Okay, now that we've looked at functions and inverse functions of sets and intersections, unions, and complements of sets, it makes sense to look at functions of intersections of sets (and the other combinations). The trick here is that functions of sets correspond to functions of points. To figure out what's going on with the sets, follow the individual points in the sets.
There are a pile of statements that look like identities involving functions of sets. It would be nice if the statements are true, but we would have to prove that first, and some statements that look like they "should be" true are not in fact true in general. For example, if f is a function that maps the set X to the set Y, and A is a subset of X and Ac is the complement of A, is it true that f(Ac)=[f(A)]c? Unfortunately, no. If f is not onto, then f(X)= the range of f, which is a subset of Y but not equal to Y. So [f(X)]c is every point in Y that is not in the range of f. But Xc=∅, so f(Xc)=f(∅)=∅. Therefore, they are not equal.
In addition, if f is not one-to-one, then the equality also doesn't hold. Choose a point y in Y such that for two points a and b in X, f(a)=f(b)=y, but a≠b. Then choose a subset A of X that includes the point a but not b. Then f(A) includes the point y, but f(Ac) also includes the point y, so f(A) cannot be the complement of f(Ac).
However, if f is both one-to-one and onto, then the equality does hold. Given a set A⊂X, choose any point y∈Y. Since f is onto, there exists a point x∈X such that f(x)=y. Since f is one-to-one, x is unique. If x∈A, then f(x)=y∈f(A). If x∉A, then f(x)=y∉f(A). If x∉A, then x∈Ac, and if y∉f(A), y∈[f(A)]c, so f(Ac)=[f(A)]c for any A⊂X.
But let's look at inverse functions. If x is a point in the set X, and A is a subset of X, and f is a function that maps X to the set Y, f(x) is a point in Y and f(A) is a subset of Y. If y is a point in Y and B is a subset of Y, f−1(y) is only a function if f is one-to-one and onto, but f−1(B) is always a function. So let's look at f−1(Bc).
Choose any point x in X. Then there is one point, y, in Y such that f(x)=y. y is either a member of B, or it is a member of Bc. If y is a member of B, then x is a member of f−1(B). If y is not a member of B, then x is not a member of f−1(B), and is therefore a member of [f−1(B)]c. Also, since y is not a member of B, y is a member of Bc, so x is a member of f−1(Bc). Therefore, [f−1(B)]c=f−1(Bc).
I won't go through the details here, but the same types of reasoning apply for unions and intersections of sets in X and Y. Take λ as an index of a collection of sets. Then Aλ is a subset of X for each λ. (From the discussion of indexes last time, the possible values of λ do not have to be organized.) Then, using the same type of reasoning as with complements, we can prove that f(∪Aλ)=∪[f(Aλ)] and f−1(∪Bλ)=∪[f−1(Bλ)].
A couple of posts ago, I introduced some definitions that will be useful here, so as a reminder, if B=f(A), we can call B the image of A, and if A=f−1(B), we can call A the preimage of B. Then in words (math jargon words, but words nonetheless), the first statement is that the image of a union of sets is equal to the union of the images of sets. The second statement is that preimage of a union of sets is equal to the union of the preimages of the sets.
Looking at intersections, we can show that f(∩Aλ)=∩[f(Aλ)] only if f is one-to-one. The image of an intersection of sets is only equal to the intersection of the images if f is one-to-one. You can verify this by taking a simple function f which is not one-to-one, so two different values in X map to the same value in Y, and defining some subsets of X where the equality does not hold. However, once again, f−1(∩Bλ)=[∩f−1(Bλ)] for any f. The preimage of an intersection of sets is always equal to the intersection of the preimages of the sets.
The conclusion here is that we can switch the order of operations for preimages and for complements, unions, or intersections, but we can only generally switch the order for images with unions. In this sense, preimages are more powerful than images.