Proposition 5.8. Assuming that:

  • (Aj)jJ, U, as usual

  • n

  • for each j we have Bj,CjAjn

Then
  • (1) [(Bj)jJ][(Cj)jJ]=[(BjCj)jJ]
  • (2) [(Bj)jJ][(Cj)jJ]=[(BjCj)jJ]
  • (3) [(Bj)jJ][(Cj)jJ]=[(BjCj)jJ]
  • (4) If n>0, k{1,,n} then
    πk([(Bj)jJ])=[(πk(Bj))jJ].