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2 added 292 characters in body

(Edit: As Mike Shulman points out, I switched my left and right in what follows. Furthermore, since I don't produce a morphism of fibrations, I probably do not even address the original question, regardless of the chirality confusion. Sorry! His answer is clearly a much better response.)

I'll write something very concrete, perhaps more in the style of Vistoli's notes or SGA 1.

How certain are you that you need a left adjoint? If you look at right adjoints instead, then the result proves itself. Unless I am misreading Finn Lawler's response, it looks as if he also switched to right adjoints, but since he doesn't mention it, perhaps I am misunderstanding. I'll write out the details to produce the right adjoint, but it is probably easier to work it out yourself than to read what follows.

I'm guessing that when you write "morphism $F\to G$," you mean a cartesian functor over $C$. (I looked up your reference to Vistoli's exposition, and this seems to be his usage.) If that guess is wrong, then what I say below may not be relevant, since I use the cartesian property repeatedly.

Let $\Phi: F \to G$ be a cartesian functor between categories fibered over $C$. Assume that for each object $U$ of $C$, the functor $\Phi_U:F_U \to G_U$ on fiber categories admits a right adjoint. We wish to show that $\Phi$ admits a right adjoint. Let $y$ be an object of $G$ over the object $U$ of $C$. We need to produce a universal arrow $y\to \Phi x$ from $y$ to $\Phi$. The obvious candidate is a universal arrow $y\to \Phi_U x = \Phi x$ to $\Phi_U$, which exists by hypothesis (after switching your "left" for my "right"). We must prove that this arrow is universal to $\Phi$.

Let $y\to \Phi x'$ be a morphism in $G$ covering $\rho: U\to U'$ in $C$. Let $\rho^* x'\to x'$ be a pullback in $F$. Since $\Phi$ is cartesian, $\Phi(\rho^* x') \to \Phi x'$ is a pullback in $G$, and so $y\to \Phi x'$ factors uniquely as the composition of a morphism $y \to \Phi(\rho^* x')$ in $G_U$ with $\Phi(\rho^* x')\to \Phi x'$. By the universal property of $y \to \Phi x$ over $U$, there is a unique morphism $x \to \rho^* x'$ in $F_U$ such that $y \to \Phi(\rho^* x')$ factors as the composition of $y\to \Phi x$ and $\Phi x\to \Phi(\rho^* x)$. Composing $x\to \rho^* x'$ and $\rho^* x'\to x$ gives us a morphism $x\to x'$ covering $\rho$ such that $y \to \Phi x'$ factors as the composition of our candidate universal arrow $y \to \Phi x$ and $\Phi x \to \Phi x'$. It remains to see that such a $x\to x'$ is unique.

Suppose we have two morphisms $x\to x'$ in $F$ such that $y\to \Phi x'$ factors as the composition of $y\to \Phi x$ with either of the corresponding $\Phi x \to \Phi x'$. Since $y\to \Phi x$ is vertical, we find that $\Phi x \to \Phi x'$ covers $\rho$. Since $\Phi$ is a functor over $C$, we find that both morphisms $x\to x'$ cover $\rho$. Thus it suffices to show that the two morphisms $x\to \rho^* x'$ through which our morphisms $x\to x'$ factor are equal. Using the fact that $\Phi(\rho^* x')\to \Phi(x')$ is cartesian (as I am assuming $\Phi$ is cartesian), one sees that the $y/\Phi(x)/\Phi(x')$ picture (with morphisms covering $\rho$) pulls back to a $y/\Phi(x)/\Phi(\rho^* x')$ picture in the fibers over $U$. Now the desired uniqueness follows from the universal property of $y\to\Phi(x)$ over $U$.

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I'll write something very concrete, perhaps more in the style of Vistoli's notes or SGA 1.

How certain are you that you need a left adjoint? If you look at right adjoints instead, then the result proves itself. Unless I am misreading Finn Lawler's response, it looks as if he also switched to right adjoints, but since he doesn't mention it, perhaps I am misunderstanding. I'll write out the details to produce the right adjoint, but it is probably easier to work it out yourself than to read what follows.

I'm guessing that when you write "morphism $F\to G$," you mean a cartesian functor over $C$. (I looked up your reference to Vistoli's exposition, and this seems to be his usage.) If that guess is wrong, then what I say below may not be relevant, since I use the cartesian property repeatedly.

Let $\Phi: F \to G$ be a cartesian functor between categories fibered over $C$. Assume that for each object $U$ of $C$, the functor $\Phi_U:F_U \to G_U$ on fiber categories admits a right adjoint. We wish to show that $\Phi$ admits a right adjoint. Let $y$ be an object of $G$ over the object $U$ of $C$. We need to produce a universal arrow $y\to \Phi x$ from $y$ to $\Phi$. The obvious candidate is a universal arrow $y\to \Phi_U x = \Phi x$ to $\Phi_U$, which exists by hypothesis (after switching your "left" for my "right"). We must prove that this arrow is universal to $\Phi$.

Let $y\to \Phi x'$ be a morphism in $G$ covering $\rho: U\to U'$ in $C$. Let $\rho^* x'\to x'$ be a pullback in $F$. Since $\Phi$ is cartesian, $\Phi(\rho^* x') \to \Phi x'$ is a pullback in $G$, and so $y\to \Phi x'$ factors uniquely as the composition of a morphism $y \to \Phi(\rho^* x')$ in $G_U$ with $\Phi(\rho^* x')\to \Phi x'$. By the universal property of $y \to \Phi x$ over $U$, there is a unique morphism $x \to \rho^* x'$ in $F_U$ such that $y \to \Phi(\rho^* x')$ factors as the composition of $y\to \Phi x$ and $\Phi x\to \Phi(\rho^* x)$. Composing $x\to \rho^* x'$ and $\rho^* x'\to x$ gives us a morphism $x\to x'$ covering $\rho$ such that $y \to \Phi x'$ factors as the composition of our candidate universal arrow $y \to \Phi x$ and $\Phi x \to \Phi x'$. It remains to see that such a $x\to x'$ is unique.

Suppose we have two morphisms $x\to x'$ in $F$ such that $y\to \Phi x'$ factors as the composition of $y\to \Phi x$ with either of the corresponding $\Phi x \to \Phi x'$. Since $y\to \Phi x$ is vertical, we find that $\Phi x \to \Phi x'$ covers $\rho$. Since $\Phi$ is a functor over $C$, we find that both morphisms $x\to x'$ cover $\rho$. Thus it suffices to show that the two morphisms $x\to \rho^* x'$ through which our morphisms $x\to x'$ factor are equal. Using the fact that $\Phi(\rho^* x')\to \Phi(x')$ is cartesian (as I am assuming $\Phi$ is cartesian), one sees that the $y/\Phi(x)/\Phi(x')$ picture (with morphisms covering $\rho$) pulls back to a $y/\Phi(x)/\Phi(\rho^* x')$ picture in the fibers over $U$. Now the desired uniqueness follows from the universal property of $y\to\Phi(x)$ over $U$.