Exercises 0.2 #2

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, s1), (x, s2) ∈ U ⇒ s1 = s2. Since α is injective, β is well-defined. If s ∈ S β(α(s)) = s ⇒ βα = 1S. Conversely, suppose α is not injective. There exists s1, s2 such that α(s1) = α(s2) and s1 ≠ s2. So s1 = β(α(s1)) = β(α(s2)) = s2, contradicting s1 ≠ s2.

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 = 1T. Conversely, suppose α is not surjective. Then there exists a t1 ≠ α(s) for every s ∈ S. Since αβ = 1T, α(β(t1)) = t1. 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 s1, s2 such that α(s1) = α(s2) and s1 ≠ s2. If (α(s1), s1) ∈ β, replace it with (α(s1), s2). If (α(s1), s2) ∈ β, replace it with (α(s1), s1). Thus β is not unique. Contradiction.

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