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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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    Related: https://math.stackexchange.com/questions/271010/2018-11-28

4 Answers 4

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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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    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).2012-01-21
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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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    @Pierre-YvesGaillard You are welcome. Very informative web site.2012-01-24
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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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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 is a limit cone.2015-02-20