# 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?

• I've just seen this is also the proof in ncatlab: ncatlab.org/nlab/show/adjoint+functor Jan 21 '12 at 14:41
• Dear Bruno: Here the symbol $\simeq$ means "is canonically isomorphic to", not just "is isomorphic to". 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... Jan 21 '12 at 14:53
• Dear Bruno: If $D\to D_i$ is one of the natural maps, then all the isomorphisms in the chain are compatible with it. Jan 21 '12 at 15:07
• 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. Jan 21 '12 at 15:37

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.

• Thank you for your answer; however, I'm confused by notation and nomenclature. I don't really understand the argument... Jan 21 '12 at 15:28
• Can you be more specific about what needs clarification? Jan 21 '12 at 15:30
• I dont't see how you conclude there is a unique natural transformation such that [...] 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). Jan 21 '12 at 15:41
• I think Bruno is right here, at that point you do use the fact that the $G(\lambda_i)_*$ are actually a limit cone to write down that equality: you consider that the isomorphism on the objects is compatible with the cone structure, which is exactly what we want to prove. Oct 9 '19 at 9:57

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 $f_i$.

• 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.. 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. 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? 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...) 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. Jan 21 '12 at 21:47

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.

A new answer was added recently, and from my perspective, none of the answers reflect what I think the proof is trying to get at, so I thought I'd add another answer. However, I do agree with Zhen Lin's answer (+1) that the proof goes wrong when it assumes already that the limit of $$GD$$ exists. The point is rather to show that $$(G\lim D,G(\alpha_i))$$ form a limiting cone if $$(\lim D,\alpha_i)$$ form a limiting cone for $$D$$.

Other answers argue that there are better proofs, and I won't necessarily agree or disagree, but the idea of this proof is sound, so let me see if I can salvage it.

Review of the Yoneda Lemma


In particular, if you have a natural isomorphism $$\C(-,L)\simeq \text{cones from - to }D,$$ the natural isomorphism corresponds to a cone from $$L$$ to $$D$$, given by the image of $$1_L\in\C(L,L)$$, which is the limiting cone.

The argument

Thus, the way this argument should go is the following: Let $$L=\lim D$$ for notational convenience. \begin{align} \C(-,GL)&\simeq \D(F-,L) \\ &\simeq \lim_i \D(F-,D_i)\\ &\simeq \lim_i \D(-,GD_i)\\ &\simeq \text{cones from - to } GD. \end{align} This natural isomorphism proves that $$GL$$ is the limit of $$GD$$ already. To see what the limiting cone is, we have to see where $$1_{GL}$$ maps under the natural isomorphism.

Under the first natural isomorphism, $$1_{GL}$$ maps to the [counit] of the adjunction, $$\epsilon_L : FGL\to L$$. The next natural isomorphism sends $$\epsilon_L$$ to $$(\alpha_i\circ \epsilon_L)_i$$, which belong to the limit of hom sets because the $$(\alpha_i)$$ do. Finally, the adjunct of $$\alpha_i\circ \epsilon_L$$ is given by $$G(\alpha_i\circ \epsilon_L)\circ \eta_{GL} = G\alpha_i \circ G\epsilon_L \circ \eta_{GL} =G\alpha_i,$$ with the last equality being the triangle identities.

Finally, the last isomorphism is just a reinterpretation of the elements of the limit as cones, since checking what it means to take the limit in $$\Set$$ of the hom-sets against what it means for elements to be cones we see that they agree. Thus $$(G\alpha_i)$$ is the limiting cone corresponding to this natural isomorphism, as desired.

TL;DR

The natural isomorphism already encodes the data of the limiting cone. All we need to do is check that we get the expected limiting cone.

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.

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

There seems to be a misunderstanding. The proof is correct without further steps.

One can read a similar version of the proof given by Tom Leinster in the book "Basic category theory", which is freely available on the arxiv: https://arxiv.org/abs/1612.09375

The proof is given on p. 159, Theorem 6.3.1 and he gives reasons for each performed step, referencing all relevant Lemmas. In the following, I will try to summarize as briefly as possible why the steps (that also appear in the question above) make the proof complete.

First of all, all the isomorphisms used are already natural in $$A$$. Second, it might be worth noting what the pre-last line

$$\lim_{\leftarrow} \text{hom}_{\mathcal{C}}(A, GD)$$

actually means: $$\text{hom}_{\mathcal{C}}(A,GD):I\to \text{Set}$$ is itself a functor because $$GD$$ takes an object $$i$$ of $$I$$ and maps it to $$\mathcal{C}$$, so $$\text{hom}_{\mathcal{C}}(A,GD)$$ takes an object $$i$$ of $$I$$ and maps it to the category of sets. As a consequence, the limit in the pre-last line is actually taken in the category of sets and can thus be given an explicit form. It is the set of all families $$(f_i), i\in I$$ such that $$f_i \in \text{hom}_{\mathcal{C}}(A, GD(i))$$ for all $$i \in I$$ and

$$GD(u)\circ f_i = f_j$$

for all $$u:i\to j$$. This means an element of $$\lim_{\leftarrow} \text{hom}_{\mathcal{C}}(A, GD)$$ is the same thing as a cone on $$GD$$ with vertex $$A$$. (cf. Leinster's book, p. 147, Lemma 6.2.1)

And now the proof outlined in the question asked tells us that all cones on $$GD$$ with vertex $$A$$ are actually in natural 1-to-1 correspondence with the maps from $$A$$ to $$G(\lim_{\leftarrow} D)=:L$$. This $$L$$ is therefore a represenation for the cones on $$GD$$ with vertex $$A$$ and therefore the isomorphisms respect the natural maps asked for by the OP, implying that $$G(\lim_{\leftarrow} D)$$ must be the limit of $$GD$$.

• It appears to me that the question the op is asking is why is $G(\alpha_i)$ a limit cone. I.e., how do you show the morphisms are correct. Your answer doesn't appear to address this.
– jgon
Apr 11 '20 at 18:03
• @jgon it does adress this because I pointed to the fact that the 1-to-1-correspondence is natural in $A$. Therefore, the morphisms turn out to be correct. In particular, $G(\lim D)$ is a representation for the cone on $GD$ with vertex $A$ which implies what you wrote in your new answer. This is also explained in the notes by Leinster I referenced. But I will edit the answer to add the represenation part, thank you. Apr 12 '20 at 17:25