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Suppose $F:C\to D$ and $G:D\to E$ are functors. Assume that $F$ is fully faithful, $G$ has a left adjoint $H:E\to D$, and for each $Z \in E$ there exists $X \in C$ such that $F(X) = H(Z)$. Does $F$ has an adjoint, either left or right? I can define a functor $L: D \to C$ by sending an object $Y \in D$ to an object $L(Y) \in C$ such that $F(L(Y)) = H(G(Y))$ and an arrow $g: Y_1 \to Y_2$ of $D$ to the unique morphism $L(f): L(Y_1) \to L(Y_2)$ of $C$ such that $F(L(f)) = H(G(f))$. There is also an obvious natural transformation $\eta: 1_D \to F H$ that maps an object $Y \in D$ to the local identity of $Y$ in $D$. But I don't see if $L$ is adjoint to $F$.

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The answer to your question is no. Here is a counterexample. Let $\mathcal{C}$ be the category of finite sets, let $\mathcal{D}$ be the category of all sets, and let $\mathcal{E}$ be the terminal category. Then, the inclusion $F : \mathcal{C} \to \mathcal{D}$ is fully faithful but has no left adjoint or right adjoint (for various reasons), and the unique functor $G : \mathcal{D} \to \mathcal{E}$ has both a left adjoint and a right adjoint; the left adjoint $H : \mathcal{E} \to \mathcal{D}$ is the constant functor with value $\emptyset$. Clearly, the condition that for every $Z$ in $\mathcal{E}$ there is an $X$ in $\mathcal{C}$ such that $F X = H Z$ is satisfied.

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Thanks for the quick answer! – user70152 Mar 30 '13 at 13:21

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