Proof of Theorem: Composite Function Inverse

AI Thread Summary
The discussion centers on proving the theorem that states if functions f and g are bijective, then the inverse of their composition (g o f) equals the composition of their inverses (inverse of f o inverse of g). A user requests assistance with the proof, prompting suggestions to define f(x) = y and g(y) = z. The hint encourages computing (g o f) o (g o f)^-1 and (g o f)^-1 o (g o f) to demonstrate that both yield the identity function. This approach is aimed at establishing the validity of the theorem through direct computation. The conversation emphasizes the importance of understanding function composition and inverses in the context of bijective functions.
irvin.b
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i really need to see the proof of this theorem:

if f and g are bijective then the inverse of (g o f) = inverse of f o inverse of g
 
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i hope you will help me..
 
what have you tried?
 
Hint: define f(x) = y and g(y) = z.
 
Just compute (g o f) o (g o f)^-1 and (g o f)^-1 o (g o f) and see that they both give you the identity.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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