Fully faithful functors and composition Let $\mathcal{C}$ be an essentially small category, and suppose that $F:\mathcal{C}\to\mathcal{D}$ and $G:\mathcal{D}\to\mathcal{E}$ are two functors whose composition $G\circ F$ is fully faithful. Assume also that $F$ is fully faithful. 
Is $G$ also fully faithful?
 A: I'm not sure why you add the essentially small condition (maybe you meant "locally small"?), but $F$ being fully faithful means $\mathcal{C}(A,B)\cong\mathcal{D}(FA,FB)$ natural in $A$ and $B$.  Thus we have, $\mathcal{C}(A,B)\cong\mathcal{E}(GFA, GFB)$ and composing gives $\mathcal{D}(FA,FB)\cong\mathcal{E}(GFA,GFB)$ so $G$ is "fully faithful on the image of $F$". However, the image of $F$ can be a quite small full subcategory of $\mathcal{D}$. For example, you could have $FA = D$ for all $A$.
Indeed, this produces a simple counter example. Let $\mathcal{C}$ be the terminal category having one object with only the identity, and let $\mathcal{D}$ and $\mathcal{E}$ have a terminal object (or you could use an initial object instead). Then if $F1 \cong 1$ and $G1 \cong 1$, i.e. both functors preserve terminal (or initial) objects, then both $F$ and $G\circ F$ will be fully faithful. They are basically completely unconstrained otherwise. 
As a bit more specific example, take any category for $\mathcal{D}$ and take $\mathcal{E}$ to have the same objects as $\mathcal{D}$ but have at most one arrow between any two objects depending on whether $\mathcal{D}$ has any arrows between those two objects. $G$ will rarely be fully faithful in such a situation. As a very simple example of this, let $\mathcal{D}$ be a category with three objects $A$, $B$, and $1$ with a single arrow each from $A$ and $B$ to $1$, making $1$ a terminal object, and whatever set of arrows you want between $A$ and $B$. $G$ will only be fully faithful if there is at most one arrow between $A$ and $B$.
