Reading Brown & Spencers papper on double groupoids and crossed modules I don't quite understand the constructed double groupoid(see http://www.numdam.org/item/CTGDC_1976__17_4_343_0.pdf page 11)
Let $(A,B,\partial)$ be a crossed module, to construct the double groupoid of this crossed module we consider $E_0=1$(just some one point) $E_1=B$ and $E_2$ is the quintlet with $E_2=(\alpha,a,b,c,d\vert \partial(\alpha)=a^{-1}bcd^{-1}, \alpha\in A)$
I don't understand what the initial and final maps from $E_2$ to the two copies of $E_1$ are suppose to be, are we choosing arbitrary inital and final maps?(or source and target maps)
The author's are a bit needlessly terse here, but if you look back to page to page 3, they represent squares labelled by $\alpha$ as quadruples: $(\partial_{0}\alpha,\epsilon_{0}\alpha,\partial_{1}\alpha,\epsilon_{1}\alpha)$ where each component labels an edge of the square starting from the left vertical edge and going around clockwise (I tried to input a diagram but I don't think MSE supports this, sorry!).
On page 11, they depict the boundary of a square $\alpha\in E_{2}$ (using such a quadruple as above) as $(a,b,c,d)$, i.e., $$\partial_{0}\alpha = a$$ and $$\partial_{1}\alpha = c$$ this is what is meant when they write that "the initial and final maps are just $\partial_{0}$ and $\partial_{1}$". This certainly makes sense too since, if we denote the "zero map for +" by $\iota$, we get that $$\partial_{0}\iota(a) = a = \partial_{1}\iota(a)$$ which is what one would hope for.