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