Exercises 0.2 #3

I) Let α: S → T. Suppose α is surjective and there exist maps β1, β2 of T into a set U such that β1 ≠ β2 but β1α = β2α. Then there is a t ∈ T such that β1(t) ≠ β2(t). And there is an s ∈ S such that α(s) = t. So β1(α(s)) ≠ β2(α(s)). Contradiction. Conversely, by contraposition, if α is not surjective define β1: T → U as the union of { (t, u0) | t ∈ im α, u0 an arbitrary fixed element in U } and { (t, u1) | t ∉ im α, u1 an arbitrary fixed element in U }. Define β2 as β1 with (t, u1) replaced by (t, u2), u1 ≠ u2 for every t ∉ im α. Thus β1α = β2α and β1 ≠ β2.

II) Suppose α is injective and there exist maps γ1, γ2 of a set U into S such that γ1 ≠ γ2 but αγ1 = αγ2. Then there is a u ∈ U such that γ1(u) ≠ γ2(u). So α(γ1(u)) ≠ α(γ2(u)). Contradiction. Conversely, by contraposition, if α is not injective there exists s1, s2 such that s1 ≠ s2 and α(s1) = α(s2). Define γ1: U → S as { (u, s1) | u ∈ U }. Define γ2: U → S as { (u, s2) | u ∈ U }. Thus γ1 ≠ γ2 but αγ1 = αγ2.

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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.

Exercises 0.2 #1

a) g(f(x1)) = g(f(x2)) ⇒ f(x1) = f(x2) ⇒ x1 = x2

b) Suppose f is not injective. There exists x1, x2 such that f( x1) = f( x2) and x1 ≠ x2. Hence g(f( x1)) = g(f( x2)) and x1 ≠ x2. Contradiction.

c) z = g(y) for some y, y = f(x) for some x ⇒ z = g(f(x))

d) Suppose g is not surjective. There exists z ≠ g(y) for every y ⇒ z ≠ g(f(x)) for every x. Contradiction.

e) X = ℕ, f: n → 2n+1, is injective and not surjective. f(n) = { 0 if n=0, n-1 otherwise } is surjective and not injective.

f) By (b) and (d), f is injective and g is surjective. Suppose f is surjective and g is not injective. There exists y1, y2 such that g( y1) = g( y2) and y1 ≠ y2. So g(f( x1)) = g(f( x2)) and [y1 = f(x1)] ≠ [y2 = f(x2)] for some x1, x2. By definition of function x1 ≠ x2, contradicting gf being bijective. Thus f is surjective ⇒ g is injective.