Let $A = F\_2$ be the free group on two generators (not sure if this is important.)


Suppose you have a semidirect product $A\rtimes C$ coming from some homomorphism $\beta: C\rightarrow \text{Aut}(A)$.


Let $G$ be a finite group, and let $f\_A: A\twoheadrightarrow G$ be a surjection such that
$$f\_A(^c\cdot) := f\_A\circ\beta(c)\equiv f\_A\mod\text{Inn}(G)\qquad\text{for all $c\in C$}$$


then, under these conditions, must there exist:


1. A homomorphism $f : A\rtimes C\twoheadrightarrow G$ such that $f|\_A = f\_A$, or equivalently...
2. A homomorphism $f\_C : C\rightarrow G$ such that we have
$$(f\_A\circ\beta(c))(a) := f\_A(^ca) = f\_C(c)f\_A(a)f\_C(c)^{-1}\qquad\text{for all $c\in C, a\in A$}$$


*Remark:* Our assumptions tell us that for every $c\in C$, there is a $g\_c\in G$ such that $f\_A(^ca) = g\_cf\_A(a)g\_c^{-1}$ for all $a\in A$, so what I'm asking is if it's always possible to pick the $g\_c$'s wisely so that the map $c\mapsto g\_c$ is a homomorphism.


This question came from me thinking about Teichmuller level structures and how they classify torsors, so in my context $1\rightarrow A\rightarrow A\rtimes C\twoheadrightarrow C\rightarrow 1$ is an exact sequence of fundamental groups associated to some punctured curve (either $\mathbb{P}^1$ minus 3 points or an elliptic curve punctured once).


*EDIT:* As Dave Witte Morris points out, this is not possible in general. Though, I wonder: Are there conditions we can put on $G$ or $f\_A$ to ensure that the statement *is* true? If $G$ is abelian, then $f\_C$ could literally be anything, but that's quite a strong condition. Can we say anything for nonabelian $G$? I'm trying to understand exactly why this fails... Is there some kind of cohomology going on?



extending homomorphisms to semidirect products
No. For a counterexample, let $G = \langle i,j \rangle$ be the quaternion group of order $8$, let $f\_A$ map the generating set $\{a,b\}$ of $F\_2$ to $\{i,j\}$, and let $C = \langle c \rangle$, where $c$ is the automorphism that fixes $a$ and inverts $b$. 


We can choose $g\_c$ to be $i$, so the hypothesis is satisfied. However, since $c$ has order 2, and the only element of order 2 in $G$ is central, there is no way to pick $g\_c$ to get a homomorphism.


