theorem Th29:
    M is proper implies (canHom q)"M is proper
    proof
      assume
A1:   M is proper;
      assume not (canHom q)"M is proper; then
      M = (canHom q).:[#]A by Th24
      .= (canHom q).:dom(canHom q) by FUNCT_2:def 1
      .= rng(canHom q) by RELAT_1:113 .= [#](A/q) by FUNCT_2:def 3;
      hence contradiction by A1;
    end;
