# 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 really do not understand inverse mapping...

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(T2

I really do not understand inverse mapping and any detailed explanation that can help me understand would be greatly appreciated.

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### 1 Answer

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}