Lemma 1.8.9 in B. Jacobs book Categorical Logic and Type Theory describes how a collection of fibre-wise adjoints of a morphism of fibrations can be promoted to a global adjoint.
Let $p: \mathbb E \to \mathbb B$ and $q: \mathbb D \to \mathbb B$ be Grothendieck fibrations. Assume $H: \mathbb E \to \mathbb D$ is a functor such that $qH = p$ and $H$ preserves cartesian morphisms. Then $H$ has a fibred left (respectively right) adjoint $K$ if and only if the following two conditions hold.
(a) Each induced $H_I: \mathbb E_I \to \mathbb D_I$ has a left (respectively right) adjoint $K_J$.
(b) The $K_J$ satisfy a Beck-Chevalley condition. For each morphism $u: I\to J$ in the base $\mathbb B$ the canonical natural transformation $K_Iu^*\Rightarrow u^* K_J$ (respectively $u^*K_J \Rightarrow K_Iu^*$) is an isomorphism.
That $K$ is fibred just means that $pK = q$ and that $K$ preserves cartesian morphism. Jacobs uses the word fibred instead of the more standard term cartesian functor.
In the proof Jacobs shows how to produce a left adjoint $K$ from the $K_J$ and states that the result for right adjoints (in brackets) can be proven in a similar way.
I am not sure that this is true. I believe that the correct dual result for right adjoints should involve cofibrations and cofibred functors instead of fibrations. At least that is the point where I get stuck when I try to dualize Jacobs proof. So my question is: Is Jacobs lemma 1.8.9 correct? Does it work out for both left and right adjoints?
The rest of the question are just context, references and more details!
Let me start by repeating Jacobs proof, so that I can point you to the exact place where I get stuck when I try to dualize it. So assume we are given fibre-wise left adjoints $K: \mathbb D_I \to \mathbb E_I$ of $H: \mathbb E_I \to \mathbb D_I$ for each object $I$ in the base category. I will denote all of them by the same letter $K$ to make the proof more readable. Consequently I will also denote all counits and units by the same letter $\varepsilon$ and $\eta$ respectively. Assume we have chosen a cleveage for both $p$ and $q$. We will show that $H$ has a total left adjoint $K: \mathbb D \to \mathbb E$ by showing that $(KX, \eta X: X \to HKX)$ is initial in $X \downarrow H$ and not just in the fibre $X \downarrow H_{qX}$. So take some $(Y,f: X \to HY)$. We need to show that there is $g: KX \to Y$ such that $f = Hg\circ \eta X$, and we need to show that there is only one such $g$. Let $u: I\to J$ be the image of $f$ in the base $\mathbb B$.
To see that there is some $g$, consider the cartesian lift $\overline uY: u^*Y \to Y$ of $u$ in $\mathbb E$ and apply $H$ to it. The resulting map $H(\overline uY)$ is cartesian in $\mathbb D$ because $H$ is fibred by assumption. We need a picture.
Because $H(\overline u Y)$ is cartesian, we get the map $X \to Hu^*Y$ which is already drawn in the picture. Now we can use the fact that $\eta X$ is initial in the fibre $\mathbb E_I$ to obtain a $k: KX \to u^*Y$ such as in the diagram. We can now set $g = \overline uY \circ k$.
I will skip over the rest of proof, because the relevant part is already there. Jacobs proceeds by showing that $g$ is unique. This yields $K$, and one doesn't need the Beck-Chevalley condition at all to produce $K$. The Beck-Chevalley condition is finally used to show that the functor $K$ is fibred (that is preserves cartesian morphisms).
Next let me show you where I get stuck when I try to perform the coproof. As before, assume that we are given fibre-wise right adjoints $K: \mathbb D_I \to \mathbb E_I$. We like to show that $H$ has a total right adjoint $K$ by showing that each $(X, \varepsilon X: HKX \to X)$ is terminal in all of $H\downarrow X$ and not just in its fibre. So take some $f: HY \to X$. Write $u: I\to J$ for the image of $f$ in the base category $\mathbb B$. We are in the following situation.
Now what? In the case of a left adjoint $K$ we proceeded by taking the pullback $u^*Y \to Y$ of $Y$, but we can't do this now. It seems like we need a cocartsian lift $\underline uY : Y \to u_*Y$. I am stuck. Am I missing something?
Relevant references: I checked out most of the references which are mentioned in chapter I of Jacobs book. I couldn't find the result anywhere else, except in B. Jacobs PhD thesis where they state it without a proof. The version of the lemma where $K$ is left adjoint appears as proposition 8.4.1. in Borceux's Handbook of Categorical Algebra 2, but they only state the result for left adjoints!
Best Answer
The Beck-Chevalley condition implies $H$ is a total right adjoint as follows.
Given a cartesian lift $u^*X\to X$, we have that $HY\to X$ factors as $HY\to HKu^*X\to u^*X\to X$ for a unique morphism $Y\to Ku^*X$. The Beck-Chevalley condition is that the unique vertical morphism $u^*KX\to Ku^*X$ for which $Hu^*KX\to HKX\to X$ factors as $Hu^*KX\to HKu^*X\to u^*X\to X$ is an isomorphism.
Its inverse is then an isomorphism $Ku^*X\to u^*KX$ such that $HKu^*X\to u^*X\to X$ is the same as $HKu^*X\to Hu^*KX\to HKX\to X$. It follows that $Y\to Ku^*X\to u^*KX\to KX$ is the unique morphism for which $HY\to X$ factors as $HY\to HKX\to X$.