Let f:A->B where A & B are nonempty let T1 & T2 be subsets of B Prove f-1(T1) - f-1(T2) = f-1(T1 -T2) Prove f-1(T1 U T2) = f-1(T1) U f-1(T2I...

For the existence  inverse function  of f: A
-->B , the given relation between the sets  must be  a
bijection.

Therefore  for any T1 a subset of B , f^-1(T1) =
A1  which is a subset of A.

So  T1 = f(A1)
.............(1)

Similarly for any  set T2 a subset of B
, f^-1(T2) = A2 which is a subset of A.

Therefore  T2 =
f(A2).........(2)

Therefore from (1)and  (2)   we get the
diffrence of sets: 

f(A1) - f(A2) = T1
-T2.

Therefore f^-1{f(A1) - f(A2} =
f^(-1)(T1-T2)

A1 -A2 =
f^-1(T1-T2).

f^-1(T1) - f^-1(T2) =
f^-1(T1-T2)

2)

Againg f^-1(T1)
= A1 . f^-1(T2) = A2, where A1 and A2 are subsets of
A.

f(A1) = T1

f(A2)
=T2.

So  f(A1) U f(A2) = T1 U
T2.

f^-1{f(A1) U f^-1(A2) } = f^-1{T1 U
T2)

A1 U A2 = f^-1{T1 U T2
}

f^-1(T1) U f^-1(T2) = f^-1{T1 U
T2}