Originally Posted by

**kalagota** i think, that is too long..

Let TS be bijective.

If S is surjective, then S in onto V, that is for every v in V, there is a u in U such that S(u)=v.

since TS is bejective, then S is also injective (from the previous item), that is S is one to one.. so if (TS)(u1) = T(S(u1)) = T(S(u2)) = (TS)(u2) for u1,u2 in U, then u1=u2 (TS is bijective) and so S(u1)=S(u2) (S is injective) and these S(u1) and S(u2) are in V. thus, T is injective..

can you do for the other one? it is just similar..