Question

Prove that if f:X→Y is any set function and U⊂X then U⊂f⁻¹(f(U)) and equality holds if f is injective

Answer 1

To prove that if f:X→Y is any set function and U⊂X, then U⊂f-1(f(U)) and equality holds if f is injective.

Step-by-step explanation:

To prove that if f:X→Y is any set function and U⊂X, then U⊂f-1(f(U)) and equality holds if f is injective.U⊂f-1(f(U)): Let x be an element of U. Since f is a set function, f(U) is a subset of Y. Therefore, f(x) is in f(U).U⊂f-1(f(U)): Let y be an element of f(U). Since f is a set function, f-1(y) is a subset of X. Therefore, f-1(y) contains an element x such that f(x)=y.Equality holds if f is injective: If f is injective, then for every x⊂1 and x⊂2 in X, if f(x⊂1)=f(x⊂2), then x⊂1=x⊂2. If U⊂f-1(f(U)), and if there exist x⊂1 and x⊂2 in U such that f(x⊂1)=f(x⊂2), then x⊂1=x⊂2. Thus, U⊂f-1(f(U)).