Inverse composite proof (wording of the proof)

[psycho2499]

Homework Statement

Let f:A->B and g:B->C be invertible mappings. Show (g o f)^-1 = f^-1 o g^-1.

Homework Equations

A mapping is invertible iff it is bijective

The Attempt at a Solution

I understand why these are equivalent statements; however, I can't figure out the wording of the proof.

The best I can think of is:

Suppose f and g are invertible mappings defined f:A->B and g:B->C. Let a, b, and c be elements of sets A, B, and C respectively such that f(a)=b and g(b)=c. Since f and g are bijective, f^-1(b)=a and g^-1(c)=b. So (g o f)^-1(c) = a = f^-1(g^-1(c)).

 

[Office_Shredder]

(g o f)^-1(c) = a

You haven't actually proven this to be the case.We want to show for all c\in C that (g\circ f)^{-1}(c) = f^{-1}(g^{-1}(c)). Let b\in B be the unique element of B such that g^{-1}(c)=b and a\in A the unique element of A such that f^{-1}(b) = a. These both exist because f and g are invertible.

Clearly f^{-1}(g^{-1}(c)) = a just by how a and b were defined. Since c is arbitrary, it suffices to prove that (g\circ f)^{-1}(c) = a as well. How can you do that?

 

[psycho2499]

"Since c is arbitrary, it suffices to prove that (g∘f)−1(c)=a as well. How can you do that?"

I don't quiet understand your question, or the statement before it.

 

[Deveno]

what other statement involving a and c can you use, here?

where did a and c come from?

can we get c from a, somehow using f and g? how?

our assumptions have been as follows:

g^-1:c-->b

f^-1:b-->a

what is f(a)?
what is g(b)?

 

[Office_Shredder]

"Since c is arbitrary, it suffices to prove that (g∘f)−1(c)=a as well. How can you do that?"

I don't quiet understand your question, or the statement before it.

To prove that two functions a(x) and b(x) are equal, you can prove that given any possible input x₀, a(x₀)=b(x₀) (this shouldn't be particularly surprising) So to prove that (g\circ f)^{-1} = f^{-1}\circ g^{-1}, you can prove that given any c\in C, (g\circ f)^{-1}(c) = f^{-1}(g^{-1}(c)). We know that the right hand side of this last equation is a by how we defined b and a in my post... how can you prove that the left hand side is equal to a as well?

 

[psycho2499]

This is the part that confuses me. Should I be saying something to the effect:
For any a\inA there exists a unique b\inB and a unique c\inC such that f(a)=b and g(b)=c. It follows that g(f(a))=c. Thus (g o f)^(-1)(c)=a