# Prove inverse image of function

1. Apr 19, 2010

### simmonj7

1. The problem statement, all variables and given/known data

When f: A -> B and S ⊆ B, we definte If(S) = { x ∈ A: f(s) ∈ S}. Let X and Y be subets of B.
a) Determine whether must equal If(X) ∪ If(Y).
b) Determine whether If(X ∩ Y) must equal If(X) ∩ If(Y).

2. Relevant equations

The definition of an inverse image is stated in the problem.

3. The attempt at a solution

So far I have tried to draw the schematic representation of the problem and didn't really get anything out of that.
Then I tried messing around with some proofs by saying "Choose an arbitrary x in If(X ∪ Y). Then, x = f(a) for some a in X ∪ Y." And then from there are I start questioning myself and don't know if I should continue/where I should go.

Thanks

2. Apr 19, 2010

### hgfalling

Well, you seem to be kind of on the right track with your second attempt.

Suppose you have an arbitrary x in If(X $$\cup$$Y). Then a = f(x) for some a in X $$\cup$$Y.

Now break that union sign apart, and see what you get.

The same type of thinking will work for the "and" part of the question too.

Think about what this all means: you have two domains. In domain #1, you have "source" points. In domain #2, you have "target" points, like X and Y.

Now suppose you have a bunch of "target" points in domain #2. The preimage If is the set of "source" points that give you those target points.

3. Apr 19, 2010

### simmonj7

Ok so I see that you break up the a∈ X∪Y to get a∈X or a∈Y but I don't see quite how to go from there.
I know when I prove this problem with images rather than inverse images, at this point I would say (assuming the problem and variables were in the right spots) f(a)∈ f(X) or f(a)∈ f(Y) and then x would be equal to f(A) but since I am doing inverse images, I don't know how to get back to x.
Hope what I'm trying to say kinda makes sense.

4. Apr 20, 2010

### hgfalling

What have you shown so far? You picked an arbitrary x value in the preimage of X∪Y (the "source"). So that means that a = f(x) for some a in X∪Y; that is, there's a "target" a in
X∪Y. That means that the target is either in X or in Y or both.

If the target is in X, then the source is in If(X), right? That's the definition of a preimage. And if the target is in Y, then the source is in If(Y).

Since we know that the target is either in X or Y or both, then the source is in the preimage of X or the preimage of Y or both. Hence we have shown that part 1 is true.

Now take a look at the second part. You can use the same ideas of sources and targets to solve that; you probably won't be surprised that part 2 is NOT true, since part 1 WAS true. Funny how problems often work that way.

5. Apr 20, 2010