Recitation 4

Previously, we have proved that for any function f the union of inverse images is the same as the inverse image of the union, i.e., for any two subsets of the range of f, say A, B, I_f(A)\cup I_f(B)=I_f(A\cup B). Similarly, I_f(A)\cap I_f(B)=I_f(A\cap B).

So the natural question one could ask is whether this works for the image of any function. Precisely, given a function f and two subsets of the domain, say A, B, is it still true that f(A)\cup f(B)=f(A\cup B) and f(A)\cap f(B)=f(A\cap B)?

It turned out that the first assertion is true, while the second is not in general.

However, if the function f is injective, which means no two elements in the domain are mapped to the same element in the range, then the second assertion holds.

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