Dividing left adjoints If I have categories $C,D,E$ with "forgetful functors" $G_1:D\to C$ and $G_2:E\to D$, with $G_3=G_1\circ G_2:E\to C$, and left adjoints $F_1:C\to D$ and $F_3:C\to E$, is it possible to deduce a left adjoint $F_2:D\to E$ to $G_2$ satisfying $F_3=F_2\circ F_1$?
I can form the functor $F'=F_3\circ G_1$ which has the right signature, but I am not certain if composing a left adjoint with a right adjoint gives anything special, and I am not a category theorist so my intuitions are lacking here.
For a concrete example, take $C={\bf Set}, D={\bf Mnd}, E={\bf Grp}$. Then $G_1$ and $G_3$ take a monoid/group to its underlying set, and $F_1$ and $F_3$ are the free monoid/group constructions. $G_2$ is the map that interprets a group as a monoid with the same operation. Can I use this information to find out the "free group over a monoid" $F_2$? (I'm not even sure what that should look like.)
 A: No, $G_2$ may not even necessarily have a left adjoint itself, in fact. Let $*$ be the terminal category, and consider that the functor $\mathsf{C} \to *$ has a left adjoint iff $\mathsf{C}$ has an initial object. So simply consider a diagram of the type $\mathsf{E} \xrightarrow{G_2} \mathsf{D} \to * = \mathsf{C}$ where $\mathsf{E}$, $\mathsf{D}$ have an initial object and $G_2$ doesn't have a left adjoint.
For a very explicit (and rather random) example, you can take $\mathsf{E} = \mathsf{Set}$ and $\mathsf{D} = \mathsf{Grp}$, and $G_2$ the free group functor. Then since the free group on the set $X \times Y$ isn't the product of the free groups on $X$ and on $Y$, $G_2$ doesn't preserve limits, so it cannot possibly have a left adjoint, but both categories have an initial object so $G_1$ and $G_3$ (both are the functor to the terminal category) do have a left adjoint.
A: This is not a definite answer (I suspect the answer to be no, but I don't have counter-examples coming to mind), but at least the example you suggest shows that even when there is a left adjoint $F_2$, it will not be $F_3\circ G_1$ in general.
Indeed, in your case $F_2$ exists : if $M$ is a monoid, take $F_M$ the free group on the underlying set of $M$ (ie $F_M = (F_3\circ G_1)(M)$) and then quotient $F_M$ by all relations that hold in $M$. This gives you a group $F_2(M)$ and a natural monoid morphism $M\to F_2(M)$ that gives an adjunction with $G_2$.
