Degree-one maps and the quotient action are well-defined

If a is N-fixed, then $n(ga)=g(g^{-1}ng)a=ga$, since N is normal. Thus ga is N-fixed. Replacing g by gn with n in N does not change ga, proving the Q-action on $A^N$. For a crossed map c into $A^N$, $c(\pi (gh))=c(\pi g)+gc(\pi h)$; hence its inflation is crossed. Principal c from a in $A^N$ inflates to $g\mapsto ga-a$. Restriction plainly preserves the crossed identity and sends the principal map from a to the principal map from the same a.

For $d\in Z^1(N,A)$, write $u=g^{-1}ng$, $w=g^{-1}mg$. Then $(g\cdot d)(nm)=g(d(u)+ud(w))=(g\cdot d)(n)+n(g\cdot d)(m)$, so the action preserves crossed maps. It sends $\delta a$ to $\delta (ga)$ and satisfies $(gh)\cdot d=g\cdot (h\cdot d)$ by substitution. For $t\in N$, using $d(t^{-1})=-t^{-1}d(t)$ gives $(t\cdot d)(n)=d(n)+nd(t)-d(t)$. Hence N acts trivially on H1, and the action there factors through Q.

For a global crossed map D on G the identical expansion gives $gD(g^{-1}ng)=D(n)+nD(g)-D(g)$. Consequently conjugation changes its restriction by the principal map $\delta (D(g))$, so the restriction class is Q-invariant. All formulas are additive in the crossed map, and therefore descend to the stated homomorphisms. The zero module and either trivial end group obey the same identities.