# Inverse composite proof (wording of the proof)

1. Nov 8, 2011

### psycho2499

1. The problem statement, all variables and given/known data
Let f:A->B and g:B->C be invertible mappings. Show (g o f)^-1 = f^-1 o g^-1.

2. Relevant equations
A mapping is invertible iff it is bijective

3. 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)).

2. Nov 9, 2011

### Office_Shredder

Staff Emeritus
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?

3. Nov 10, 2011

### 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.

4. Nov 10, 2011

### 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)?

5. Nov 10, 2011

### Office_Shredder

Staff Emeritus

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?

6. Nov 11, 2011

### psycho2499

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