking adjoints, we get a cone to $D$, namely, $\left \langle FX,\mu _{i}^{*}\right \rangle$. Since $\left \langle L,\lambda _{i} \right \rangle$ is a limit cone, we get an arrow $\phi ^{*}:FX\rightarrow L$ unique with property that $\lambda _{i}\circ \phi ^{*}=\mu ^{*}_{i}$. Taking adjoints again, it is easy to see that $\phi :X\rightarrow GL$ is the unique arrow making the required triangle commute, for

if we write \begin{array}{ccc} \hom( X,GL) & \rightarrow & \hom (FX,L) \\ \downarrow & & \downarrow \\ \hom(X,GD _{i}) & \rightarrow & \hom(FX,D_{1} ) \end{array} and follow $\phi$ around the square, we see that $\lambda _{i}\circ \phi ^{*} =(G\lambda _{i}\circ \phi )^{*}$. But LHS of this is just $\mu ^{*}_{i}$, so that $G\lambda _{i}\circ \phi =\mu_{i}$ and $\phi$ is unique because $\phi ^{*}$ is.

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