My question is partly motivated by trying to solve this one. Let $E$ be the semidirect product of groups $G$ and $H$. Then, we have an exact sequence:
$$G \hookrightarrow^\iota G \rtimes_\phi H \twoheadrightarrow^\pi H$$
where $\phi\colon H \to \text{Aut}(G)$ is the action. I then formed the diagram:
$$\begin{array}{c} F_G &\twoheadrightarrow& G \\ \downarrow_{\iota'} & & \downarrow_\iota\\ F_{G\rtimes H} & \twoheadrightarrow &G \rtimes H \\ \downarrow_{\pi'} & & \downarrow_\pi \\ F_H & \twoheadrightarrow & H \end{array}$$
Each line is simply a group and the free group on it's generators. For instance, $F_G$ is the free group on the generators of $G$. The vertical short sequence on the right is given.
The left vertical sequence is constructed by the universal property of the free group. For instance, the map $\iota\colon G \to G\rtimes H$ defines a funtion from the basis of $F_G$ to $F_{G\rtimes H}$, and therefore the homomorphism $\iota'\colon F_G \to F_{G \rtimes H}$. Same thing for $\pi'$. It makes each square on the diagram to be commutative, and it also makes the left vertical sequence a short exact sequence.
But know, if I have a section $s\colon H \to G\rtimes H$, this would lift to a section $s'\colon F_H \to F_{G\rtimes H}$.
This would mean that $F_{G \rtimes H}$ is isomorphic to $F_G \rtimes F_H$, which seems very wrong, once free groups can't have relations! I'm once again verifying all the steps, but I still couldn't figure out where I got it wrong.
