Does this proof work to show that is a bijection if the composite ◦ is a bijection? We are given that is a function from A → A.
Injective: Let x, y ∈ A. Assume (x) = (y).
By the composite function being a bijection, if ((x)) = ((y)), x = y. Then each z = ((x)) = ((y)) ∈ A has a y = (x) = (y) ∈ A, which has x = y ∈ A. Thus, when (x) = (y), x = y.
Surjective: Let x, y ∈ A. Since the composite function is a bijection, each z = ((x)) ∈ A has a corresponding x ∈ A such that ((x)) = z. Then each z = ((x)) also has a y ∈ A = (x) ∈ A for which y = (x). Thus for the function , each y ∈ A has an x such that y = (x), and is a mapping onto A.
Since is both injective and surjective, it is a bijection. Does this proof make sense?