# 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.