I) Let X ⊂ T be the set of all elements not in the range of α. Define β: T → S as the union of { (α(s), s) | s ∈ S } and U = { (x, s) | x ∈ X, s ∈ S }, (x, s_{1}), (x, s_{2}) ∈ U ⇒ s_{1} = s_{2}. Since α is injective, β is well-defined. If s ∈ S β(α(s)) = s ⇒ βα = 1_{S}. Conversely, suppose α is not injective. There exists s_{1}, s_{2} such that α(s_{1}) = α(s_{2}) and s_{1} ≠ s_{2}. So s_{1} = β(α(s_{1})) = β(α(s_{2})) = s_{2}, contradicting s_{1} ≠ s_{2}.

II) Since α is surjective, choose one and only one s in the fiber (section 0.3) over t ∈ im α such that (t, s) ∈ β for every t ∈ T. β is clearly a map and αβ = 1_{T}. Conversely, suppose α is not surjective. Then there exists a t_{1} ≠ α(s) for every s ∈ S. Since αβ = 1_{T}, α(β(t_{1})) = t_{1}. Contradiction.

III) For part (I) if β is unique, every element t ∈ T must be in the range of α. Otherwise the image under β, of each t not in the range of α, can be made arbitrarily different. Thus contradicting the uniqueness of β.

For part (II) suppose β is unique and α is not injective. Then there exists s_{1}, s_{2} such that α(s_{1}) = α(s_{2}) and s_{1} ≠ s_{2}. If (α(s_{1}), s_{1}) ∈ β, replace it with (α(s_{1}), s_{2}). If (α(s_{1}), s_{2}) ∈ β, replace it with (α(s_{1}), s_{1}). Thus β is not unique. Contradiction.