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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