Working with the inverses of composite functions

[Esran]

Let f:X->Y and g:Y->Z be functions. Suppose A is a subset of Z. I'm wondering whether (g o f)^-1(A)=f^-1(g^-1(A)).

I'm lost as to where to go with this problem. I know I need to do something using images, but manipulating the inverses of composite functions is proving to be very confusing; I don't know how to properly translate these functions into terms that match up with my definitions, which are as follows:

1) Let f:X->Y be a function: for a subset A of X, the set f(A)={y in Y: y=f(x) for some x in A}.

2) For a subset C of Y, the set f^-1(C) = {x in X: f(x) is in C}.

Can you please point me in the right direction? I'm sure that if I could just grasp how to prove this theorem, I'd understand inverses so much better.

Thank you for your time!

 

[tiny-tim]

Let f:X->Y and g:Y->Z be functions. Suppose A is a subset of Z. I'm wondering whether (g o f)^-1(A)=f^-1(g^-1(A)).

Hi Esran!

(have an exists: ∃ and an in/epsilon: ε )

Hint: x ε f^-1(g^-1(A))

means ∃ y ε g^-1(A) with f(y) = x

means ∃ z ε A with … ?

 

[Esran]

x ε f^-1(g^-1(A))

means ∃ y ε g^-1(A) with f(y) = x (shouldn't this be f^-1(y) = x?)

means ∃ z ε A with g(y) = z.

I need to show that somehow, the above implies x ε (g o f)^-1(A), and vice versa.

So, x ε (g o f)^-1(A)

means ∃ z ε A with (g o f)(x) = z or g(f(x)) = z?

means ∃ y ε f(x) with g(y) = z?

I'm lost again on how to translate this statement. What do I do when the inverse is on the outside of a composite function?

 

[tiny-tim]

x ε f^-1(g^-1(A))

means ∃ y ε g^-1(A) with f(y) = x (shouldn't this be f^-1(y) = x?)

oops!

yes … I meant to type: f(x) = y.

means ∃ z ε A with g(y) = z.

I need to show that somehow, the above implies x ε (g o f)^-1(A), and vice versa.

So, x ε (g o f)^-1(A)

means ∃ z ε A with (g o f)(x) = z or g(f(x)) = z?

That's right! , just stop there …

you have two expressions, (g o f)^-1(A) and f^-1(g^-1(A)), and you want to prove that they're the same.

So you rewrite each in the form "given x ε …"

You have:
i] ∃ z ε A with g(f(x)) = z

ii] ∃ y ε g-1(A) with y = f(x)
and ∃ z ε A with g(y) = z

Don't you see, that once you write it out clearly like that, you have the answer?

 

[Esran]

Alright! I see it now. Thank you.