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Preimage and function composition?

  • Thread starter SMA_01
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Homework Statement



Let f:S→T and let A[itex]\subseteq[/itex]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[itex]\subseteq[/itex]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.
 

Answers and Replies

  • #2
jbunniii
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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.
.
It's not exactly function composition, because [itex]f^{-1}[/itex] isn't necessarily a function. This is because, for a given [itex]y[/itex] in the image of [itex]f[/itex], there may be more than one [itex]x[/itex] satisfying [itex]y = f(x)[/itex].

The way to interpret the expression in red is that [itex]f^{-1}(B)[/itex] is the set of all [itex]x[/itex] such that [itex]f(x) \in B[/itex]. And f(f^{-1}(B)) is the image of that set under [itex]f[/itex].
 
  • #3
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Homework Statement



Let f:S→T and let A[itex]\subseteq[/itex]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[itex]\subseteq[/itex]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.
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
218
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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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