A. Bahat said:The function f might not be surjective. If it were not, there would be an x in E such that f-1({x})=∅. Then f(f-1({x}))=∅, which is a proper subset of {x}.
The reason why f(f^-1(E)) $\subset$ E is because the function f^-1(E) maps elements of the set E back to their original elements in the domain of f. When we apply the function f to these elements, it simply transforms them back to their original form, thus resulting in a subset of E.
f(f^-1(E)) $\subset$ E is a direct result of the definition of an inverse function. The inverse function f^-1(E) is defined as the function that maps elements of E back to their original elements in the domain of f. Thus, when we apply f to these elements, it will result in a subset of E.
Let's say we have a function f(x) = 2x, where x is any real number. The inverse function of this would be f^-1(x) = x/2. Now, let's take the set E = {2, 4, 6}. If we apply the inverse function f^-1 to this set, we get {1, 2, 3}. And when we apply the original function f to this set, we get {2, 4, 6}, which is a subset of our original set E.
Understanding the concept of f(f^-1(E)) $\subset$ E is important because it helps us to understand the relationship between a function and its inverse. It also helps in solving equations involving inverse functions and in proving mathematical theorems related to functions.
Yes, f(f^-1(E)) $\subset$ E is always true for any function f and set E. This is because the definition of an inverse function guarantees that the elements of the image of f^-1(E) will always be a subset of the original set E.