# Questions regarding function operations on sets.

1. Sep 19, 2008

### Esran

Let F:X->Y be an arbitrary function over sets X and Y.

Why is F-1(Y) = X always true?

Suppose B1 and B2 are some subsets of Y. Why is F-1(B1 $$\bigcap$$ B2) = F-1(B1) $$\bigcap$$ F-1(B2) always true?

These aren't homework questions. I'm just curious. I saw these statements the other day, and I already know that F(X) = Y is not necessarily true, and neither is (for some subsets A1 and A2 of X) F(A1 $$\bigcap$$ A2) = F(A1) $$\bigcap$$ F(A2).

Why would the other statements always be true? It seems to me that inverse functions are just like any other functions when it comes to the two theorems I already know. Why are they different?

2. Sep 19, 2008

### Moo Of Doom

The idea behind these is that for bijections (i.e. when an inverse exists), the same is true for the image as for the inverse image. When we have an arbitrary function (not necessarily injective or surjective) these properties go away for one direction but not the other. Why is that?

Because a function $f : X \to Y$ is defined to satisfy
(i) Each member of $X$ is mapped to an element of $Y$.
(ii) This element of $Y$ is unique for each $x$.

Don't these sound similar to surjective and injective?
(sur) Each member of $Y$ is mapped to by an element of $X$.
(inj) This element of $X$ is unique for each $y$.

What you need to know is that $f(X) = Y$ if $f$ is surjective, and $f(A \cap B) = f(A) \cap f(B)$ if $f$ is injective.

I hope this gives you an intuitive reason why the inverse image behaves nicer in general than the image.

3. Sep 20, 2008

### jostpuur

It seems you are confusing preimages with images of inverse functions. $$f^{-1}(B)$$ is not necessarily image of any mapping $$Y\to X$$.