# Preimage and function composition?

1. Sep 18, 2012

### SMA_01

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

Let f:S→T and let A$\subseteq$T. Define the preimage of A as f-1(A)={x in S: f(x) is in A}.
Demonstrate that for any such map f and B$\subseteq$ran(f), f(f-1(B)) = B.

I am going to use set inclusion to prove this, but can I use function composition in the portion in red? I was going to say an element y is in f(f-1(B)) and then was thinking to apply function composition so as to map an element x back to S then apply f to it and map the new element to B.
Does this seem right?

Thanks.

2. Sep 18, 2012

### jbunniii

It's not exactly function composition, because $f^{-1}$ isn't necessarily a function. This is because, for a given $y$ in the image of $f$, there may be more than one $x$ satisfying $y = f(x)$.

The way to interpret the expression in red is that $f^{-1}(B)$ is the set of all $x$ such that $f(x) \in B$. And f(f^{-1}(B)) is the image of that set under $f$.

3. Sep 19, 2012

### SteveL27

For what it's worth, it's good to remember that the notation f-1 is actually being abused here. This is the normal use, but it's good to remember what it means.

f is a function from S -> T; and at some other time in class we may have occasion to talk about its inverse function f-1 : T -> S.

However here we're not doing that. f-1 is being defined as a function from the power set of T to the power set of S; that is,

f-1 : P(T) -> P(S)

For each subset of T, we define P(T) as a particular subset of S.

You have to be careful to remember that f goes left to right from points to points, and f-1 goes from right to left from subsets to subsets.

4. Sep 19, 2012

### SMA_01

I see, thank you! I was a little hesitant with using composition, but now I know to just use the definition of preimage and I think that should do the trick.

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