The kernel of transgression is the image of restriction

If d is the restriction of a global crossed map c, its graph $C=\{(c(g),g)\}$ is a subgroup of $A⋊G$ and $C\cap (A⋊N)=D_d$. The latter is normal in C because N is normal in G. Thus $C\subseteq L_d$, and $C\mathrm{/}{D}_d\cong Q$ is a subgroup of $L_d\mathrm{/}{D}_d$ meeting the kernel $A^N$ trivially and mapping onto Q. It gives a homomorphic section. Therefore Tra[d]=0. If only the cohomology classes agree, replace the restriction by its principal-equivalent representative; F1 says Tra is unchanged.

Conversely if Tra[d]=0, let $\bar{C}\le L_d\mathrm{/}{D}_d$ be the image of a splitting, supplied by F3. Take its full inverse image C in $L_d$. It contains $D_d$. It meets A trivially: an element of $C\cap A$ lies in $A^N$ by F1 and has class in $\bar{C}\cap A^N=1$, so lies in $D_d\cap A=1$. The projection $C\to G$ is onto: given g, select one member of C whose projection has quotient pi(g), then multiply it by the unique element of $D_d$ correcting its projection to g. This is an elementwise existence proof, not a family of selections. Hence $C\to G$ is an isomorphism and its inverse is the graph of a uniquely defined function c:G to A. The subgroup law gives $c(gh)=c(g)+gc(h)$, and $C\cap (A⋊N)=D_d$ gives $c{\mathrm{\mid }}_N=d$. Thus [d] is in the restriction image.