I'm reading this part of Borceux Handbook of Categorical Algebra, and I have a problem with the equation on the last but one line in the snippet:
$$\Xi_{A'}\ast F\alpha=G\alpha\ast \Xi_A.$$
My problem would be solved if someone gave me a commutative square with natural (understandable) horizontal and vertical pairs of arrows and easy to see 4 corners in that diagram.
I even do not know of what type these data should be for the equation in question: are they 0-,1- or 2-cells (both, for corners and for arrows in the diagram)? I'd like to understand this Whiskering in some detail.
Borceux has written $\alpha$ for both a 2-natural transformation and a 2-morphism here. Let's write $\gamma:f\to g$ instead for an arbitrary 2-morphism in $\mathcal A$. Then $G\gamma:Gf\to Gg:GA\to GA'$ while $\Xi_A:\alpha_A\to \beta_A:FA\to GA$. The repeated colon notation indicates the domain and codomain of a 2-morphism followed by the domain and codomain of the domain and codomain.
Horizontal composition of these 2-morphisms gives $$G\gamma*\Xi_A: Gf\circ \alpha_A\to Gg\circ \beta_A: FA\to GA'.$$ The other side of the equation is constructed in the same way: from $F\gamma: Ff\to Fg$ and $\Xi_{A'}:\alpha_{A'}\to \beta_{A'}$ we get $$\Xi_{A'}*F\gamma:\alpha_{A'}\circ Ff\to \beta_{A'}\circ Fg.$$ Now the naturality condition on $\alpha$ and $\beta$ shows we have $\alpha_{A'}\circ Ff=Gf\circ\alpha_A$ and $\beta_{A'}\circ Fg=Gg\circ \beta_A$, so that the two 2-morphisms assumed to be equal have the same domain and codomain. $\ \ \ \ $