Assume (1) commutes
exactly when (2) does.
WTS (3) commutes;
fix c,d,d′,k,f. Let
c′=c and
h=idc and
g=k∘f; then (1) trivially commutes so (2) commutes entailing that
G(k)∘β(f)=β(k∘f), which is what we wanted to show; done.
WTS (4) commutes;
fix c,c′,d,h,g. Let
d=d′ and
k=idd and
f=g∘F(h); then (1) trivially commutes so (2) commutes entailing that
β(g)∘h=β(g∘F(h)) which is what we wanted to show; done.