Right adjoints preserve limits

Question

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?

Answer 1

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.

Answer 2

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 $f_i$.