FALSE: quotient-copy conjugacy is the equivalence relation behind first cohomology

For every integer $m$, the map $z_m:C_2\to \mathbb{Z}$ given by $z_m(1)=0$ and $z_m(t)=m$ is a crossed homomorphism. Conjugation by the kernel element $(1,1)$ changes the graph of $z_1$ to the graph of $z_3$, since the corresponding principal cocycle has value $t\cdot (-1)-(-1)=2$ at $t$. Hence $z_1$ and $z_3$ represent the same class under the kernel-conjugacy relation of [L1].

The quotient copy has only the elements $(0,1)$ and $(0,t)$. Conjugating the graph of $z_m$ by $(0,t)$ gives the graph of $z_{-m}$, so quotient-copy conjugacy sends ${\mathrm{\Gamma }}_{z_1}$ only to itself or to ${\mathrm{\Gamma }}_{z_{-1}}$, never to ${\mathrm{\Gamma }}_{z_3}$. Thus quotient-copy conjugacy misses a pair that [L1] identifies, and it is not the equivalence relation defining $H^1$.