Crossed homomorphism/derivation on free group

Question

Let $F$ be a free group with basis $S$ and $A$ an (additive) abelian group on which the group $F$ acts.

A map $\delta:F\rightarrow A$ is called a crossed homomorphism (derivation) if for all $x,y\in F$ $$\delta(xy)=\delta(x)+\delta(y)^x.$$ The last term denotes action of $x\in F$ on $\delta(y)\in A$.

My question is simple, an analogue for a theorem on free group (its property).

Given a free group $F$ on a set $S$ and an abelian group $A$ with an action of $F$. Then any map $f:S\rightarrow A$ extends uniquely to a derivation $\hat{f}:F\rightarrow A$.

Proof. Let $A\rtimes F$ denotes semidirect product formed w.r.t given action of $F$ on $A$.

The set map $f:S\rightarrow A$ gives a set map $S\rightarrow A\rtimes F$, $s\mapsto (f(s),s).$

Since $F$ is free, this map extends to a homomorphism from $F$ to $A\rtimes F$.

So if $w$ is any word in $F$ then $F(w)=(\hat{f}(w),w)$ for some $\hat{f}(w)\in A$.

This $w\mapsto \hat{f}(w)$ is a derivation which is extension of $f$.

Question. Is the quoted statement and its proof correct?

Answer 1

This is basically correct. You have not actually proved that $\hat{f}$ is a derivation or the unique derivation extending $S$, though. These are easy enough to check though: the fact that $\hat{f}$ is a derivation is immediate from the fact that $F$ is a homomorphism and the definition of multiplication in the semidirect product, and for uniqueness it is easy to see that since $S$ generates $F$ a derivation is determined by what it does on $S$.

Answer 2

You may be interested in looking at the link between these ideas and Fow's differential calculus. The quickest route is in the old book by Crowell and Fox. The uses of this in Knot Theory are fun. There are links also with augmentation ideals of free groups worth chasing up.